Solution of Fundamentals of Electric Circuits

Chapter 1, Solution 1 (a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C (b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.1986...

Author: Charles Alexander | Matthew Sadiku

1052 downloads 5634 Views 17MB Size Report

This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!

Report copyright / DMCA form

DOWNLOAD PDF

Chapter 1, Solution 1 (a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C (b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C (c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C (d) q = 1.628x1020 x [-1.602x10-19 C] = -26.08 C Chapter 1, Solution 2 (a) (b) (c) (d) (e)

i = dq/dt = 3 mA i = dq/dt = (16t + 4) A i = dq/dt = (-3e-t + 10e-2t) nA i=dq/dt = 1200π cos 120π t pA i =dq/dt = − e −4t (80 cos 50 t + 1000 sin 50 t ) µ A

Chapter 1, Solution 3 (a) q(t) = ∫ i(t)dt + q(0) = (3t + 1) C (b) q(t) = ∫ (2t + s) dt + q(v) = (t 2 + 5t) mC

(c) q(t) = ∫ 20 cos (10t + π / 6 ) + q(0) = (2sin(10t + π / 6) + 1) µ C

(d)

10e -30t ( −30 sin 40t - 40 cos t) 900 + 1600 = − e - 30t (0.16cos40 t + 0.12 sin 40t) C

q(t) = ∫ 10e -30t sin 40t + q(0) =

Chapter 1, Solution 4

q = ∫ idt = ∫ =

10

−5 5sin 6 π t dt = cos 6π t 6π 0

5 (1 − cos 0.06π ) = 4.698 mC 6π

Chapter 1, Solution 5

q = ∫ idt = ∫ =

1 e dt mC = - e -2t 2

1 (1 − e 4 ) mC = 490 µC 2

Chapter 1, Solution 6

(a) At t = 1ms, i =

dq 80 = = 40 mA dt 2

(b) At t = 6ms, i =

dq = 0 mA dt

(c) At t = 10ms, i =

dq 80 = = - 20 mA 4 dt

Chapter 1, Solution 7 25A, dq  i= = - 25A, dt   25A,

2

-2t

0
which is sketched below:

0

Chapter 1, Solution 8

q = ∫ idt =

10 × 1 + 10 × 1 = 15 µC 2

Chapter 1, Solution 9 1

(a) q = ∫ idt = ∫ 10 dt = 10 C 0

3 5 ×1  q = ∫ idt = 10 × 1 + 10 −  + 5 ×1 0 (b) 2   = 15 + 10 − 25 = 22.5 C

5

(c) q = ∫ idt = 10 + 10 + 10 = 30 C 0

Chapter 1, Solution 10

q = ixt = 8 x10 3 x15 x10 − 6 = 120 µ C

Chapter 1, Solution 11

q = it = 85 x10-3 x 12 x 60 x 60 = 3,672 C E = pt = ivt = qv = 3672 x1.2 = 4406.4 J

Chapter 1, Solution 12

For 0 < t < 6s, assuming q(0) = 0, t

∫

t

∫

q (t ) = idt + q (0 ) = 3tdt + 0 = 1.5t 2 0

0

At t=6, q(6) = 1.5(6)2 = 54 For 6 < t < 10s,

t

t

∫

∫

q (t ) = idt + q (6 ) = 18 dt + 54 = 18 t − 54 6

6

At t=10, q(10) = 180 – 54 = 126 For 10
∫

t

∫

q (t ) = idt + q (10 ) = ( −12)dt + 126 = −12t + 246 10

10

At t=15, q(15) = -12x15 + 246 = 66 For 15
∫

q (t ) = 0 dt + q (15) =66 15

Thus,  1.5t 2 C, 0 < t < 6s   18 t − 54 C, 6 < t < 10s q (t ) =  −12t + 246 C, 10 < t < 15s  66 C, 15 < t < 20s 

The plot of the charge is shown below. 140 120 100

q(t)

80 60 40 20 0

0

5

10 t

15

20

Chapter 1, Solution 13 2

2

w = ∫ vidt = ∫ 1200 cos 2 4 t dt 0

0

2

= 1200 ∫ ( 2 cos 8t - 1)dt (since, cos 2 x = 2 cos 2x - 1) 0

2

2  1  = 1200 sin 8t − t  = 1200 sin 16 − 2  8 0 4  = - 2.486 kJ

Chapter 1, Solution 14

q = ∫ idt = ∫ 10(1 - e -0.5t )dt = 10(t + 2e -0.5t ) 1

(a)

(b)

0

= 10(1 + 2e

-0.5

− 2 ) = 2.131 C

1 0

p(t) = v(t)i(t) p(1) = 5cos2 ⋅ 10(1- e-0.5) = (-2.081)(3.935) = -8.188 W

Chapter 1, Solution 15

(a)

q = ∫ idt = ∫

2

0

− 3 2t 3e dt = e 2

2

-2t

= −1.5(e − 1) = 1.297 C

0

-2

(b)

5di = −6e 2t ( 5) = −30e -2t dt p = vi = − 90 e − 4 t W v=

3

(c) w = ∫ pdt = -90∫ e -4t dt = 0

3

− 90 -4t e = − 22.5 J −4 0

Chapter 1, Solution 16  0
0< t <1  10t V  v(t) = 10 V 1< t < 3  40 - 10t V 3 < t < 4 2

3

4

2

3

w = ∫ v(t)i(t)dt = ∫ 10 + (25t)dt + ∫ 10( 25t)dt + ∫ 10(100 − 25t)dt + ∫ ( 40 − 10t)(100 - 25t)mJ 0

1

=

1

3

2

4  250 3 250 t2  t + + 250 4 t -  + ∫ 250( 4 − t) 2 dt 3 2 1 2 2 3 0  4

 250 250 9 t2    2 ( 3) + 25012 − − 8 + 2  + 25016 t - 4t +  = + 3 2 2 3 3    = 916.7 mJ

Chapter 1, Solution 17

Σ p = 0 → -205 + 60 + 45 + 30 + p3 = 0 p3 = 205 – 135 = 70 W Thus element 3 receives 70 W. Chapter 1, Solution 18

p1 = 30(-10) = -300 W p2 = 10(10) = 100 W p3 = 20(14) = 280 W p4 = 8(-4) = -32 W p5 = 12(-4) = -48 W

Chapter 1, Solution 19

∑p=0

 →

−4 I s − 2 x6 − 13 x 2 + 5 x10 = 0

 →

Is = 3 A

Chapter 1, Solution 20

Since Σ p = 0 -30×6 + 6×12 + 3V0 + 28 + 28×2 - 3×10 = 0 72 + 84 + 3V0 = 210 or 3V0 = 54 V0 = 18 V

Chapter 1, Solution 21 i= =

∆q  photon  1  electron  = 4 × 1011   ⋅ 1. 6 × 1019 ( C / electron)  ⋅  ∆t  sec  8  photon  4 × 1011 × 1. 6 × 10 −19 C/s = 0.8 × 10 -8 C/s = 8 nA 8

Chapter 1, Solution 22

It should be noted that these are only typical answers. (a) (b) (c) (d) (e) (f) (g) (h)

Light bulb Radio set TV set Refrigerator PC PC printer Microwave oven Blender

60 W, 100 W 4W 110 W 700 W 120 W 18 W 1000 W 350 W

Chapter 1, Solution 23 (a) i =

p 1500 = = 12.5 W v 120

(b) w = pt = 1. 5 × 103 × 45 × 60 ⋅ J = 1.5 × (c) Cost = 1.125 × 10 = 11.25 cents

45 kWh = 1.125 kWh 60

Chapter 1, Solution 24

p = vi = 110 x 8 = 880 W

Chapter 1, Solution 25

4 Cost = 1.2 kW × hr × 30 × 9 cents/kWh = 21.6 cents 6

Chapter 1, Solution 26

0. 8A ⋅ h = 80 mA 10h (b) p = vi = 6 × 0.08 = 0.48 W (c) w = pt = 0.48 × 10 Wh = 0.0048 kWh (a) i =

Chapter 1, Solution 27 (a) Let T = 4h = 4 × 36005 T

q = ∫ idt = ∫ 3dt = 3T = 3 × 4 × 3600 = 43.2 kC 0

T T 0 . 5t   ( b) W = ∫ pdt = ∫ vidt = ∫ ( 3) 10 + dt 0 0 3600   4×3600

 0. 25t 2   = 310t + 3600  0  = 475.2 kJ

( c)

= 3[40 × 3600 + 0. 25 × 16 × 3600]

W = 475.2 kWs, (J = Ws) 475.2 Cost = kWh × 9 cent = 1.188 cents 3600

Chapter 1, Solution 28

(a) i =

P 30 = = 0.25 A V 120

( b) W = pt = 30 × 365 × 24 Wh = 262.8 kWh Cost = $0.12 × 262.8 = $31.54

Chapter 1, Solution 29

(20 + 40 + 15 + 45)  30  hr + 1.8 kW  hr 60  60  = 2.4 + 0.9 = 3.3 kWh Cost = 12 cents × 3.3 = 39.6 cents w = pt = 1. 2kW

Chapter 1, Solution 30

Energy = (52.75 – 5.23)/0.11 = 432 kWh

Chapter 1, Solution 31

Total energy consumed = 365(4 +8) W Cost = $0.12 x 365 x 12 = $526.60

Chapter 1, Solution 32

(20 + 40 + 15 + 45)  30  hr + 1.8 kW  hr 60  60  = 2.4 + 0.9 = 3.3 kWh Cost = 12 cents × 3.3 = 39.6 cents w = pt = 1. 2kW

Chapter 1, Solution 33 i=

dq → q = ∫ idt = 2000 × 3 × 10 3 = 6 C dt

Chapter 1, Solution 34

(b) Energy =

∑ pt

= 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2

= 10,000 kWh (c) Average power = 10,000/24 = 416.67 W

Chapter 1, Solution 35

( a) W = ∫ p( t ) dt = 400 × 6 + 1000 × 2 + 200 × 12 × 1200 × 2 + 400 × 2 = 7200 + 2800 = 10.4 kWh ( b)

10.4 kW = 433.3 W/h 24 h

Chapter 1, Solution 36

160A ⋅ h =4A 40 160Ah 160, 000h ( b) t = = = 6,667 days 0.001A 24h / day (a)

i=

Chapter 1, Solution 37

q = 5 × 10 20 (− 1. 602 × 10 −19 ) = −80. 1 C W = qv = −80. 1 × 12 = − 901.2 J

Chapter 1, Solution 38

P = 10 hp = 7460 W W = pt = 7460 × 30 × 60 J = 13.43 × 106 J

Chapter 1, Solution 39

p = vi → i =

p 2 × 10 3 = = 16.667 A v 120

Chapter 2, Solution 1 v = iR

i = v/R = (16/5) mA = 3.2 mA

Chapter 2, Solution 2 p = v2/R →

R = v2/p = 14400/60 = 240 ohms

Chapter 2, Solution 3 R = v/i = 120/(2.5x10-3) = 48k ohms Chapter 2, Solution 4 (a) (b)

i = 3/100 = 30 mA i = 3/150 = 20 mA

Chapter 2, Solution 5 n = 9; l = 7; b = n + l – 1 = 15 Chapter 2, Solution 6 n = 12;

l = 8;

b = n + l –1 = 19

Chapter 2, Solution 7 7 elements or 7 branches and 4 nodes, as indicated. 30 V 1

20 Ω

2

3 ++++ -

2A

30 Ω

60 Ω

4

40 Ω

10 Ω

Chapter 2, Solution 8 12 A a

i1 b

8A

i3

i2

12 A c At node a, At node c, At node d,

9A d

8 = 12 + i1 9 = 8 + i2 9 = 12 + i3

i1 = - 4A i2 = 1A i3 = -3A

Chapter 2, Solution 9 Applying KCL, i1 + 1 = 10 + 2 1 + i2 = 2 + 3 i2 = i3 + 3

i1 = 11A i2 = 4A i3 = 1A

Chapter 2, Solution 10 2

4A 1

-2A i2

i1

3

3A At node 1, At node 3,

4 + 3 = i1 3 + i2 = -2

i1 = 7A i2 = -5A

Chapter 2, Solution 11 Applying KVL to each loop gives -8 + v1 + 12 = 0 -12 - v2 + 6 = 0 10 - 6 - v3 = 0 -v4 + 8 - 10 = 0

v1 = 4v v2 = -6v v3 = 4v v4 = -2v

Chapter 2, Solution 12 + 15v -

loop 2 – 25v + + 20v -

+ 10v + v1 -

loop 1

For loop 1, For loop 2, For loop 3,

+ v2 -

loop 3

-20 -25 +10 + v1 = 0 -10 +15 -v2 = 0 -v1 +v2 +v3 = 0

+ v3 -

v1 = 35v v2 = 5v v3 = 30v

Chapter 2, Solution 13 2A

1

I2

7A 2

3

I4

4 4A

I1

3A

I3

At node 2, 3 + 7 + I2 = 0

 →

I 2 = −10 A

At node 1, I1 + I 2 = 2

 →

I 1 = 2 − I 2 = 12 A

At node 4, 2 = I4 + 4

 →

I 4 = 2 − 4 = −2 A

At node 3, 7 + I4 = I3

 →

I3 = 7 − 2 = 5 A

Hence, I 1 = 12 A,

I 2 = −10 A,

I 3 = 5 A,

I 4 = −2 A

Chapter 2, Solution 14 + 3V -

+ I3

4V

+

V3 -

 →

V4 = 7V

For mesh 2, +4 + V3 + V4 = 0

 →

V3 = −4 − 7 = −11V

 →

V1 = V3 + 3 = −8V

 →

V2 = −V1 − 2 = 6V

For mesh 3, −3 + V1 − V3 = 0

For mesh 4, −V1 − V2 − 2 = 0

Thus, V1 = −8V ,

V2 = 6V ,

+

- V4

For mesh 1, −V4 + 2 + 5 = 0

V1

V3 = −11V ,

I4 2V -

+

I2

+

-

V4 = 7V

+

V2

+ I1

5V -

Chapter 2, Solution 15

+ +

+

12V -

1 - 8V +

v2 -

v1 -

3

+

2 v3

10V +

-

For loop 1, 8 − 12 + v2 = 0

 →

v2 = 4V

For loop 2, − v3 − 8 − 10 = 0

 →

v3 = −18V

 →

v1 = −6V

For loop 3, − v1 + 12 + v3 = 0

Thus, v1 = −6V ,

v2 = 4V ,

v3 = −18V

Chapter 2, Solution 16

+ v1 -

6V

loop 1

+

-

+-

12V

10V +-

+ v1 -

loop 2 + v2 -

Applying KVL around loop 1, –6 + v1 + v1 – 10 – 12 = 0

v1 = 14V

Applying KVL around loop 2, 12 + 10 – v2 = 0

v2 = 22V

Chapter 2, Solution 17 + v1 -

24V

+

loop 1

-

+ v3 -

-

v2 +

loop 2 -+ 12V It is evident that v3 = 10V Applying KVL to loop 2, v2 + v3 + 12 = 0

v2 = -22V

Applying KVL to loop 1, -24 + v1 - v2 = 0

v1 = 2V

Thus, v1 = 2V, v2 = -22V, v3 = 10V Chapter 2, Solution 18 Applying KVL, -30 -10 +8 + I(3+5) = 0 8I = 32

I = 4A

-Vab + 5I + 8 = 0

Vab = 28V

+

-

10V

Chapter 2, Solution 19 Applying KVL around the loop, we obtain -12 + 10 - (-8) + 3i = 0

i = -2A

Power dissipated by the resistor: p 3Ω = i2R = 4(3) = 12W Power supplied by the sources: p12V = 12 (- -2) = 24W p10V = 10 (-2) = -20W p8V = (- -2) = -16W Chapter 2, Solution 20 Applying KVL around the loop, -36 + 4i0 + 5i0 = 0

i0 = 4A

Chapter 2, Solution 21 Apply KVL to obtain 10 Ω

-45 + 10i - 3V0 + 5i = 0

+ v0 -

But v0 = 10i, -45 + 15i - 30i = 0 P3 = i2R = 9 x 5 = 45W

i = -3A

45V

+

+

-

5Ω

3v0

Chapter 2, Solution 22 4Ω + v0 6Ω

10A

2v0

At the node, KCL requires that v0 + 10 + 2 v 0 = 0 4

v0 = –4.444V

The current through the controlled source is i = 2V0 = -8.888A and the voltage across it is v = (6 + 4) i0 = 10

v0 = −11.111 4

Hence, p2 vi = (-8.888)(-11.111) = 98.75 W Chapter 2, Solution 23 8//12 = 4.8, 3//6 = 2, (4 + 2)//(1.2 + 4.8) = 6//6 = 3 The circuit is reduced to that shown below. ix

1Ω +

6A

2Ω

vx

3Ω

Applying current division, ix =

2 (6 A) = 2 A, 2 + 1+ 3

v x = 1i x = 2V

The current through the 1.2- Ω resistor is 0.5ix = 1A. The voltage across the 12- Ω resistor is 1 x 4.8 = 4.8 V. Hence the power is p=

v 2 4.8 2 = = 1.92W 12 R

Chapter 2, Solution 24 (a)

I0 =

Vs R1 + R2

V0 = −α I0 (R3 R4 ) = −

αV0 R1 + R 2

⋅

R3 R4 R3 + R4

V0 − αR3 R4 = Vs (R1 + R2 )(R3 + R4 ) (b)

If R1 = R2 = R3 = R4 = R, V0 α R α = ⋅ = = 10 VS 2R 2 4

Chapter 2, Solution 25 V0 = 5 x 10-3 x 10 x 103 = 50V Using current division, I20 =

5 (0.01x50) = 0.1 A 5 + 20

V20 = 20 x 0.1 kV = 2 kV p20 = I20 V20 = 0.2 kW

α = 40

Chapter 2, Solution 26 V0 = 5 x 10-3 x 10 x 103 = 50V Using current division, I20 =

5 (0.01x50) = 0.1 A 5 + 20

V20 = 20 x 0.1 kV = 2 kV p20 = I20 V20 = 0.2 kW Chapter 2, Solution 27

Using current division, i1 =

4 (20) = 8 A 4+6

i2 =

6 (20) = 12 A 4+6

Chapter 2, Solution 28

We first combine the two resistors in parallel 15 10 = 6 Ω

We now apply voltage division, v1 =

14 (40) = 20 V 14 + 6

v2 = v3 = Hence,

6 (40) = 12 V 14 + 6

v1 = 28 V, v2 = 12 V, vs = 12 V

Chapter 2, Solution 29

The series combination of 6 Ω and 3 Ω resistors is shorted. Hence i2 = 0 = v2 v1 = 12, i1 =

12 = 3A 4

Hence v1 = 12 V, i1 = 3 A, i2 = 0 = v2 Chapter 2, Solution 30 8Ω i1

i 9A

By current division, i =

6Ω

+ v -

4Ω

12 (9) = 6 A 6 + 12

i1 = 9 − 6 = 3A, v = 4i1 = 4 x 3 = 12 V

p6 = 12R = 36 x 6 = 216 W Chapter 2, Solution 31

The 5 Ω resistor is in series with the combination of 10 (4 + 6) = 5Ω . Hence by the voltage division principle, v=

5 (20V) = 10 V 5+5

by ohm's law, i=

v 10 = = 1A 4 + 6 4+ 6

pp = i2R = (1)2(4) = 4 W

Chapter 2, Solution 32

We first combine resistors in parallel. 20 30 =

20 x30 = 12 Ω 50

10 40 =

10x 40 = 8Ω 50

Using current division principle, 8 12 i1 + i 2 = (20) = 8A, i 3 + i 4 = (20) = 12A 8 + 12 20 i1 =

20 (8) = 3.2 A 50

i2 =

30 (8) = 4.8 A 50

i3 =

10 (12) = 2.4A 50

i4 =

40 (12) = 9.6 A 50

Chapter 2, Solution 33

Combining the conductance leads to the equivalent circuit below i + v -

9A

1S

i

4S

4S

6x3 = 25 and 25 + 25 = 4 S 9 Using current division, 6 S 3S =

i=

1 1 1+ 2

(9) = 6 A, v = 3(1) = 3 V

9A

+ v -

1S

2S

Chapter 2, Solution 34

By parallel and series combinations, the circuit is reduced to the one below:

Thus i1 =

8Ω

i1

10 x15 = 6Ω 10 ( 2 + 13 ) = 25 15 x15 15 (4 + 6) = = 6Ω 25 12 (6 + 6) = 6Ω

28V

+ v1 -

+

-

6Ω

28 = 2 A and v1 = 6i1 = 12 V 8+6

We now work backward to get i2 and v2. i1 = 2A

8Ω

6Ω

1A

1A

28V

+ 12V -

+

-

8Ω

i1 = 2A

6Ω

+ 6V -

12 Ω

6Ω

4Ω

1A

0.6A

1A

28V

Thus, v2 =

+ 12V -

+

-

12 Ω

+ 6V -

+ 15 Ω

3.6V

v 13 (3 ⋅ 6) = 3 ⋅ 12, i2 = 2 = 0.24 13 15

p2 = i2R = (0.24)2 (2) = 0.1152 W i1 = 2 A, i2 = 0.24 A, v1 = 12 V, v2 = 3.12 V, p2 = 0.1152 W Chapter 2, Solution 35 i 70 Ω 50V

+

-

a

+ V1 i1 -

30 Ω I0 +

20 Ω i2

b

V0 5 Ω -

-

6Ω

Combining the versions in parallel, 70 30 =

i=

70x30 = 21Ω , 100

20 15 =

20x 5 =4 Ω 25

50 =2 A 21 + 4

vi = 21i = 42 V, v0 = 4i = 8 V v v i1 = 1 = 0.6 A, i2 = 2 = 0.4 A 70 20 At node a, KCL must be satisfied i1 = i2 + I0

0.6 = 0.4 + I0

I0 = 0.2 A

Hence v0 = 8 V and I0 = 0.2A Chapter 2, Solution 36

The 8-Ω resistor is shorted. No current flows through the 1-Ω resistor. Hence v0 is the voltage across the 6Ω resistor. I0 =

4 4 = =1 A 2 + 3 16 4

v0 = I0 (3 6 ) = 2I 0 = 2 V

Chapter 2, Solution 37

Let I = current through the 16Ω resistor. If 4 V is the voltage drop across the 6 R combination, then 20 - 4 = 16 V in the voltage drop across the 16Ω resistor. 16 Hence, I = = 1 A. 16 20 6R R = 12 Ω 4= 6R= But I = =1 6+R 16 + 6 R

Chapter 2, Solution 38

Let I0 = current through the 6Ω resistor. Since 6Ω and 3Ω resistors are in parallel. 6I0 = 2 x 3

R0 = 1 A

The total current through the 4Ω resistor = 1 + 2 = 3 A. Hence vS = (2 + 4 + 2 3 ) (3 A) = 24 V I=

vS = 2.4 A 10

Chapter 2, Solution 39 (a) Req = R 0 = 0

(b)

(c) (d)

(e)

R R + = R 2 2 Req = (R + R ) (R + R ) = 2R 2R = R Req = R R + R R =

1 Req = 3R (R + R R ) = 3R (R + R ) 2 3 3Rx R 2 =R = 3 3R + R 2  R ⋅ 2R  Req = R 2R 3R = 3R    3R  2 3Rx R 2 3 = 6R = 3R R= 2 11 3 3R + R 3

Chapter 2, Solution 40

Req = 3 + 4 (2 + 6 3) = 3 + 2 = 5Ω I=

10 10 = = 2A Re q 5

Chapter 2, Solution 41

Let R0 = combination of three 12Ω resistors in parallel 1 1 1 1 = + + R o 12 12 12

Ro = 4

R eq = 30 + 60 (10 + R 0 + R ) = 30 + 60 (14 + R )

50 = 30 +

60(14 + R ) 74 + R

74 + R = 42 + 3R

or R = 16 Ω Chapter 2, Solution 42

5x 20 = 4Ω 25

(a)

Rab = 5 (8 + 20 30) = 5 (8 + 12) =

(b)

Rab = 2 + 4 (5 + 3) 8 + 5 10 (6 + 4) = 2 + 4 4 + 5 5 = 2 + 2 + 2.5 = 6.5 Ω

Chapter 2, Solution 43

5x 20 400 + = 4 + 8 = 12 Ω 25 50

(a)

Rab = 5 20 + 10 40 =

(b)

1 1   1 60 20 30 =  + +   60 20 30  Rab = 80 (10 + 10) =

−1

=

60 = 10Ω 6

80 + 20 = 16 Ω 100

Chapter 2, Solution 44

(a) Convert T to Y and obtain 20 x 20 + 20 x10 + 10 x 20 800 = = 80 Ω 10 10 800 R2 = = 40 Ω = R3 20

R1 =

The circuit becomes that shown below.

R1

a

R3

R2

5Ω

b R1//0 = 0,

R3//5 = 40//5 = 4.444 Ω

Rab = R2 / /(0 + 4.444) = 40 / /4.444 = 4Ω

(b) 30//(20+50) = 30//70 = 21 Ω Convert the T to Y and obtain

20 x10 + 10 x 40 + 40 x 20 1400 = = 35Ω 40 40 1400 1400 R2 = = 70 Ω , R3 = = 140 Ω 20 10 The circuit is reduced to that shown below. 15Ω R1 =

R1

11 Ω R2

R3 21 Ω

30 Ω 21 Ω Combining the resistors in parallel

R1//15 =35//15=10.5, 30//R2=30//70 = 21 leads to the circuit below. 10.5 Ω

11 Ω

21 Ω

140 Ω 21 Ω

21 Ω

Coverting the T to Y leads to the circuit below. 10.5 Ω

11 Ω

R4 R5

R6

R4 =

21x140 + 140 x 21 + 21x 21 6321 = = 301Ω = R6 21 21

R5 =

6321 = 45.15 140

21 Ω

10.5//301 = 10.15, 301//21 = 19.63 R5//(10.15 +19.63) = 45.15//29.78 = 17.94 Rab = 11 + 17 .94 = 28.94Ω

Chapter 2, Solution 45 (a) 10//40 = 8, 20//30 = 12, 8//12 = 4.8

Rab = 5 + 50 + 4.8 = 59.8 Ω (b) 12 and 60 ohm resistors are in parallel. Hence, 12//60 = 10 ohm. This 10 ohm and 20 ohm are in series to give 30 ohm. This is in parallel with 30 ohm to give 30//30 = 15 ohm. And 25//(15+10) = 12.5. Thus Rab = 5 + 12.8 + 15 = 32.5Ω

Chapter 2, Solution 46

(a)

30x 70 60 + 20 + 40 + 100 80

Rab = 30 70 + 40 + 60 20 =

= 21 + 40 + 15 = 76 Ω

(b)

The 10-Ω, 50-Ω, 70-Ω, and 80-Ω resistors are shorted. 20 30 =

20x30 = 12Ω 50

40 60 =

40x 60 = 24 100

Rab = 8 + 12 + 24 + 6 + 0 + 4 = 54 Ω Chapter 2, Solution 47 5 20 =

6 3=

5x 20 = 4Ω 25 6x3 = 2Ω 9 10 Ω

8Ω a

b

4Ω

Rab = 10 + 4 + 2 + 8 = 24 Ω

2Ω

Chapter 2, Solution 48 R 1 R 2 + R 2 R 3 + R 3 R 1 100 + 100 + 100 = = 30 R3 10 Ra = Rb = Rc = 30 Ω

(a)

Ra =

(b)

Ra =

30x 20 + 30x50 + 20x 50 3100 = = 103.3Ω 30 30 3100 3100 Rb = = 155Ω, R c = = 62Ω 20 50

Ra = 103.3 Ω, Rb = 155 Ω, Rc = 62 Ω Chapter 2, Solution 49

(a)

(b)

R1 =

RaRc 12 + 12 = = 4Ω Ra + Rb + Rc 36 R1 = R2 = R3 = 4 Ω

60x30 = 18Ω 60 + 30 + 10 60 x10 R2 = = 6Ω 100 30x10 R3 = = 3Ω 100 R1 =

R1 = 18Ω, R2 = 6Ω, R3 = 3Ω Chapter 2, Solution 50

Using R ∆ = 3RY = 3R, we obtain the equivalent circuit shown below:

30mA

3R

3R 3R

R R

30mA

3R

3R/2

3RxR 3 = R 4R 4 3R (3RxR ) /(4R ) = 3 /(4R ) 3R R =

3 3Rx R 3  3 3 2 3R  R + R  = 3R R = 3 4  2 4 3R + R = R 2 800 x 10-3 = (30 x 10-3)2 R P = I2 R R = 889 Ω Chapter 2, Solution 51 30 30 = 15Ω and 30 20 = 30 x 20 /(50) = 12Ω

(a)

Rab = 15 (12 + 12) = 15x 24 /(39) = 9.31 Ω a

a 30 Ω

30 Ω

30 Ω 30 Ω

b

20 Ω

12 Ω 15 Ω

12 Ω

20 Ω b

Converting the T-subnetwork into its equivalent ∆ network gives

(b)

Ra'b' = 10x20 + 20x5 + 5x10/(5) = 350/(5) = 70 Ω Rb'c' = 350/(10) = 35Ω, Ra'c' = 350/(20) = 17.5 Ω Also

30 70 = 30 x 70 /(100) = 21Ω and 35/(15) = 35x15/(50) = 10.5

Rab = 25 + 17.5 (21 + 10.5) = 25 + 17.5 31.5 Rab = 36.25 Ω 30 Ω

30 Ω

a

25 Ω

10 Ω 5Ω

b

20 Ω

a 15 Ω

25 Ω

a’

17.5 Ω

b

70 Ω

b’

35 Ω

c’

15 Ω

c’

Chapter 2, Solution 52 (a) We first convert from T to ∆ . 100 Ω

a

100 Ω

a 100 Ω 100 Ω 100 Ω 100 Ω

b

100 Ω

200 Ω

100 Ω 100 Ω

200 Ω

b

R1 =

100 Ω 100 Ω

100 Ω

100 Ω 100 Ω

100 Ω

R3 R2

100x 200 + 200x 200 + 200 x100 80000 = = 800Ω 100 100

R2 = R3 = 80000/(200) = 400 100x 400 But 100 400 = = 80Ω 500 We connect the ∆ to Y. 100 Ω

a

b

100 Ω

a

100 Ω 100 Ω

80 Ω

100 Ω 100 Ω

80 Ω

800 Ω

b

100 Ω 100 Ω

Rb

100 Ω 100 Ω

Rc

80 x800 64,000 400 = = Ω 80 + 80 + 800 960 3 80x80 20 = Ω Rb = 960 3

Ra = Rc =

We convert T to ∆ . a

500/3 Ω 100 Ω 320/3 Ω

b

100 Ω 500/3 Ω

500/3 Ω

a

R2’ R1’ R3’

b

Ra

500/3 Ω

R1

R 1' =

320 320 + 100 x 3 3 = 94,000 /(3) = 293.75Ω 320 320 /(3) 3

100 x100 + 100 x

R '2 = R 13 =

94,000 /(3) = 313.33 100

940 /(30) 500 /(3) =

940 /(3) x500 /(3) = 108.796 1440 /(3)

Rab = 293.75 (2 x108.796) =

293.75x 217.6 = 125 Ω 511.36

Converting the Ts to ∆ s, we have the equivalent circuit below.

(b)

100 Ω

100 Ω

a 100 Ω

300 Ω 300 Ω

100 Ω

300 Ω

100 Ω

100 Ω

100 Ω 100 Ω

a

100 Ω

300 Ω

300 Ω

b

300 Ω

100 Ω 100 Ω

100 Ω 100 Ω

b

100 Ω

300 100 = 300 x100 /(400) = 75, 300 (75 + 75) = 300 x150 /(450) = 100

Rab = 100 + 100 300 + 100 = 200 + 100 x 300 /(400) Rab = 2.75 Ω 100 Ω 300 Ω

100 Ω

300 Ω 300 Ω

100 Ω

100 Ω

Chapter 2, Solution 53

(a)

Converting one ∆ to T yields the equivalent circuit below: 30 Ω

a’

20 Ω

a

60 Ω

b’

b

4Ω

20 Ω

c’

5Ω

80 Ω

40 x10 10 x50 40x50 = 4Ω, R b 'n = = 5Ω, R c 'n = = 20Ω 40 + 10 + 50 100 100 Rab = 20 + 80 + 20 + (30 + 4) (60 + 5) = 120 + 34 65 Ra'n =

Rab = 142.32 Ω (a) We combine the resistor in series and in parallel. 30 (30 + 30) =

30x 60 = 20Ω 90

We convert the balanced ∆ s to Ts as shown below: a

30 Ω

30 Ω

a 10 Ω

30 Ω 30 Ω

20 Ω

10 Ω

30 Ω

b 30 Ω

10 Ω

10 Ω 10 Ω

10 Ω

b

Rab = 10 + (10 + 10) (10 + 20 + 10) + 10 = 20 + 20 40 Rab = 33.33 Ω Chapter 2, Solution 54

(a) Rab = 50 + 100 / /(150 + 100 + 150 ) = 50 + 100 / /400 = 130 Ω (b) Rab = 60 + 100 / /(150 + 100 + 150 ) = 60 + 100 / /400 = 140 Ω

20 Ω

Chapter 2, Solution 55

We convert the T to ∆ .

I0

a

24 V +

-

I0 20 Ω 40 Ω

60 Ω

10 Ω 20 Ω

50 Ω

a 140 Ω

60 Ω

24 V +

35 Ω

-

70 Ω

70 Ω

b b

Req

Req R R + R 2 R 3 + R 3 R 1 20 x 40 + 40 x10 + 10 x 20 1400 Rab = 1 2 = = = 35Ω R3 40 40 Rac = 1400/(10) = 140Ω, Rbc = 1400/(40) = 35Ω 70 70 = 35 and 140 160 = 140x60/(200) = 42

Req = 35 (35 + 42) = 24.0625Ω I0 = 24/(Rab) = 0.9774A Chapter 2, Solution 56

We need to find Req and apply voltage division. We first tranform the Y network to ∆ . 30 Ω

+ 100 V

-

16 Ω

15 Ω

35 Ω

12 Ω

30 Ω 16 Ω

10 Ω 20 Ω

Req

15x10 + 10x12 + 12x15 450 = = 37.5Ω 12 12 Rac = 450/(10) = 45Ω, Rbc = 450/(15) = 30Ω

Rab =

Combining the resistors in parallel,

+ 100 V

35 Ω

Req

a

37.5 Ω 30 Ω 45 Ω

c

b 20 Ω

30||20 = (600/50) = 12 Ω, 37.5||30 = (37.5x30/67.5) = 16.667 Ω 35||45 = (35x45/80) = 19.688 Ω Req = 19.688||(12 + 16.667) = 11.672Ω By voltage division, v =

11.672 100 = 42.18 V 11.672 + 16

Chapter 2, Solution 57 4Ω a

2Ω 27 Ω 1Ω

18 Ω

b d 10 Ω

36 Ω

c e

7Ω

14 Ω

28 Ω

f

6x12 + 12x8 + 8x 6 216 = = 18 Ω 12 12 Rac = 216/(8) = 27Ω, Rbc = 36 Ω 4x 2 + 2x8 + 8x 4 56 Rde = = 7Ω 8 8 Ref = 56/(4) = 14Ω, Rdf = 56/(2) = 28 Ω Rab =

Combining resistors in parallel, 280 36x 7 = 7.368Ω, 36 7 = = 5.868Ω 38 43 27 x 3 27 3 = = 2.7Ω 30

10 28 =

4Ω

4Ω

18 Ω 5.868 Ω

7.568 Ω

1.829 Ω

2.7 Ω

3.977 Ω

0.5964 Ω

14 Ω 7.568 Ω

14 Ω

18x 2.7 18x 2.7 = = 1.829 Ω 18 + 2.7 + 5.867 26.567 18x5.868 = = 3.977 Ω 26.567 5.868x 2.7 = = 0.5904 Ω 26.567 = 4 + 1.829 + (3.977 + 7.368) (0.5964 + 14)

R an = R bn R cn R eq

= 5.829 + 11.346 14.5964 = 12.21 Ω

i = 20/(Req) = 1.64 A Chapter 2, Solution 58

The resistor of the bulb is 120/(0.75) = 160Ω 40 Ω

2.25 A

+ 90 V - 0.75 A VS

+

-

160 Ω

1.5 A

+

80 Ω

120

Once the 160Ω and 80Ω resistors are in parallel, they have the same voltage 120V. Hence the current through the 40Ω resistor is 40(0.75 + 1.5) = 2.25 x 40 = 90 Thus vs = 90 + 120 = 210 V

Chapter 2, Solution 59 Total power p = 30 + 40 + 50 + 120 W = vi or i = p/(v) = 120/(100) = 1.2 A Chapter 2, Solution 60 p = iv i = p/(v) i30W = 30/(100) = 0.3 A i40W = 40/(100) = 0.4 A i50W = 50/(100) = 0.5 A Chapter 2, Solution 61 There are three possibilities (a)

Use R1 and R2: R = R 1 R 2 = 80 90 = 42.35Ω p = i2R i = 1.2A + 5% = 1.2 ± 0.06 = 1.26, 1.14A p = 67.23W or 55.04W, cost = $1.50

(b)

Use R1 and R3: R = R 1 R 3 = 80 100 = 44.44 Ω p = I2R = 70.52W or 57.76W, cost = $1.35

(c)

Use R2 and R3: R = R 2 R 3 = 90 100 = 47.37Ω p = I2R = 75.2W or 61.56W, cost = $1.65

Note that cases (b) and (c) give p that exceed 70W that can be supplied. Hence case (a) is the right choice, i.e. R1 and R2 Chapter 2, Solution 62 pA = 110x8 = 880 W,

pB = 110x2 = 220 W

Energy cost = $0.06 x 360 x10 x (880 + 220)/1000 = $237.60

Chapter 2, Solution 63 Use eq. (2.61), Im 2 x10 −3 x100 Rn = = 0.04Ω Rm = I − Im 5 − 2 x10 −3 In = I - Im = 4.998 A p = I 2n R = (4.998) 2 (0.04) = 0.9992 ≅ 1 W Chapter 2, Solution 64 When Rx = 0, i x = 10A

R=

When Rx is maximum, ix = 1A

110 = 11 Ω 10

R + Rx =

i.e., Rx = 110 - R = 99 Ω Thus, R = 11 Ω, Rx = 99 Ω

110 = 110 Ω 1

Chapter 2, Solution 65 Rn =

Vfs 50 − Rm = − 1 kΩ = 4 kΩ 10mA I fs

Chapter 2, Solution 66

20 kΩ/V = sensitivity =

1 I fs

1 kΩ / V = 50 µA 20 V The intended resistance Rm = fs = 10(20kΩ / V) = 200kΩ I fs V 50 V (a) R n = fs − R m = − 200 kΩ = 800 kΩ i fs 50 µA i.e., Ifs =

(b)

p = I fs2 R n = (50 µA) 2 (800 kΩ) = 2 mW

Chapter 2, Solution 67

(a)

By current division, i0 = 5/(5 + 5) (2 mA) = 1 mA V0 = (4 kΩ) i0 = 4 x 103 x 10-3 = 4 V

(b)

4k 6k = 2.4kΩ. By current division,

5 (2mA) = 1.19 mA 1 + 2.4 + 5 v '0 = (2.4 kΩ)(1.19 mA) = 2.857 V

i '0 =

v 0 − v '0 1.143 x 100% = x100 = 28.57% v0 4

(c)

% error =

(d)

4k 30 kΩ = 3.6 kΩ. By current division,

5 (2mA) = 1.042mA 1 + 3.6 + 5 v '0 (3.6 kΩ)(1.042 mA) = 3.75V

i '0 =

% error =

v − v '0 0.25x100 = 6.25% x100% = v0 4

Chapter 2, Solution 68

(a)

40 = 24 60Ω

(b)

4 = 0.1 A 16 + 24 4 i' = = 0.09756 A 16 + 1 + 24 0.1 − 0.09756 % error = x100% = 2.44% 0.1 i=

(c)

Chapter 2, Solution 69

With the voltmeter in place, R2 Rm V0 = VS R1 + R S + R 2 R m where Rm = 100 kΩ without the voltmeter, R2 VS V0 = R1 + R 2 + R S 100 kΩ 101

(a)

When R2 = 1 kΩ, R m R 2 =

(b)

100 V0 = 101 (40) = 1.278 V (with) 100 101 + 30 1 V0 = (40) = 1.29 V (without) 1 + 30 1000 When R2 = 10 kΩ, R 2 R m = = 9.091kΩ 110 9.091 V0 = (40) = 9.30 V (with) 9.091 + 30 10 V0 = (40) = 10 V (without) 10 + 30 When R2 = 100 kΩ, R 2 R m = 50kΩ

(c)

50 (40) = 25 V (with) 50 + 30 100 V0 = (40) = 30.77 V (without) 100 + 30 V0 =

Chapter 2, Solution 70

(a) Using voltage division, 12 (25) = 15V 12 + 8 10 vb = (25) = 10V 10 + 15 = va − vb = 15 − 10 = 5V va =

vab

(b) + 25 V -

15k Ω

8k Ω a

b 10k Ω

12k Ω o va = 0 ,

vb = 10V ,

vab = va − vb = 0 − 10 = −10V

Chapter 2, Solution 71 R1 iL

Vs + −

RL

Given that vs = 30 V, R1 = 20 Ω, IL = 1 A, find RL.

v s = i L ( R1 + R L )

 →

RL =

vs 30 − R1 = − 20 = 10Ω iL 1

Chapter 2, Solution 72

The system can be modeled as shown. 12A

+ 9V -

R

R

R

•••

The n parallel resistors R give a combined resistance of R/n. Thus,

9 = 12 x

R n

 →

n=

12 xR 12 x15 = = 20 9 9

Chapter 2, Solution 73

By the current division principle, the current through the ammeter will be one-half its previous value when R = 20 + Rx 65 = 20 + Rx

Rx = 45 Ω

Chapter 2, Solution 74

With the switch in high position, 6 = (0.01 + R3 + 0.02) x 5

R3 = 1.17 Ω

At the medium position, 6 = (0.01 + R2 + R3 + 0.02) x 3

R2 + R3 = 1.97

or R2 = 1.97 - 1.17 = 0.8 Ω At the low position, 6 = (0.01 + R1 + R2 + R3 + 0.02) x 1 R1 = 5.97 - 1.97 = 4 Ω

R1 + R2 + R3 = 5.97

Chapter 2, Solution 75

100 Ω

R

VS

M 12 Ω

+

-

(a) When Rx = 0, then t Im = Ifs = R + Rm

R2 =

E2 2 − Rm= − 100 = 19.9kΩ I fs 0.1x10 3

I fs = 0.05mA 2 E 2 − (R + R m ) = − 20kΩ = 20 kΩ Rx = Im 0.05x10 −3

(b) For half-scale deflection, Im = Im =

E R + Rm + Rx

Chapter 2, Solution 76

For series connection, R = 2 x 0.4Ω = 0.8Ω V 2 (120) 2 p= = = 18 kΩ (low) R 0.8 For parallel connection, R = 1/2 x 0.4Ω = 0.2Ω V 2 (120) 2 p= = = 72 kW (high) R 0.2 Chapter 2, Solution 77

(a)

5 Ω = 10 10 = 20 20 20 20 i.e., four 20 Ω resistors in parallel.

(b)

311.8 = 300 + 10 + 1.8 = 300 + 20 20 + 1.8 i.e., one 300Ω resistor in series with 1.8Ω resistor and a parallel combination of two 20Ω resistors.

(c)

40kΩ = 12kΩ + 28kΩ = 24 24k + 56k 50k i.e., Two 24kΩ resistors in parallel connected in series with two 50kΩ resistors in parallel.

(d)

42.32kΩ = 42l + 320 = 24k + 28k = 320 = 24k = 56k 56k + 300 + 20 i.e., A series combination of 20Ω resistor, 300Ω resistor, 24kΩ resistor and a parallel combination of two 56kΩ resistors.

Chapter 2, Solution 78 The equivalent circuit is shown below: R

VS

V0 =

+

+

V0

-

(1-α)R

-

(1 − α)R VS = (1 − α )R 0 VS R + (1 − α)R

V0 = (1 − α)R VS

Chapter 2, Solution 79 Since p = v2/R, the resistance of the sharpener is R = v2/(p) = 62/(240 x 10-3) = 150Ω I = p/(v) = 240 mW/(6V) = 40 mA Since R and Rx are in series, I flows through both. IRx = Vx = 9 - 6 = 3 V Rx = 3/(I) = 3/(40 mA) = 3000/(40) = 75 Ω Chapter 2, Solution 80 The amplifier can be modeled as a voltage source and the loudspeaker as a resistor:

V

+

V

R1

-

Case 1

Hence p =

V 2 p2 R1 , = R p1 R 2

+

R2

-

Case 2

p2 =

R1 10 p1 = (12) = 30 W 4 R2

Chapter 2, Solution 81 Let R1 and R2 be in kΩ. R eq = R 1 + R 2 5

(1)

5 R2 V0 = VS 5 R 2 + R 1

(2)

From (1) and (2), 0.05 =

5 R1

2 = 5 R2 =

40

From (1), 40 = R1 + 2

5R 2 or R2 = 3.33 kΩ 5+ R2

R1 = 38 kΩ

Thus R1 = 38 kΩ, R2 = 3.33 kΩ Chapter 2, Solution 82 (a)

10 Ω 40 Ω 10 Ω 80 Ω 1

2 R12

R12 = 80 + 10 (10 + 40) = 80 +

50 = 88.33 Ω 6

(b)

3 10 Ω 10 Ω

20 Ω 40 Ω R13

80 Ω 1

R13 = 80 + 10 (10 + 40) + 20 = 100 + 10 50 = 108.33 Ω

4

(c)

20 Ω 10 Ω R14 10 Ω

40 Ω

80 Ω 1

R14 = 80 + 0 (10 + 40 + 10) + 20 = 80 + 0 + 20 = 100 Ω Chapter 2, Solution 83

The voltage across the tube is 2 x 60 mV = 0.06 V, which is negligible compared with 24 V. Ignoring this voltage amp, we can calculate the current through the devices. p1 45mW = = 5mA V1 9V p 480mW I2 = 2 = = 20mA V2 24

I1 =

60 mA

i2 = 20 mA iR1

24 V

R1

+

-

i1 = 5 mA

R2 iR2

By applying KCL, we obtain

I R1 = 60 − 20 = 40 mA and I R 2 = 40 − 5 = 35 mA Hence, I R1 R1 = 24 - 9 = 15 V

I R 2 R 2 = 9V

R2 =

R1 =

15V = 375 Ω 40mA

9V = 257.14 Ω 35mA

Chapter 3, Solution 1. 40 Ω

v1

v2

8Ω

6A

2Ω

10 A

At node 1, 6 = v1/(8) + (v1 - v2)/4

48 = 3v1 - 2v2

(1)

40 = v1 - 3v2

(2)

At node 2, v1 - v2/4 = v2/2 + 10 Solving (1) and (2), v1 = 9.143V, v2 = -10.286 V

v12 (9.143)2 P8Ω = = = 10.45 W 8 8 P4Ω =

(v 1 − v 2 )2 4

= 94.37 W

v12 (= 10.286)2 = = 52.9 W P2Ω = 2 2 Chapter 3, Solution 2

At node 1, v − v2 − v1 v1 − = 6+ 1 10 5 2 At node 2, v2 v − v2 = 3+ 6+ 1 4 2 Solving (1) and (2), v1 = 0 V, v2 = 12 V

60 = - 8v1 + 5v2

36 = - 2v1 + 3v2

(1)

(2)

Chapter 3, Solution 3

Applying KCL to the upper node, 10 =

v0 vo vo v + + +2+ 0 10 20 30 60

i1 =

v0 v v v = 4 A , i2 = 0 = 2 A, i3 = 0 = 1.33 A, i4 = 0 = 67 mA 10 20 30 60

v0 = 40 V

Chapter 3, Solution 4 2A

v1 i1 4A

5Ω

i2

v2 i3

10 Ω

10 Ω

i4 5Ω

At node 1, 4 + 2 = v1/(5) + v1/(10)

v1 = 20

At node 2, 5 - 2 = v2/(10) + v2/(5)

v2 = 10

i1 = v1/(5) = 4 A, i2 = v1/(10) = 2 A, i3 = v2/(10) = 1 A, i4 = v2/(5) = 2 A Chapter 3, Solution 5

Apply KCL to the top node. 30 − v 0 20 − v 0 v 0 + = 2k 6k 4k

v0 = 20 V

5A

Chapter 3, Solution 6

i1 + i2 + i3 = 0

v 2 − 12 v 0 v 0 − 10 + + =0 4 6 2

or v0 = 8.727 V Chapter 3, Solution 7

At node a, 10 − Va Va Va − Vb (1) = +  → 10 = 6Va − 3Vb 30 15 10 At node b, Va − Vb 12 − Vb − 9 − Vb + + =0  → 24 = 2Va − 7Vb 10 20 5 Solving (1) and (2) leads to Va = -0.556 V, Vb = -3.444V

(2)

Chapter 3, Solution 8 3Ω

i1

v1

i3

5Ω

i2 + V0

3V 2Ω

+

–

+ 4V0 –

– 1Ω

v1 v1 − 3 v1 − 4 v 0 + + =0 5 1 5 2 8 v 0 = v1 so that v1 + 5v1 - 15 + v1 - v1 = 0 5 5 or v1 = 15x5/(27) = 2.778 V, therefore vo = 2v1/5 = 1.1111 V

i1 + i2 + i3 = 0 But

Chapter 3, Solution 9 3Ω

i1

v1

+ v0 – 12V

6Ω

i3 i2

+

+

v1

–

8Ω

–

+ –

2v0

At the non-reference node, 12 − v1 v1 v1 − 2 v 0 = + 3 8 6

(1)

But -12 + v0 + v1 = 0

v0 = 12 - v1

(2)

Substituting (2) into (1), 12 − v1 v1 3v1 − 24 = + 3 8 6

v0 = 3.652 V

Chapter 3, Solution 10

At node 1, v 2 − v1 v = 4+ 1 1 8

32 = -v1 + 8v2 - 8v0 1Ω

4A v1 8Ω

i0

2i0 v0

v2 2Ω

4Ω

(1)

At node 0, 4=

v0 v + 2I 0 and I 0 = 1 8 2

16 = 2v0 + v1

(2)

v2 = v1

(3)

At node 2, 2I0 =

v 2 − v1 v 2 v + and I 0 = 1 1 4 8

From (1), (2) and (3), v0 = 24 V, but from (2) we get v 4− o 2 = 2 − 24 = 2 − 6 = - 4 A io = 4 2 Chapter 3, Solution 11 4Ω

i1 v i2

3Ω

i3 10 V

+

–

5A

6Ω

Note that i2 = -5A. At the non-reference node 10 − v v +5= 4 6 i1 =

v = 18

10 − v = -2 A, i2 = -5 A 4

Chapter 3, Solution 12 10 Ω

v1

20 Ω

50 Ω

v2

i3 24 V

+

–

40 Ω

5A

At node 1,

24 − v 1 v − v 2 v1 − 0 = 1 + 10 20 40

At node 2, 5 +

v1 − v 2 v 2 = 20 50

96 = 7v1 - 2v2

500 = -5v1 + 7v2

(1)

(2)

Solving (1) and (2) gives, v1 = 42.87 V, v2 = 102.05 V v v i1 = 1 = 1.072 A, v2 = 2 = 2.041 A 40 50

Chapter 3, Solution 13

At node number 2, [(v2 + 2) – 0]/10 + v2/4 = 3 or v2 = 8 volts But, I = [(v2 + 2) – 0]/10 = (8 + 2)/10 = 1 amp and v1 = 8x1 = 8volts Chapter 3, Solution 14

5A

v0

v1 1Ω

8Ω

2Ω 4Ω

40 V

20 V

–

+

+

–

At node 1,

40 − v 0 v1 − v 0 +5= 1 2

At node 0,

v1 − v 0 v v + 20 +5= 0 + 0 2 4 8

Solving (1) and (2), v0 = 20 V

v1 + v0 = 70

4v1 - 7v0 = -20

(1)

(2)

Chapter 3, Solution 15

5A

v0

v1 1Ω

2Ω 4Ω

40 V

8Ω

20 V

+

–

Nodes 1 and 2 form a supernode so that v1 = v2 + 10 At the supernode, 2 + 6v1 + 5v2 = 3 (v3 - v2) At node 3, 2 + 4 = 3 (v3 - v2)

(1) 2 + 6v1 + 8v2 = 3v3

v3 = v2 + 2

2 + 6v2 + 60 + 8v2 = 3v2 + 6

v2 =

54 11

i0 = 6vi = 29.45 A 2

P65 =

v12  54  = v12 G =   6 = 144.6 W R  11  2

 − 56  P55 = v G =   5 = 129.6 W  11  2 2

P35 = (v L − v 3 ) G = (2) 2 3 = 12 W 2

(2) (3)

Substituting (1) and (3) into (2),

v1 = v2 + 10 =

–

+

− 56 11

Chapter 3, Solution 16 2S

i0 2A

8S

v2

v1

+

1S

v0

4S

v3 13 V

–

+

–

At the supernode, 2 = v1 + 2 (v1 - v3) + 8(v2 – v3) + 4v2, which leads to 2 = 3v1 + 12v2 - 10v3

(1)

But v1 = v2 + 2v0 and v0 = v2. Hence v1 = 3v2 v3 = 13V

(2) (3)

Substituting (2) and (3) with (1) gives, v1 = 18.858 V, v2 = 6.286 V, v3 = 13 V

Chapter 3, Solution 17 i0 4Ω

2Ω 10 Ω

60 V 60 V

8Ω +

–

3i0

60 − v1 v1 v1 − v 2 = + 4 8 2 60 − v 2 v1 − v 2 At node 2, 3i0 + + =0 10 2

At node 1,

120 = 7v1 - 4v2

(1)

60 − v1 . 4

But i0 = Hence

3(60 − v1 ) 60 − v 2 v1 − v 2 + + =0 4 10 2

1020 = 5v1 - 12v2

Solving (1) and (2) gives v1 = 53.08 V. Hence i0 =

60 − v1 = 1.73 A 4

(2)

Chapter 3, Solution 18

–+

v2

v1 2Ω

5A

v3 2Ω 8Ω

4Ω

10 V

+

+

v1

v3

–

(a)

At node 2, in Fig. (a), 5 =

At the supernode,

–

(b)

v 2 − v1 v 2 − v3 + 2 2

10 = - v1 + 2v2 - v3

v 2 − v1 v 2 − v 3 v1 v 3 + = + 2 2 4 8

From Fig. (b), - v1 - 10 + v3 = 0

v3 = v1 + 10

Solving (1) to (3), we obtain v1 = 10 V, v2 = 20 V = v3

40 = 2v1 + v3

(1)

(2) (3)

Chapter 3, Solution 19

At node 1, V1 − V3 V1 − V2 V1 + + 2 8 4 At node 2,

5 = 3+

V1 − V2 V2 V2 − V3 = + 8 2 4 At node 3,

12 − V3

 →

 →

+

 7 − 1 − 4  V1   16        − 1 7 − 2 V2  =  0  4 2 − 7  V3   − 36   Using MATLAB,  10  V = A −1 B =  4.933  12.267 

 →

 →

(1)

0 = −V1 + 7V2 − 2V3

V1 − V3 V2 − V3 + =0 8 2 4 From (1) to (3),

3+

16 = 7V1 − V2 − 4V3

 →

(2)

− 36 = 4V1 + 2V2 − 7V3 (3)

AV = B

V1 = 10 V, V2 = 4.933 V, V3 = 12.267 V

Chapter 3, Solution 20

Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence V1 V2 V3 + + =0  → V1 + 4V2 + V3 = 0 (1) 4 1 4 . V1

4Ω

.

V2

1Ω

2Ω

V3

4Ω

Between nodes 1 and 3, − V1 + 12 + V3 = 0  → V3 = V1 − 12 Similarly, between nodes 1 and 2, V1 = V2 + 2i But i = V3 / 4 . Combining this with (2) and (3) gives . V2

= 6 + V1 / 2

(2) (3) (4)

Solving (1), (2), and (4) leads to V1 = −3V, V2 = 4.5V, V3 = −15V

Chapter 3, Solution 21 4 kΩ

v1

2 kΩ

v3

3v0 +

3v0 v2

+ v0

3 mA

–

1 kΩ

+

+

+

v3

v2

–

–

(b)

(a)

Let v3 be the voltage between the 2kΩ resistor and the voltage-controlled voltage source. At node 1, v − v 2 v1 − v 3 3x10 −3 = 1 12 = 3v1 - v2 - 2v3 (1) + 4000 2000 At node 2, v1 − v 2 v1 − v 3 v 2 3v1 - 5v2 - 2v3 = 0 (2) + = 4 2 1 Note that v0 = v2. We now apply KVL in Fig. (b) - v3 - 3v2 + v2 = 0 From (1) to (3), v1 = 1 V, v2 = 3 V

v3 = - 2v2

(3)

Chapter 3, Solution 22

At node 1,

12 − v 0 v1 v − v0 = +3+ 1 2 4 8

At node 2, 3 +

24 = 7v1 - v2

(1)

v1 − v 2 v 2 + 5v 2 = 8 1

But, v1 = 12 - v1 Hence, 24 + v1 - v2 = 8 (v2 + 60 + 5v1) = 4 V 456 = 41v1 - 9v2

(2)

Solving (1) and (2), v1 = - 10.91 V, v2 = - 100.36 V Chapter 3, Solution 23

At the supernode, 5 + 2 =

v1 v 2 + 10 5

70 = v1 + 2v2

(1)

v2 = v1 + 8

(2)

Considering Fig. (b), - v1 - 8 + v2 = 0 Solving (1) and (2), v1 = 18 V, v2 = 26 V v1

v2

5A

–+ 2A

10 Ω

5Ω

8V

+

+

v1

v2

–

(a)

–

(b)

Chapter 3, Solution 24

6mA 1 kΩ

2 kΩ V1

+ 30V -

3 kΩ V2

io

-

4 kΩ

5 kΩ

At node 1, 30 −V 1 V V − V2 =6+ 1 + 1  → 96 = 7V1 − 2V2 1 4 2 At node 2, (−15 −V 2) V2 V2 − V1 6+ = +  → 30 = −15V1 + 31V2 3 5 2 Solving (1) and (2) gives V1=16.24. Hence io = V1/4 = 4.06 mA Chapter 3, Solution 25

20V

+

i0

(2)

v0

2Ω

10V

–

1Ω

(1)

40V

+

–

4Ω

+

–

2Ω

Using nodal analysis, 20 − v 0 40 − v 0 10 − v 0 v −0 + + = 0 1 2 2 4

i0 =

20 − v 0 = 0A 1

v0 = 20V

15V +

Chapter 3, Solution 26 At node 1, V − V3 V1 − V2 15 − V1 = 3+ 1 +  → − 45 = 7V1 − 4V2 − 2V3 20 10 5 At node 2, V1 − V2 4 I o − V2 V2 − V3 + = 5 5 5 V − V3 . Hence, (2) becomes But I o = 1 10 0 = 7V1 − 15V2 + 3V3 At node 3, V − V3 − 10 − V3 V2 − V3 3+ 1 + + =0  → − 10 = V1 + 2V2 − 5V3 10 5 5 Putting (1), (3), and (4) in matrix form produces

 7 − 4 − 2  V1   − 45        → AV = B  7 − 15 3 V2  =  0  1 2 − 5  V3   − 10   Using MATLAB leads to  − 9.835    −1 V = A B =  − 4.982   − 1.96    Thus, V1 = −9.835 V, V2 = −4.982 V, V3 = −1.95 V

Chapter 3, Solution 27 At node 1, 2 = 2v1 + v1 – v2 + (v1 – v3)4 + 3i0, i0 = 4v2. Hence, At node 2,

2 = 7v1 + 11v2 – 4v3

v1 – v2 = 4v2 + v2 – v3

(1) 0 = – v1 + 6v2 – v3

At node 3, 2v3 = 4 + v2 – v3 + 12v2 + 4(v1 – v3)

(2)

(1)

(2)

(3) (4)

or

– 4 = 4v1 + 13v2 – 7v3

(3)

In matrix form, 7 11 − 4  v 1   2  1 − 6 1   v  =  0   2     4 13 − 7  v 3   − 4 7

11

∆ = 1 −6 4

13 7

2

−4

2

11

1 = 176, ∆ 1 = 0

−6

−7

−4

−4

7

∆2 = 1 0 1 = 66, 4 −4 −7

v1 =

13 11

−4 1 = 110 −7 2

∆ 3 = 1 − 6 0 = 286 4 13 − 4

∆ 1 110 ∆ 66 = = 0.625V, v2 = 2 = = 0.375V ∆ ∆ 176 176

v3 =

∆3 286 = = 1.625V. ∆ 176

v1 = 625 mV, v2 = 375 mV, v3 = 1.625 V. Chapter 3, Solution 28 At node c, Vd − Vc Vc − Vb Vc = +  → 0 = −5Vb + 11Vc − 2Vd (1) 10 4 5 At node b, Va + 45 − Vb Vc − Vb Vb + =  → − 45 = Va − 4Vb + 2Vc (2) 8 4 8 At node a, Va − 30 − Vd Va Va + 45 − Vb + + =0  → 30 = 7Va − 2Vb − 4Vd (3) 4 16 8 At node d, Va − 30 − Vd Vd Vd − Vc = +  → 150 = 5Va + 2Vc − 7Vd (4) 4 20 10 In matrix form, (1) to (4) become

 0 − 5 11 − 2  Va   0        1 − 4 2 0  Vb   − 45   →  7 − 2 0 − 4  V  =  30  c       5 0 2 − 7 V   150    d    We use MATLAB to invert A and obtain

AV = B

 − 10.14     7.847  −1 V = A B= − 1.736     − 29.17    Thus, Va = −10.14 V, Vb = 7.847 V, Vc = −1.736 V, Vd = −29.17 V

Chapter 3, Solution 29 At node 1, 5 + V1 − V4 + 2V1 + V1 − V2 = 0  → − 5 = 4V1 − V2 − V4 At node 2, V1 − V2 = 2V2 + 4(V2 − V3 ) = 0  → 0 = −V1 + 7V2 − 4V3 At node 3, 6 + 4(V2 − V3 ) = V3 − V4  → 6 = −4V2 + 5V3 − V4 At node 4, 2 + V3 − V4 + V1 − V4 = 3V4  → 2 = −V1 − V3 + 5V4 In matrix form, (1) to (4) become  4 − 1 0 − 1 V1   − 5        − 1 7 − 4 0 V2   0  AV = B  →  0 − 4 5 − 1 V  =  6  3       − 1 0 − 1 5 V   2    4    Using MATLAB,

 − 0.7708     1.209  −1 V = A B= 2.309     0.7076    i.e. V1 = −0.7708 V, V2 = 1.209 V, V3 = 2.309 V, V4 = 0.7076 V

(1)

(2) (3) (4)

Chapter 3, Solution 30 v2 40 Ω

I0 v1 10 Ω 100 V

+

120 V

20 Ω

v0

1

2 4v0

–

–+

+ –

2I0

80 Ω

At node 1, v 1 − v 2 100 − v 1 4 v o − v 1 = + 40 10 20

But, vo = 120 + v2

(1)

v2 = vo – 120. Hence (1) becomes 7v1 – 9vo = 280

(2)

At node 2, Io + 2Io =

vo − 0 80

 v + 120 − v o  v o 3 1 = 40  80  or

6v1 – 7vo = -720

from (2) and (3),

7 − 9  v 1   280  6 − 7  v  =  − 720    o  

∆=

∆1 =

(3)

7 −9 = −49 + 54 = 5 6 −7

280 − 9 = −8440 , − 720 − 7

∆2 =

7 280 = −6720 6 − 720

v1 =

∆ ∆1 − 8440 − 6720 = = −1688, vo = 2 = − 1344 V ∆ ∆ 5 5

Io = -5.6 A Chapter 3, Solution 31 1Ω

+ v0 – v2

v1 1A

2v0

v3

2Ω

i0 4Ω

1Ω

10 V

4Ω

At the supernode, 1 + 2v0 =

v1 v 2 v1 − v 3 + + 4 1 1

(1)

But vo = v1 – v3. Hence (1) becomes, 4 = -3v1 + 4v2 +4v3 At node 3, 2vo + or

10 − v 3 v2 = v1 − v 3 + 4 2

20 = 4v1 + v2 – 2v3

At the supernode, v2 = v1 + 4io. But io = v2 = v1 + v3 Solving (2) to (4) leads to, v1 = 4 V, v2 = 4 V, v3 = 0 V.

(2)

(3)

v3 . Hence, 4

(4)

+

–

Chapter 3, Solution 32 5 kΩ

v1

v3

v2

+ 10 kΩ

4 mA

10 V

20 V

–+

+–

+

loop 1

v1

12 V

–

+

loop 2

–

v3

–

(b)

(a)

We have a supernode as shown in figure (a). It is evident that v2 = 12 V, Applying KVL to loops 1and 2 in figure (b), we obtain, -v1 – 10 + 12 = 0 or v1 = 2 and -12 + 20 + v3 = 0 or v3 = -8 V Thus,

v1 = 2 V, v2 = 12 V, v3 = -8V.

Chapter 3, Solution 33 (a) This is a non-planar circuit because there is no way of redrawing the circuit with no crossing branches. (b) This is a planar circuit. It can be redrawn as shown below.

4Ω

3Ω

12 V

+

5Ω

–

1Ω

2Ω

Chapter 3, Solution 34

(a)

This is a planar circuit because it can be redrawn as shown below, 7Ω 2Ω

1Ω

3Ω

6Ω 10 V

5Ω

+

–

4Ω

(b)

This is a non-planar circuit.

Chapter 3, Solution 35

30 V

20 V

+

–

+

–

i1 2 kΩ

+

i2

v0

–

5 kΩ

4 kΩ

Assume that i1 and i2 are in mA. We apply mesh analysis. For mesh 1, -30 + 20 + 7i1 – 5i2 = 0 or 7i1 – 5i2 = 10

(1)

For mesh 2, -20 + 9i2 – 5i1 = 0 or -5i1 + 9i2 = 20 Solving (1) and (2), we obtain, i2 = 5. v0 = 4i2 = 20 volts.

(2)

Chapter 3, Solution 36 10 V

4Ω i1

12 V

+–

i2

I1

+

I2

6Ω

–

i3

2Ω

Applying mesh analysis gives, 12 = 10I1 – 6I2 -10 = -6I1 + 8I2  6   5 − 3  I 1   − 5 =  − 3 4   I   2    

or

∆=

5 −3 6 −3 5 6 = 11, ∆1 = = 9, ∆ 2 = = −7 −3 4 −5 4 −3 −5 I1 =

∆1 9 I = ∆2 = − 7 = , 2 ∆ 11 ∆ 11

i1 = -I1 = -9/11 = -0.8181 A, i2 = I1 – I2 = 10/11 = 1.4545 A. vo = 6i2 = 6x1.4545 = 8.727 V. Chapter 3, Solution 37 3Ω

3V

+ v0

–

5Ω

2Ω

i1 1Ω

+

–

i2

4v0

+ –

Applying mesh analysis to loops 1 and 2, we get, 6i1 – 1i2 + 3 = 0 which leads to i2 = 6i1 + 3

(1)

-1i1 + 6i2 – 3 + 4v0 = 0

(2)

But, v0 = -2i1

(3)

Using (1), (2), and (3) we get i1 = -5/9. Therefore, we get v0 = -2i1 = -2(-5/9) = 1.111 volts Chapter 3, Solution 38 3Ω

6Ω

+ v0 – 12 V

+

–

i1

8Ω

2v0

i2

+ –

We apply mesh analysis. 12 = 3 i1 + 8(i1 – i2) which leads to 12 = 11 i1 – 8 i2

(1)

-2 v0 = 6 i2 + 8(i2 – i1) and v0 = 3 i1 or i1 = 7 i2

(2)

From (1) and (2), i1 = 84/69 and v0 = 3 i1 = 3x89/69 v0 = 3.652 volts

Chapter 3, Solution 39

For mesh 1, − 10 − 2 I x + 10 I 1 − 6 I 2 = 0 But I x = I 1 − I 2 . Hence, 10 = −12 I 1 + 12 I 2 + 10 I 1 − 6 I 2  → 5 = 4 I 1 − 2 I 2 For mesh 2, 12 + 8I 2 − 6 I 1 = 0  → 6 = 3I 1 − 4 I 2 Solving (1) and (2) leads to I 1 = 0.8 A, I 2 = -0.9A

(1) (2)

Chapter 3, Solution 40 2 kΩ

30V

+

i2 2 kΩ

i1

–

6 kΩ

6 kΩ

i3

4 kΩ

4 kΩ

Assume all currents are in mA and apply mesh analysis for mesh 1. 30 = 12i1 – 6i2 – 4i3

15 = 6i1 – 3i2 – 2i3

(1)

0 = -3i1 + 7i2 – i3

(2)

0 = -2i1 – i2 + 5i3

(3)

for mesh 2, 0 = - 6i1 + 14i2 – 2i3 for mesh 2, 0 = -4i1 – 2i2 + 10i3 Solving (1), (2), and (3), we obtain, io = i1 = 4.286 mA. Chapter 3, Solution 41 10 Ω

i1

6V

2Ω

+– 1Ω

i2 4Ω 8V

i3

+

–

i i2

i3 0

5Ω

For loop 1, 6 = 12i1 – 2i2

3 = 6i1 – i2

(1)

For loop 2, -8 = 7i2 – 2i1 – i3

(2)

For loop 3, -8 + 6 + 6i3 – i2 = 0

2 = 6i3 – i2

We put (1), (2), and (3) in matrix form, 6 − 1 0  i1   3  2 − 7 1  i  =  8    2    0 − 1 6 i 3  2 6

−1 0

6 3 0

∆ = 2 − 7 1 = −234, ∆ 2 = 2 8 1 = −240 0

−1 6

0 2 6

6

−1 3

∆ 3 = 2 − 7 8 = −38 0 −1 2

At node 0, i + i2 = i3 or i = i3 – i2 =

∆3 − ∆2 − 38 − 240 = = 1.188 A − 234 ∆

(3)

Chapter 3, Solution 42

For mesh 1, − 12 + 50 I 1 − 30 I 2 = 0  → 12 = 50 I 1 − 30 I 2 (1) For mesh 2, − 8 + 100 I 2 − 30 I 1 − 40 I 3 = 0  → 8 = −30 I 1 + 100 I 2 − 40 I 3 For mesh 3, (3) − 6 + 50 I 3 − 40 I 2 = 0  → 6 = −40 I 2 + 50 I 3 Putting eqs. (1) to (3) in matrix form, we get 0  I 1  12   50 − 30       − 30 100 − 40  I 2  =  8   0 − 40 50  I 3   6  

 →

(2)

AI = B

Using Matlab,  0.48    I = A B =  0.40   0.44    −1

i.e. I1 = 0.48 A, I2 = 0.4 A, I3 = 0.44 A

Chapter 3, Solution 43 20 Ω a 80 V

+

i1

–

30 Ω

+

i3

30 Ω

20 Ω 80 V

+

i2

–

20 Ω

30 Ω

Vab

– b

For loop 1, 80 = 70i1 – 20i2 – 30i3

8 = 7i1 – 2i2 – 3i3

(1)

For loop 2, 80 = 70i2 – 20i1 – 30i3

8 = -2i1 + 7i2 – 3i3

(2)

0 = -30i1 – 30i2 + 90i3

0 = i1 + i2 – 3i3

(3)

For loop 3,

Solving (1) to (3), we obtain i3 = 16/9 Io = i3 = 16/9 = 1.778 A Vab = 30i3 = 53.33 V. Chapter 3, Solution 44 6V +

2Ω

i3

4Ω

i2

1Ω 6V 5Ω

i1 3A i1

i2

Loop 1 and 2 form a supermesh. For the supermesh, 6i1 + 4i2 - 5i3 + 12 = 0

(1)

For loop 3,

-i1 – 4i2 + 7i3 + 6 = 0

(2)

Also,

i2 = 3 + i1

(3)

Solving (1) to (3), i1 = -3.067, i3 = -1.3333; io = i1 – i3 = -1.7333 A

+

–

Chapter 3, Solution 45 4Ω

30V

+

i3

i4

2Ω

6Ω

i1

–

8Ω

3Ω

i2

1Ω

For loop 1,

30 = 5i1 – 3i2 – 2i3

(1)

For loop 2,

10i2 - 3i1 – 6i4 = 0

(2)

For the supermesh,

6i3 + 14i4 – 2i1 – 6i2 = 0

(3)

But

i4 – i3 = 4 which leads to i4 = i3 + 4

(4)

Solving (1) to (4) by elimination gives i = i1 = 8.561 A. Chapter 3, Solution 46 For loop 1, − 12 + 11i1 − 8i2 = 0  → For loop 2, − 8i1 + 14i2 + 2vo = 0 But vo = 3i1 ,

11i1 − 8i2 = 12

(1)

− 8i1 + 14i2 + 6i1 = 0  → i1 = 7i2 (2) Substituting (2) into (1), 77i2 − 8i2 = 12  → i 2 = 0.1739 A and i1 = 7i2 = 1.217 A

Chapter 3, Solution 47 First, transform the current sources as shown below. - 6V +

2Ω

V1

8Ω

4Ω

V2

I3

4Ω + 20V -

I1

2Ω

V3

8Ω I2

+ 12V -

For mesh 1, − 20 + 14 I 1 − 2 I 2 − 8I 3 = 0  → 10 = 7 I 1 − I 2 − 4 I 3 For mesh 2, 12 + 14 I 2 − 2 I 1 − 4 I 3 = 0  → − 6 = − I 1 + 7 I 2 − 2 I 3 For mesh 3, − 6 + 14 I 3 − 4 I 2 − 8I 1 = 0  → 3 = −4 I 1 − 2 I 2 + 7 I 3 Putting (1) to (3) in matrix form, we obtain  7 − 1 − 4  I 1   10        → AI = B  − 1 7 − 2  I 2  =  − 6   − 4 − 2 7  I   3    3   

Using MATLAB,  2  −1 I = A B = 0.0333 1.8667  But

 → I 1 = 2.5, I 2 = 0.0333, I 3 = 1.8667

20 − V  → V1 = 20 − 4 I1 = 10 V 4 V2 = 2( I1 − I 2 ) = 4.933 V Also, V − 12 I2 = 3  → V3 = 12 + 8I 2 = 12.267V 8 I1 =

(1)

(2) (3)

Chapter 3, Solution 48

We apply mesh analysis and let the mesh currents be in mA. 3k Ω I4 4k Ω

2k Ω Io

1k Ω + 12 V -

I1

5k Ω

I2 + 8V -

I3

10k Ω

6V +

For mesh 1, − 12 + 8 + 5I 1 − I 2 − 4 I 4 = 0  → 4 = 5I 1 − I 2 − 4 I 4 (1) For mesh 2, − 8 + 13I 2 − I 1 − 10 I 3 − 2 I 4 = 0  → 8 = − I 1 + 13I 2 − 10 I 3 − 2 I 4 (2) For mesh 3, (3) − 6 + 15I 3 − 10 I 2 − 5I 4 = 0  → 6 = −10 I 2 + 15I 3 − 5I 4 For mesh 4, − 4 I 1 − 2 I 2 − 5I 3 + 14 I 4 = 0 (4) Putting (1) to (4) in matrix form gives −1 − 4  I 1   4  0  5       − 1 13 − 10 − 2  I 2   8  AI = B  →  0 − 10 15 − 5  I  =  6  3       − 4 − 2 − 5 14  I   0    4    Using MATLAB,  7.217    8 . 087   I = A −1 B =  7.791     6    The current through the 10k Ω resistor is Io= I2 – I3 = 0.2957 mA

Chapter 3, Solution 49 3Ω

i3 2Ω

1Ω

i1

2Ω

16 V

i2

+

–

2i0 i1

i2

0 (a)

2Ω

1Ω

2Ω

+ i1

v0

+ or

–

v0

–

i2

16V +

–

(b)

For the supermesh in figure (a), 3i1 + 2i2 – 3i3 + 16 = 0

(1)

At node 0,

i2 – i1 = 2i0 and i0 = -i1 which leads to i2 = -i1

(2)

For loop 3,

-i1 –2i2 + 6i3 = 0 which leads to 6i3 = -i1

(3)

Solving (1) to (3), i1 = (-32/3)A, i2 = (32/3)A, i3 = (16/9)A i0 = -i1 = 10.667 A, from fig. (b), v0 = i3-3i1 = (16/9) + 32 = 33.78 V.

Chapter 3, Solution 50

i1

4Ω

2Ω

i3

10 Ω 8Ω 60 V

+

i2

–

3i0 i3

i2

For loop 1,

16i1 – 10i2 – 2i3 = 0 which leads to 8i1 – 5i2 – i3 = 0

(1)

For the supermesh, -60 + 10i2 – 10i1 + 10i3 – 2i1 = 0 or

-6i1 + 5i2 + 5i3 = 30

(2)

Also, 3i0 = i3 – i2 and i0 = i1 which leads to 3i1 = i3 – i2

(3)

Solving (1), (2), and (3), we obtain i1 = 1.731 and i0 = i1 = 1.731 A Chapter 3, Solution 51 5A

i1 8Ω 2Ω

i3

1Ω

i2 40 V

+

–

4Ω

+ v0

20V

–

+

For loop 1,

i1 = 5A

(1)

For loop 2,

-40 + 7i2 – 2i1 – 4i3 = 0 which leads to 50 = 7i2 – 4i3

(2)

For loop 3,

-20 + 12i3 – 4i2 = 0 which leads to 5 = - i2 + 3 i3

(3)

Solving with (2) and (3), And,

i2 = 10 A, i3 = 5 A

v0 = 4(i2 – i3) = 4(10 – 5) = 20 V.

Chapter 3, Solution 52

+ v0 2 Ω

i2

– VS

+

–

8Ω

3A

i2

i1

i3 4Ω

i3

+ –

2V0

For mesh 1, 2(i1 – i2) + 4(i1 – i3) – 12 = 0 which leads to 3i1 – i2 – 2i3 = 6

(1)

For the supermesh, 2(i2 – i1) + 8i2 + 2v0 + 4(i3 – i1) = 0 But v0 = 2(i1 – i2) which leads to -i1 + 3i2 + 2i3 = 0

(2)

For the independent current source, i3 = 3 + i2

(3)

Solving (1), (2), and (3), we obtain, i1 = 3.5 A, i2 = -0.5 A, i3 = 2.5 A.

Chapter 3, Solution 53

+ v0 2 Ω

i2

– VS

+

–

8Ω

3A

i2

i1

i3 4Ω

i3

+ –

2V0

For mesh 1, 2(i1 – i2) + 4(i1 – i3) – 12 = 0 which leads to 3i1 – i2 – 2i3 = 6

(1)

For the supermesh, 2(i2 – i1) + 8i2 + 2v0 + 4(i3 – i1) = 0 But v0 = 2(i1 – i2) which leads to -i1 + 3i2 + 2i3 = 0

(2)

For the independent current source, i3 = 3 + i2

(3)

Solving (1), (2), and (3), we obtain, i1 = 3.5 A, i2 = -0.5 A, i3 = 2.5 A.

Chapter 3, Solution 54 Let the mesh currents be in mA. For mesh 1, − 12 + 10 + 2 I 1 − I 2 = 0  → 2 = 2I1 − I 2 (1) For mesh 2, − 10 + 3I 2 − I 1 − I 3 = 0  → 10 = − I 1 + 3I 2 − I 3 For mesh 3, (3) − 12 + 2 I 3 − I 2 = 0  → 12 = − I 2 + 2 I 3 Putting (1) to (3) in matrix form leads to  2 − 1 0  I 1   2        − 1 3 − 1 I 2  = 10   0 − 1 2  I  12    3    Using MATLAB,

 5.25  I = A B =  8.5  10.25 −1

 →

10 V

I2 i1

4A

AI = B

 → I 1 = 5.25 mA, I 2 = 8.5 mA, I 3 = 10.25 mA

Chapter 3, Solution 55 b

c

+

1A

I2

6Ω

1A I1

i2

I4

2Ω

i3

12 Ω a

I3

d

I4 4A

(2)

4Ω +– 8V

I3 0

It is evident that I1 = 4 For mesh 4,

12(I4 – I1) + 4(I4 – I3) – 8 = 0

For the supermesh At node c,

(1) (2)

6(I2 – I1) + 10 + 2I3 + 4(I3 – I4) = 0 or -3I1 + 3I2 + 3I3 – 2I4 = -5

(3)

I2 = I3 + 1

(4)

Solving (1), (2), (3), and (4) yields, I1 = 4A, I2 = 3A, I3 = 2A, and I4 = 4A At node b,

i1 = I2 – I1 = -1A

At node a,

i2 = 4 – I4 = 0A

At node 0,

i3 = I4 – I3 = 2A

Chapter 3, Solution 56 + v1 – 2Ω 2Ω

12 V

+

–

i2

2Ω

2Ω

i1

2Ω

i3

+ v2

–

For loop 1, 12 = 4i1 – 2i2 – 2i3 which leads to 6 = 2i1 – i2 – i3

(1)

For loop 2, 0 = 6i2 –2i1 – 2 i3 which leads to 0 = -i1 + 3i2 – i3

(2)

For loop 3, 0 = 6i3 – 2i1 – 2i2 which leads to 0 = -i1 – i2 + 3i3

(3)

In matrix form (1), (2), and (3) become,  2 − 1 − 1  i1  6  − 1 3 − 1 i  = 0   2     − 1 − 1 3  i 3  0 2

−1 −1

2

6 −1

∆ = − 1 3 − 1 = 8, ∆2 = − 1 3 − 1 = 24 −1 −1 3 −1 0 3 2

−1 6

∆3 = − 1 3 0 = 24 , therefore i2 = i3 = 24/8 = 3A, −1 −1 0 v1 = 2i2 = 6 volts, v = 2i3 = 6 volts

Chapter 3, Solution 57

Assume R is in kilo-ohms. V2 = 4kΩx18mA = 72V , V1 = 100 − V2 = 100 − 72 = 28V Current through R is 3 3 28 = iR = io , V1 = i R R  → (18) R 3+ R 3+ R This leads to R = 84/26 = 3.23 k Ω

Chapter 3, Solution 58 30 Ω

i2 30 Ω

10 Ω

i1

10 Ω

i3

+

–

120 V

30 Ω

For loop 1, 120 + 40i1 – 10i2 = 0, which leads to -12 = 4i1 – i2

(1)

For loop 2, 50i2 – 10i1 – 10i3 = 0, which leads to -i1 + 5i2 – i3 = 0

(2)

For loop 3, -120 – 10i2 + 40i3 = 0, which leads to 12 = -i2 + 4i3

(3)

Solving (1), (2), and (3), we get, i1 = -3A, i2 = 0, and i3 = 3A Chapter 3, Solution 59 40 Ω

I0

–+ 120 V

i2

10 Ω 20 Ω

i1

100V +

4v0

–

+

i3

+ –

v0

80 Ω

– 2I0

i2 For loop 1, -100 + 30i1 – 20i2 + 4v0 = 0, where v0 = 80i3 or 5 = 1.5i1 – i2 + 16i3

i3

(1)

For the supermesh, 60i2 – 20i1 – 120 + 80i3 – 4 v0 = 0, where v0 = 80i3 or 6 = -i1 + 3i2 – 12i3

(2)

Also, 2I0 = i3 – i2 and I0 = i2, hence, 3i2 = i3

(3)

From (1), (2), and (3),

3

∆ = −1 0

−2

32

 3 − 2 32   − 1 3 − 12   3 − 1   0 3

10

32

 i1  10 i  =  6   2   i 3   0  3

− 2 10

3

− 12 = 5, ∆2 = − 1

6

− 12 = −28, ∆3 = − 1

3

6 = −84

3

−1

0

−1

3

0

0

I0 = i2 = ∆2/∆ = -28/5 = -5.6 A v0 = 8i3 = (-84/5)80 = -1344 volts

0

Chapter 3, Solution 60 0.5i0

4Ω

v1

10 V

1Ω

10 V

8Ω

v2 2Ω

+

–

i0

At node 1, (v1/1) + (0.5v1/1) = (10 – v1)/4, which leads to v1 = 10/7 At node 2, (0.5v1/1) + ((10 – v2)/8) = v2/2 which leads to v2 = 22/7 P1Ω = (v1)2/1 = 2.041 watts, P2Ω = (v2)2/2 = 4.939 watts P4Ω = (10 – v1)2/4 = 18.38 watts, P8Ω = (10 – v2)2/8 = 5.88 watts Chapter 3, Solution 61 v1 is

20 Ω

v2

10 Ω i0

+ v0

–

30 Ω

– + 5v0

At node 1, is = (v1/30) + ((v1 – v2)/20) which leads to 60is = 5v1 – 3v2 But v2 = -5v0 and v0 = v1 which leads to v2 = -5v1 Hence, 60is = 5v1 + 15v1 = 20v1 which leads to v1 = 3is, v2 = -15is i0 = v2/50 = -15is/50 which leads to i0/is = -15/50 = -0.3

40 Ω

(1)

Chapter 3, Solution 62 4 kΩ

100V +

8 kΩ

A

i1

–

B

i2

2 kΩ

i3

+

–

40 V

We have a supermesh. Let all R be in kΩ, i in mA, and v in volts. For the supermesh, -100 +4i1 + 8i2 + 2i3 + 40 = 0 or 30 = 2i1 + 4i2 + i3

(1)

At node A,

i1 + 4 = i2

(2)

At node B,

i2 = 2i1 + i3

(3)

Solving (1), (2), and (3), we get i1 = 2 mA, i2 = 6 mA, and i3 = 2 mA. Chapter 3, Solution 63

10 Ω

A

5Ω 50 V

+

–

i1

i2 + –

For the supermesh, -50 + 10i1 + 5i2 + 4ix = 0, but ix = i1. Hence, 50 = 14i1 + 5i2 At node A, i1 + 3 + (vx/4) = i2, but vx = 2(i1 – i2), hence, i1 + 2 = i2 Solving (1) and (2) gives i1 = 2.105 A and i2 = 4.105 A vx = 2(i1 – i2) = -4 volts and ix = i2 – 2 = 4.105 amp

(1) (2)

4ix

Chapter 3, Solution 64 i1

50 Ω

i2 10 Ω + −

A

i0 i1

10 Ω

i2

+ –

4i0

i3

40 Ω

100V +

–

2A

0.2V0

i1

B

i3

20i2 – 10i1 + 4i0 = 0

For mesh 2,

(1)

But at node A, io = i1 – i2 so that (1) becomes i1 = (7/12)i2

(2)

For the supermesh, -100 + 50i1 + 10(i1 – i2) – 4i0 + 40i3 = 0 or

50 = 28i1 – 3i2 + 20i3

(3)

At node B,

i3 + 0.2v0 = 2 + i1

(4)

But,

v0 = 10i2 so that (4) becomes i3 = 2 – (17/12)i2

(5)

Solving (1) to (5), i2 = -0.674, v0 = 10i2 = -6.74 volts,

i0 = i1 - i2 = -(5/12)i2 = 0.281 amps

Chapter 3, Solution 65

For mesh 1, For mesh 2, For mesh 3, For mesh 4, For mesh 5,

12 = 12 I 1 − 6 I 2 − I 4 0 = −6 I 1 + 16 I 2 − 8I 3 − I 4 − I 5

9 = −8I 2 + 15I 3 − I 5 6 = − I1 − I 2 + 5I 4 − 2 I 5 10 = − I 2 − I 3 − 2 I 4 + 8I 5

(1) (2) (3) (4) (5)

Casting (1) to (5) in matrix form gives 1 0  I 1  12   12 − 6 0       − 6 16 − 8 − 1 − 1  I 2   0   0 − 8 15 0 − 1  I  =  9   → AI = B  3     5 − 2  I 4   6   −1 −1 0  0 − 1 − 1 − 2 8  I  10   5     Using MATLAB leads to  1.673     1.824  I = A −1 B =  1.733     2.864   2.411    Thus, I 1 = 1.673 A, I 2 = 1.824 A, I 3 = 1.733 A, I 4 = 1.864 A, I 5 = 2.411 A

Chapter 3, Solution 66

Consider the circuit below. 2 kΩ + 20V -

2 kΩ

1 kΩ

I1 1 kΩ

+ 10V -

I2 1 kΩ

Io 2 kΩ

1 kΩ I3

12V +

We use mesh analysis. Let the mesh currents be in mA. For mesh 1, 20 = 4 I 1 − I 2 − I 3 For mesh 2, − 10 = − I 1 + 4 I 2 − I 4 For mesh 3, 12 = − I 1 + 4 I 3 − I 4 For mesh 4, − 12 = − I 2 − I 3 + 4 I 4

2 kΩ I4

(1) (2) (3) (4)

In matrix form, (1) to (4) become  4 − 1 − 1 0  I 1   20        − 1 4 0 − 1 I 2   − 10  = −1 0 4 − 1 I 3   12        0 − 1 − 1 4  I   − 12    4    Using MATLAB,

 →

AI = B

 5.5     − 1.75  −1 I = A B= 3.75     − 2.5    Thus, I o = − I 3 = − 3.75 mA Chapter 3, Solution 67

G11 = (1/1) + (1/4) = 1.25, G22 = (1/1) + (1/2) = 1.5 G12 = -1 = G21, i1 = 6 – 3 = 3, i2 = 5-6 = -1 Hence, we have,

1.25 − 1  − 1 1.5  

−1

=

1.25 − 1  v 1   3   − 1 1.5  v  =  − 1   2   

1 1.5 1  , where ∆ = [(1.25)(1.5)-(-1)(-1)] = 0.875 ∆  1 1.25

 v 1  1.7143 1.1429  3  3(1.7143) − 1(1.1429)  4  v  = 1.1429 1.4286  − 1 = 3(1.1429) − 1(1.4286) = 2        2 

Clearly v1 = 4 volts and v2 = 2 volts

Chapter 3, Solution 68

By inspection, G11 = 1 + 3 + 5 = 8S, G22 = 1 + 2 = 3S, G33 = 2 + 5 = 7S G12 = -1, G13 = -5, G21 = -1, G23 = -2, G31 = -5, G32 = -2 i1 = 4, i2 = 2, i3 = -1 We can either use matrix inverse as we did in Problem 3.51 or use Cramer’s Rule. Let us use Cramer’s rule for this problem. First, we develop the matrix relationships.  8 − 1 − 5  v 1   4   − 1 3 − 2  v  =  2    2     − 5 − 2 7   v 3   − 1 8 ∆ = −1

−1 − 5 3

−5 −2

8

4

4

− 2 = 34, ∆ 1 = 2 7

−5

−1 −5 3

−1 − 2

8

−1

− 2 = 85 7

4

∆ 2 = − 1 2 − 2 = 109, ∆ 3 = − 1 3 2 = 87 − 5 −1 7 − 5 − 2 −1

v1 = ∆1/∆ = 85/34 = 3.5 volts, v2 = ∆2/∆ = 109/34 = 3.206 volts v3 = ∆3/∆ = 87/34 = 2.56 volts

Chapter 3, Solution 69

Assume that all conductances are in mS, all currents are in mA, and all voltages are in volts. G11 = (1/2) + (1/4) + (1/1) = 1.75, G22 = (1/4) + (1/4) + (1/2) = 1, G33 = (1/1) + (1/4) = 1.25, G12 = -1/4 = -0.25, G13 = -1/1 = -1, G21 = -0.25, G23 = -1/4 = -0.25, G31 = -1, G32 = -0.25 i1 = 20, i2 = 5, and i3 = 10 – 5 = 5 The node-voltage equations are: − 1   v 1  20  1.75 − 0.25  − 0.25 1 − 0.25  v 2  =  5       − 0.25 1.25   v 3   5   − 1

Chapter 3, Solution 70

G11 = G1 + G2 + G4, G12 = -G2, G13 = 0, G22 = G2 + G3, G21 = -G2, G23 = -G3, G33 = G1 + G3 + G5, G31 = 0, G32 = -G3 i1 = -I1, i2 = I2, and i3 = I1 Then, the node-voltage equations are: G 1 + G 2 + G 4  − G2  0 

− G2 G1 + G 2 − G3

  v 1  − I 1  v  =  I  − G3  2   2  G 1 + G 3 + G 5   v 3   I 1  0

Chapter 3, Solution 71

R11 = 4 + 2 = 6, R22 = 2 + 8 + 2 = 12, R33 = 2 + 5 = 7, R12 = -2, R13 = 0, R21 = -2, R23 = -2, R31 = 0, R32 = -2 v1 = 12, v2 = -8, and v3 = -20 Now we can write the matrix relationships for the mesh-current equations.  6 − 2 0   i 1   12   − 2 12 − 2 i  =  − 8    2     0 − 2 7  i 3   − 20

Now we can solve for i2 using Cramer’s Rule. 6 ∆ = −2 0

−2 12 −2

0

6

− 2 = 452, ∆ 2 = − 2 7

0

12 −8

0 − 2 = −408

− 20

7

i2 = ∆2/∆ = -0.9026, p = (i2)2R = 6.52 watts Chapter 3, Solution 72

R11 = 5 + 2 = 7, R22 = 2 + 4 = 6, R33 = 1 + 4 = 5, R44 = 1 + 4 = 5, R12 = -2, R13 = 0 = R14, R21 = -2, R23 = -4, R24 = 0, R31 = 0, R32 = -4, R34 = -1, R41 = 0 = R42, R43 = -1, we note that Rij = Rji for all i not equal to j. v1 = 8, v2 = 4, v3 = -10, and v4 = -4 Hence the mesh-current equations are: 0   i1   8   7 −2 0  − 2 6 − 4 0  i   4    2  =    0 − 4 5 − 1 i 3   − 10      0 − 1 5  i 4   − 4   0

Chapter 3, Solution 73

R11 = 2 + 3 +4 = 9, R22 = 3 + 5 = 8, R33 = 1 + 4 = 5, R44 = 1 + 1 = 2, R12 = -3, R13 = -4, R14 = 0, R23 = 0, R24 = 0, R34 = -1 v1 = 6, v2 = 4, v3 = 2, and v4 = -3 Hence,  9 − 3 − 4 0   i1   6  − 3 8 0 0  i 2   4   = − 4 0 6 − 1 i3   2       0 − 1 2  i 4  − 3 0

Chapter 3, Solution 74

R11 = R1 + R4 + R6, R22 = R2 + R4 + R5, R33 = R6 + R7 + R8, R44 = R3 + R5 + R8, R12 = -R4, R13 = -R6, R14 = 0, R23 = 0, R24 = -R5, R34 = -R8, again, we note that Rij = Rji for all i not equal to j.  V1  − V  2 The input voltage vector is =   V3     − V4  R 1 + R 4 + R 6  − R4   − R6  0 

− R4

− R6

R2 + R4 + R5

0

0 − R5

R6 + R7 + R8 − R8

  i 1   V1   i   − V  − R5 2  2  =  − R8  i 3   V3      R 3 + R 5 + R 8  i 4   − V4  0

Chapter 3, Solution 75 * Schematics Netlist * R_R4 R_R2 R_R1 R_R3 R_R5 V_V4 v_V3 v_V2 v_V1

$N_0002 $N_0001 30 $N_0001 $N_0003 10 $N_0005 $N_0004 30 $N_0003 $N_0004 10 $N_0006 $N_0004 30 $N_0003 0 120V $N_0005 $N_0001 0 0 $N_0006 0 0 $N_0002 0

i3

i1

i2

Clearly, i1 = -3 amps, i2 = 0 amps, and i3 = 3 amps, which agrees with the answers in Problem 3.44.

Chapter 3, Solution 76 * Schematics Netlist * I_I2 R_R1 R_R3 R_R2 F_F1 VF_F1 R_R4 R_R6 I_I1 R_R5

0 $N_0001 DC 4A $N_0002 $N_0001 0.25 $N_0003 $N_0001 1 $N_0002 $N_0003 1 $N_0002 $N_0001 VF_F1 3 $N_0003 $N_0004 0V 0 $N_0002 0.5 0 $N_0001 0.5 0 $N_0002 DC 2A 0 $N_0004 0.25

Clearly, v1 = 625 mVolts, v2 = 375 mVolts, and v3 = 1.625 volts, which agrees with the solution obtained in Problem 3.27.

Chapter 3, Solution 77 * Schematics Netlist * R_R2 I_I1 I_I3 R_R3 R_R1 I_I2

0 $N_0001 4 $N_0001 0 DC 3A $N_0002 $N_0001 DC 6A 0 $N_0002 2 $N_0001 $N_0002 1 0 $N_0002 DC 5A

Clearly, v1 = 4 volts and v2 = 2 volts, which agrees with the answer obtained in Problem 3.51.

Chapter 3, Solution 78

The schematic is shown below. When the circuit is saved and simulated the node voltages are displaced on the pseudocomponents as shown. Thus, V1 = −3V, V2 = 4.5V, V3 = −15V,

.

Chapter 3, Solution 79

The schematic is shown below. When the circuit is saved and simulated, we obtain the node voltages as displaced. Thus, Va = −5.278 V, Vb = 10.28 V, Vc = 0.6944 V, Vd = −26.88 V

Chapter 3, Solution 80 * Schematics Netlist * H_H1 VH_H1 I_I1 V_V1 R_R4 R_R1 R_R2 R_R5 R_R3

$N_0002 $N_0003 VH_H1 6 0 $N_0001 0V $N_0004 $N_0005 DC 8A $N_0002 0 20V 0 $N_0003 4 $N_0005 $N_0003 10 $N_0003 $N_0002 12 0 $N_0004 1 $N_0004 $N_0001 2

Clearly, v1 = 84 volts, v2 = 4 volts, v3 = 20 volts, and v4 = -5.333 volts Chapter 3, Solution 81

Clearly, v1 = 26.67 volts, v2 = 6.667 volts, v3 = 173.33 volts, and v4 = -46.67 volts which agrees with the results of Example 3.4.

This is the netlist for this circuit. * Schematics Netlist * R_R1 R_R2 R_R3 R_R4 R_R5 I_I1 V_V1 E_E1

0 $N_0001 2 $N_0003 $N_0002 6 0 $N_0002 4 0 $N_0004 1 $N_0001 $N_0004 3 0 $N_0003 DC 10A $N_0001 $N_0003 20V $N_0002 $N_0004 $N_0001 $N_0004 3

Chapter 3, Solution 82

2i0

+ v0 – 3 kΩ

1

2 kΩ

2

+

3v0

3

6 kΩ

4

4A 4 kΩ

8 kΩ

0

This network corresponds to the Netlist.

100V +

–

Chapter 3, Solution 83 The circuit is shown below. 20 Ω

1

70 Ω

2i02

3

+ v0 – 20 V

50 Ω

+

–

2 kΩ

1

30 Ω

2A

3 kΩ

2

3v0

+

3

0

4

6 kΩ

4A 4 kΩ

8 kΩ

100V +

0

When the circuit is saved and simulated, we obtain v2 = -12.5 volts Chapter 3, Solution 84 From the output loop, v0 = 50i0x20x103 = 106i0

(1)

From the input loop, 3x10-3 + 4000i0 – v0/100 = 0

(2)

From (1) and (2) we get, i0 = 0.5µA and v0 = 0.5 volt.

Chapter 3, Solution 85

The amplifier acts as a source. Rs + Vs -

RL

For maximum power transfer, R L = Rs = 9Ω

–

Chapter 3, Solution 86 Let v1 be the potential across the 2 k-ohm resistor with plus being on top. Then, [(0.03 – v1)/1k] + 400i = v1/2k

(1)

Assume that i is in mA. But, i = (0.03 – v1)/1

(2)

Combining (1) and (2) yields, v1 = 29.963 mVolts and i = 37.4 nA, therefore, v0 = -5000x400x37.4x10-9 = -74.8 mvolts

Chapter 3, Solution 87 v1 = 500(vs)/(500 + 2000) = vs/5 v0 = -400(60v1)/(400 + 2000) = -40v1 = -40(vs/5) = -8vs, Therefore, v0/vs = -8 Chapter 3, Solution 88 Let v1 be the potential at the top end of the 100-ohm resistor. (vs – v1)/200 = v1/100 + (v1 – 10-3v0)/2000

(1)

For the right loop, v0 = -40i0(10,000) = -40(v1 – 10-3)10,000/2000, or, v0 = -200v1 + 0.2v0 = -4x10-3v0 Substituting (2) into (1) gives, (vs + 0.004v1)/2 = -0.004v0 + (-0.004v1 – 0.001v0)/20 This leads to 0.125v0 = 10vs or (v0/vs) = 10/0.125 = -80

(2)

Chapter 3, Solution 89 vi = VBE + 40k IB

(1)

5 = VCE + 2k IC

(2)

If IC = βIB = 75IB and VCE = 2 volts, then (2) becomes 5 = 2 +2k(75IB) which leads to IB = 20 µA. Substituting this into (1) produces, vi = 0.7 + 0.8 = 1.5 volts. 2 kΩ

IB

40 kΩ

vi

+ VBE

+

-

–

5v

+

-

Chapter 3, Solution 90 1 kΩ

100 kΩ

vs

i1

i2

+ + VBE

+

-

IB

500 Ω

IE

VCE –

–

+

18V

+

-

V0

–

For loop 1, -vs + 10k(IB) + VBE + IE (500) = 0 = -vs + 0.7 + 10,000IB + 500(1 + β)IB which leads to vs + 0.7 = 10,000IB + 500(151)IB = 85,500IB But, v0 = 500IE = 500x151IB = 4 which leads to IB = 5.298x10-5 Therefore, vs = 0.7 + 85,500IB = 5.23 volts

Chapter 3, Solution 91 We first determine the Thevenin equivalent for the input circuit. RTh = 6||2 = 6x2/8 = 1.5 kΩ and VTh = 2(3)/(2+6) = 0.75 volts 5 kΩ

IC 1.5 kΩ

0.75 V

IB + VBE

+

-

i1

i2

+ VCE

–

–

9V

+

400 Ω

+

-

V0

IE

–

For loop 1, -0.75 + 1.5kIB + VBE + 400IE = 0 = -0.75 + 0.7 + 1500IB + 400(1 + β)IB IB = 0.05/81,900 = 0.61 µA v0 = 400IE = 400(1 + β)IB = 49 mV For loop 2, -400IE – VCE – 5kIC + 9 = 0, but, IC = βIB and IE = (1 + β)IB VCE = 9 – 5kβIB – 400(1 + β)IB = 9 – 0.659 = 8.641 volts

Chapter 3, Solution 92 I1

5 kΩ

10 kΩ

VC

IB

IC

+ + VBE 4 kΩ

IE

VCE –

–

+ V0

–

12V

+

-

I1 = IB + IC = (1 + β)IB and IE = IB + IC = I1 Applying KVL around the outer loop, 4kIE + VBE + 10kIB + 5kI1 = 12 12 – 0.7 = 5k(1 + β)IB + 10kIB + 4k(1 + β)IB = 919kIB IB = 11.3/919k = 12.296 µA Also, 12 = 5kI1 + VC which leads to VC = 12 – 5k(101)IB = 5.791 volts Chapter 3, Solution 93 1Ω 4Ω

v1 i1

24V

+

–

3v0

i 2Ω

2Ω

+ 8Ω

2Ω

v2 i3

3v0

i

i2 4Ω

+

+

+

+

v0

v1

v2

–

–

(a)

–

(b)

From (b), -v1 + 2i – 3v0 + v2 = 0 which leads to i = (v1 + 3v0 – v2)/2 At node 1 in (a), ((24 – v1)/4) = (v1/2) + ((v1 +3v0 – v2)/2) + ((v1 – v2)/1), where v0 = v2 or 24 = 9v1 which leads to v1 = 2.667 volts At node 2, ((v1 – v2)/1) + ((v1 + 3v0 – v2)/2) = (v2/8) + v2/4, v0 = v2 v2 = 4v1 = 10.66 volts Now we can solve for the currents, i1 = v1/2 = 1.333 A, i2 = 1.333 A, and i3 = 2.6667 A.

Chapter 4, Solution 1. 1Ω

+ −

1V

8 (5 + 3) = 4Ω , i =

io =

5Ω

i

io

8Ω

3Ω

1 1 = 1+ 4 5

1 1 i= = 0.1A 2 10

Chapter 4, Solution 2. 6 (4 + 2) = 3Ω, i1 = i 2 =

io =

1 A 2

1 1 i1 = , v o = 2i o = 0.5V 2 4

5Ω

4Ω

i1

io

i2 1A

8Ω

6Ω

2Ω

If is = 1µA, then vo = 0.5µV Chapter 4, Solution 3. R 3R io 3R Vs

3R

+ −

+ R

vo

1V

+ −

3R

− (a)

(b)

1.5R

(a) We transform the Y sub-circuit to the equivalent ∆ . R 3R =

3R 2 3 3 3 3 = R, R + R = R 4R 4 4 4 2

vs independent of R 2 io = vo/(R) vo =

When vs = 1V, vo = 0.5V, io = 0.5A (b) (c)

When vs = 10V, vo = 5V, io = 5A When vs = 10V and R = 10Ω, vo = 5V, io = 10/(10) = 500mA

Chapter 4, Solution 4.

If Io = 1, the voltage across the 6Ω resistor is 6V so that the current through the 3Ω resistor is 2A. 2Ω

2A 1A

2Ω

3A

3A

i1 +

3Ω

6Ω

4Ω

Is

2Ω

4Ω

v1

− (a) 3 6 = 2Ω , vo = 3(4) = 12V, i1 =

(b) vo = 3A. 4

Hence Is = 3 + 3 = 6A If

Is = 6A Is = 9A

Io = 1 Io = 6/(9) = 0.6667A

Is

Chapter 4, Solution 5. 2Ω

Vs

If vo = 1V,

If vs =

10 3

3Ω

v1

+ −

vo

6Ω

6Ω

1 V1 =   + 1 = 2V 3 10 2 Vs = 2  + v1 = 3 3

vo = 1

Then vs = 15

vo =

3 x15 = 4.5V 10

Chapter 4, Solution 6

Let RT = R2 // R3 =

R2 R3 RT , then Vo = Vs RT +R1 R2 + R3

R2 R3 V R2 + R3 R2 R3 RT k= o = = = R2 R3 Vs RT + R1 R1 R2 + R2 R3 + R3 R1 + R1 R2 + R3

6Ω

Chapter 4, Solution 7

We find the Thevenin equivalent across the 10-ohm resistor. To find VTh, consider the circuit below. 3Vx

5Ω

5Ω +

+ 15 Ω

4V -

VTh

6Ω

+

Vx

-

From the figure, 15 (4) = 3V 15 + 5 consider the circuit below:

V x = 0,

To find RTh,

VTh =

3Vx

5Ω

5Ω V1

V2

+ 4V -

15 Ω

+ At node 1, V V − V2 4 − V1 = 3V x + 1 + 1 , 5 15 5 At node 2,

1A

6Ω

Vx

V x = 6 x1 = 6

-

 →

258 = 3V2 − 7V1

(1)

V1 − V2 =0  → V1 = V2 − 95 5 Solving (1) and (2) leads to V2 = 101.75 V 2 V V 9 RTh = 2 = 101.75Ω, p max = Th = = 22.11 mW 1 4 RTh 4 x101.75

1 + 3V x +

(2)

Chapter 4, Solution 8.

Let i = i1 + i2, where i1 and iL are due to current and voltage sources respectively. 6Ω i2

i1 6Ω

4Ω 5A

20V

+ −

4Ω

(a)

i1 =

(b)

6 20 (5) = 3A, i 2 = = 2A 6+4 6+4

Thus i = i1 + i2 = 3 + 2 = 5A

Chapter 4, Solution 9.

Let i x = i x1 + i x 2 where i x1 is due to 15V source and i x 2 is due to 4A source, 12 Ω

i ix1

15V

+ −

10 Ω

(a)

40Ω

-4A

ix2 12Ω

10Ω

(b)

40Ω

For ix1, consider Fig. (a). 10||40 = 400/50 = 8 ohms, i = 15/(12 + 8) = 0.75 ix1 = [40/(40 + 10)]i = (4/5)0.75 = 0.6 For ix2, consider Fig. (b). 12||40 = 480/52 = 120/13 ix2 = [(120/13)/((120/13) + 10)](-4) = -1.92 ix = 0.6 – 1.92 = -1.32 A p = vix = ix2R = (-1.32)210 = 17.43 watts Chapter 4, Solution 10.

Let vab = vab1 + vab2 where vab1 and vab2 are due to the 4-V and the 2-A sources respectively. 3vab1

10 Ω

10 Ω

+−

3vab2

+− +

4V

+ −

vab1

+ 2A

− (a)

− (b)

For vab1, consider Fig. (a). Applying KVL gives, - vab1 – 3 vab1 + 10x0 + 4 = 0, which leads to vab1 = 1 V For vab2, consider Fig. (b). Applying KVL gives, vab = 1 + 5 = 6 V

vab2

vab2 – 3vab2 + 10x2 = 0, which leads to vab2 = 5

Chapter 4, Solution 11.

Let i = i1 + i2, where i1 is due to the 12-V source and i2 is due to the 4-A source. 6Ω

io i1

12V

+ −

2Ω

3Ω

(a)

4A

i2 6Ω

2Ω

3Ω

ix2 2Ω

4A

2Ω

(b)

For i1, consider Fig. (a). 2||3 = 2x3/5 = 6/5, io = 12/(6 + 6/5) = 10/6 i1 = [3/(2 + 3)]io = (3/5)x(10/6) = 1 A For i2, consider Fig. (b),

6||3 = 2 ohm, i2 = 4/2 = 2 A i = 1+2 = 3A

Chapter 4, Solution 12.

Let vo = vo1 + vo2 + vo3, where vo1, vo2, and vo3 are due to the 2-A, 12-V, and 19-V sources respectively. For vo1, consider the circuit below. 2A

5Ω

6Ω

3Ω

+ vo1 −

2A 4Ω

12 Ω

io 5 Ω + vo1 − 5Ω

6||3 = 2 ohms, 4||12 = 3 ohms. Hence, io = 2/2 = 1, vo1 = 5io = 5 V For vo2, consider the circuit below. 6Ω

12V

+ −

5Ω

4Ω

6Ω

+ vo2 − 3Ω

12 Ω

12V

+ −

5Ω +

+ vo2 −

v1

3Ω

3Ω

−

3||8 = 24/11, v1 = [(24/11)/(6 + 24/11)]12 = 16/5 vo2 = (5/8)v1 = (5/8)(16/5) = 2 V For vo3, consider the circuit shown below. 5Ω

4Ω

+ vo3 − 6Ω

3Ω

12 Ω

5Ω

+ −

+ vo3 − 19V

2Ω

12 Ω

4Ω + v2

+ − 19V

−

7||12 = (84/19) ohms, v2 = [(84/19)/(4 + 84/19)]19 = 9.975 v = (-5/7)v2 = -7.125 vo = 5 + 2 – 7.125 = -125 mV

Chapter 4, Solution 13

Let io = i1 + i2 + i3 , where i1, i2, and i3 are the contributions to io due to 30-V, 15-V, and 6-mA sources respectively. For i1, consider the circuit below.

1 kΩ

2 kΩ

+ 30V -

3 kΩ

i1 4 kΩ

5 kΩ

3//5 = 15/8 = 1.875 kohm, 2 + 3//5 = 3.875 kohm, 1//3.875 = 3.875/4.875 = 0.7949 kohm. After combining the resistors except the 4-kohm resistor and transforming the voltage source, we obtain the circuit below. i1

30 mA

4 kΩ

0.7949 k Ω

Using current division, i1 =

0.7949 (30mA) = 4.973 mA 4.7949

For i2, consider the circuit below. 1 kΩ

2 kΩ

3 kΩ

i2

-

4 kΩ

5 kΩ

15V +

After successive source transformation and resistance combinations, we obtain the circuit below: 2.42mA i2 4 kΩ

0.7949 k Ω

Using current division, i2 = −

0.7949 (2.42mA) = −0.4012 mA 4.7949

For i3, consider the circuit below. 6mA 1 kΩ

2 kΩ

3 kΩ

i3 4 kΩ

5 kΩ

After successive source transformation and resistance combinations, we obtain the circuit below: 3.097mA i3 4 kΩ

i3 = −

0.7949 k Ω

0.7949 (3.097mA) = −0.5134 mA 4.7949

Thus, io = i1 + i2 + i3 = 4.058 mA Chapter 4, Solution 14.

Let vo = vo1 + vo2 + vo3, where vo1, vo2 , and vo3, are due to the 20-V, 1-A, and 2-A sources respectively. For vo1, consider the circuit below. 6Ω 4Ω

2Ω +

+ − 20V

vo1

3Ω

−

6||(4 + 2) = 3 ohms, vo1 = (½)20 = 10 V

For vo2, consider the circuit below. 6Ω 4Ω

6Ω 4V

2Ω

2Ω

−+

+ 1A

4Ω

+

3Ω

vo2

vo2

−

3Ω

−

3||6 = 2 ohms, vo2 = [2/(4 + 2 + 2)]4 = 1 V For vo3, consider the circuit below. 6Ω 2A 4Ω

2A

2Ω

3Ω + 3Ω

3Ω

vo3

−

− vo3 +

6||(4 + 2) = 3, vo3 = (-1)3 = -3 vo = 10 + 1 – 3 = 8 V Chapter 4, Solution 15.

Let i = i1 + i2 + i3, where i1 , i2 , and i3 are due to the 20-V, 2-A, and 16-V sources. For i1, consider the circuit below. io 20V

+ −

1Ω i1

2Ω

3Ω

4Ω

4||(3 + 1) = 2 ohms, Then io = [20/(2 + 2)] = 5 A, i1 = io/2 = 2.5 A For i3, consider the circuit below. + 2Ω

vo ’

1Ω

4Ω

i3

− +

3Ω

16V

−

2||(1 + 3) = 4/3, vo’ = [(4/3)/((4/3) + 4)](-16) = -4 i3 = vo’/4 = -1 For i2, consider the circuit below.

2Ω

1Ω

2A (4/3)Ω i2

4Ω

3Ω

2||4 = 4/3, 3 + 4/3 = 13/3 Using the current division principle. i2 = [1/(1 + 13/2)]2 = 3/8 = 0.375 i = 2.5 + 0.375 - 1 = 1.875 A p = i2R = (1.875)23 = 10.55 watts

1Ω

2A i2 3Ω

Chapter 4, Solution 16.

Let io = io1 + io2 + io3, where io1, io2, and io3 are due to the 12-V, 4-A, and 2-A sources. For io1, consider the circuit below. 4Ω

io1 12V

+ −

3Ω

10 Ω

2Ω

5Ω

10||(3 + 2 + 5) = 5 ohms, io1 = 12/(5 + 4) = (12/9) A 4A

For io2, consider the circuit below.

3Ω

io2 4Ω

2Ω

5Ω

10Ω

i1

2 + 5 + 4||10 = 7 + 40/14 = 69/7 i1 = [3/(3 + 69/7)]4 = 84/90, io2 =[-10/(4 + 10)]i1 = -6/9 For io3, consider the circuit below.

3Ω

io3

2Ω

i2 4Ω

10 Ω

5Ω 2A

3 + 2 + 4||10 = 5 + 20/7 = 55/7 i2 = [5/(5 + 55/7)]2 = 7/9, io3 = [-10/(10 + 4)]i2 = -5/9 io = (12/9) – (6/9) – (5/9) = 1/9 = 111.11 mA

Chapter 4, Solution 17.

Let vx = vx1 + vx2 + vx3, where vx1,vx2, and vx3 are due to the 90-V, 6-A, and 40-V sources. For vx1, consider the circuit below. 30 Ω

10 Ω

+ 90V

+ −

vx1

60 Ω

20 Ω

−

30 Ω

io 10 Ω + − vx1 3A

20 Ω

12 Ω

20||30 = 12 ohms, 60||30 = 20 ohms By using current division, io = [20/(22 + 20)]3 = 60/42, vx1 = 10io = 600/42 = 14.286 V For vx2, consider the circuit below. 10 Ω i ’ o

+ 30 Ω

10 Ω i ’ o

vx2 −

+ vx2 −

60 Ω 6A

30 Ω

20 Ω

6A

20 Ω

12 Ω

io’ = [12/(12 + 30)]6 = 72/42, vx2 = -10io’ = -17.143 V For vx3, consider the circuit below. 10 Ω

+ 30 Ω

60 Ω

vx3

10 Ω

10 Ω

−

30 Ω

+ 40V

+ −

vx3

20 Ω

io”

− 7.5Ω

io” = [12/(12 + 30)]2 = 24/42, vx3 = -10io” = -5.714 vx = 14.286 – 17.143 – 5.714 = -8.571 V

4A

Chapter 4, Solution 18.

Let ix = i1 + i2, where i1 and i2 are due to the 10-V and 2-A sources respectively. To obtain i1, consider the circuit below. 2Ω 1Ω

i1 10V

+ −

5i1

4Ω

i1

1Ω

+ −

10V

10i1

2Ω

+− 4Ω

-10 + 10i1 + 7i1 = 0, therefore i1 = (10/17) A For i2, consider the circuit below. i2

1Ω

2Ω i o

+− 10i2

2A

4Ω

io

1Ω

+ −

2V

10i2

2Ω

+− 4Ω

-2 + 10i2 + 7io = 0, but i2 + 2 = io. Hence, -2 + 10i2 +7i2 + 14 = 0, or i2 = (-12/17) A vx = 1xix = 1(i1 + i2) = (10/17) – (12/17) = -2/17 = -117.6 mA Chapter 4, Solution 19.

Let vx = v1 + v2, where v1 and v2 are due to the 4-A and 6-A sources respectively. v1

ix

ix

v2

+ 2Ω

4A

8Ω

v1

+ 2Ω

6A 8Ω

−

−+

−+

4ix

4ix (a)

(b)

v2

−

To find v1, consider the circuit in Fig. (a). v1/8 = 4 + (-4ix – v1)/2 -ix = (-4ix – v1)/2 and we have -2ix = v1. Thus,

But,

v1/8 = 4 + (2v1 – v1)/8, which leads to v1 = -32/3 To find v2, consider the circuit shown in Fig. (b). v2/2 = 6 + (4ix – v2)/8 But ix = v2/2 and 2ix = v2. Therefore, v2/2 = 6 + (2v2 – v2)/8 which leads to v2 = -16 Hence,

vx = –(32/3) – 16 = -26.67 V

Chapter 4, Solution 20.

Transform the voltage sources and obtain the circuit in Fig. (a). Combining the 6-ohm and 3-ohm resistors produces a 2-ohm resistor (6||3 = 2). Combining the 2-A and 4-A sources gives a 6-A source. This leads to the circuit shown in Fig. (b). i

i 6Ω

2A

2Ω

3Ω 4A

2Ω

(a)

From Fig. (b),

2Ω

6A

(b)

i = 6/2 = 3 A

Chapter 4, Solution 21.

To get io, transform the current sources as shown in Fig. (a). io

6Ω

3Ω

+ − 12V

+ −

i 6V 2 A

6Ω

3Ω

+ vo 2 A

− (a)

(b)

From Fig. (a),

-12 + 9io + 6 = 0, therefore io = 666.7 mA

To get vo, transform the voltage sources as shown in Fig. (b). i = [6/(3 + 6)](2 + 2) = 8/3 vo = 3i = 8 V Chapter 4, Solution 22.

We transform the two sources to get the circuit shown in Fig. (a). 5Ω

− + 10V

5Ω

4Ω

10Ω

2A

(a)

i 1A

10Ω

4Ω

10Ω

2A

(b) We now transform only the voltage source to obtain the circuit in Fig. (b).

10||10 = 5 ohms, i = [5/(5 + 4)](2 – 1) = 5/9 = 555.5 mA

Chapter 4, Solution 23

If we transform the voltage source, we obtain the circuit below. 8Ω

10 Ω

6Ω

3Ω

5A

3A

3//6 = 2-ohm. Convert the current sources to voltages sources as shown below. 10 Ω

8Ω

+

2Ω

+ 10V -

30V -

Applying KVL to the loop gives − 30 + 10 + I (10 + 8 + 2) = 0  → p = VI = I 2 R = 8 W

I = 1A

Chapter 4, Solution 24

Convert the current source to voltage source. 16 Ω

1Ω

4Ω 5Ω

+

+

48 V

10 Ω + 12 V -

-

Vo -

Combine the 16-ohm and 4-ohm resistors and convert both voltages sources to current Sources. We obtain the circuit below. 1Ω

20 Ω

2.4A

5Ω

2.4A

10 Ω

Combine the resistors and current sources. 20//5 = (20x5)/25 = 4 Ω , 2.4 + 2.4 = 4.8 A Convert the current source to voltage source. We obtain the circuit below. 4Ω + 19.2V Using voltage division, Vo =

10 (19.2) = 12.8 V 10 + 4 + 1

1Ω + Vo -

10 Ω

Chapter 4, Solution 25.

Transforming only the current source gives the circuit below. 18 V

9Ω

−+ 12V

+ −

5Ω i

4Ω

vo

+

− +

−

30 V

+−

2Ω

30 V

Applying KVL to the loop gives, (4 + 9 + 5 + 2)i – 12 – 18 – 30 – 30 = 0 20i = 90 which leads to i = 4.5 vo = 2i = 9 V Chapter 4, Solution 26.

Transform the voltage sources to current sources. The result is shown in Fig. (a), 30||60 = 20 ohms,

30||20 = 12 ohms 10 Ω

+ vx − 3A

30Ω

60Ω

30Ω

6A

20Ω

(a) 20 Ω

+ −

10 Ω + vx −

60V

i (b)

12 Ω

+ −

96V

2A

Combining the resistors and transforming the current sources to voltage sources, we obtain the circuit in Fig. (b). Applying KVL to Fig. (b), 42i – 60 + 96 = 0, which leads to i = -36/42 vx = 10i = -8.571 V Chapter 4, Solution 27.

Transforming the voltage sources to current sources gives the circuit in Fig. (a). 10||40 = 8 ohms Transforming the current sources to voltage sources yields the circuit in Fig. (b). Applying KVL to the loop, -40 + (8 + 12 + 20)i + 200 = 0 leads to i = -4 vx 12i = -48 V 12 Ω + vx − 5A

10Ω

40Ω

8A

20Ω

2A

(a)

8Ω

+ −

12 Ω + vx −

40V

i (b)

20 Ω

+ −

200V

Chapter 4, Solution 28. Transforming only the current sources leads to Fig. (a). Continuing with source transformations finally produces the circuit in Fig. (d).

io 12 V

+ −

4Ω

3Ω

12 V

2Ω

10 V

5Ω

+−

+−

10Ω

(a) io + −

4Ω

12V

10 Ω + −

10Ω

22 V

(b) io + −

4Ω

12V

io 10Ω

10Ω

(c)

2.2A

+ −

12V

4Ω

5Ω io

(d)

Applying KVL to the loop in fig. (d), -12 + 9io + 11 = 0, produces io = 1/9 = 111.11 mA

11V

+ −

Chapter 4, Solution 29. Transform the dependent voltage source to a current source as shown in Fig. (a). 2||4 = (4/3) k ohms 4 kΩ 2 kΩ

2vo

(4/3) kΩ

−+

1.5vo

3 mA

1 kΩ

i

3 mA

+

1 kΩ

+ vo

−

vo

− (a)

(b)

It is clear that i = 3 mA which leads to vo = 1000i = 3 V If the use of source transformations was not required for this problem, the actual answer could have been determined by inspection right away since the only current that could have flowed through the 1 k ohm resistor is 3 mA. Chapter 4, Solution 30 Transform the dependent current source as shown below. ix

+ 12V -

24 Ω

60 Ω

30 Ω

10 Ω

+ -

7ix

Combine the 60-ohm with the 10-ohm and transform the dependent source as shown below.

24 Ω

ix

+ 12V -

30 Ω

70 Ω

0.1ix

Combining 30-ohm and 70-ohm gives 30//70 = 70x30/100 = 21-ohm. Transform the dependent current source as shown below. 24 Ω

ix

21 Ω

+ 12V -

+ -

2.1ix

Applying KVL to the loop gives

45i x − 12 + 2.1i x = 0

 →

ix =

12 = 254.8 mA 47.1

Chapter 4, Solution 31.

Transform the dependent source so that we have the circuit in Fig. (a). 6||8 = (24/7) ohms. Transform the dependent source again to get the circuit in Fig. (b). 3Ω + 12V

+ −

vx

− 8Ω

vx/3

6Ω

(a) 3Ω

+ 12V

+ −

vx

(24/7) Ω

− i

(b)

+

(8/7)vx

From Fig. (b), vx = 3i, or i = vx/3. Applying KVL, -12 + (3 + 24/7)i + (24/21)vx = 0 12 = [(21 + 24)/7]vx/3 + (8/7)vx, leads to vx = 84/23 = 3.625 V Chapter 4, Solution 32.

As shown in Fig. (a), we transform the dependent current source to a voltage source, 15 Ω

10 Ω

5ix

−+ 60V

+ −

50 Ω

40 Ω

(a) 15 Ω

60V

+ −

50 Ω

50 Ω

0.1ix

(b)

ix 60V

+ −

15 Ω

25 Ω

ix

(c)

−

2.5ix

In Fig. (b), 50||50 = 25 ohms. Applying KVL in Fig. (c), -60 + 40ix – 2.5ix = 0, or ix = 1.6 A Chapter 4, Solution 33.

(a)

RTh = 10||40 = 400/50 = 8 ohms VTh = (40/(40 + 10))20 = 16 V

(b)

RTh = 30||60 = 1800/90 = 20 ohms 2 + (30 – v1)/60 = v1/30, and v1 = VTh 120 + 30 – v1 = 2v1, or v1 = 50 V VTh = 50 V

Chapter 4, Solution 34.

To find RTh, consider the circuit in Fig. (a).

10 Ω

10 Ω

20 Ω RTh

40 Ω

3A

+ −

v1

20 Ω

v2

+ 40V

40 Ω

VTh

(a)

(b) RTh = 20 + 10||40 = 20 + 400/50 = 28 ohms To find VTh, consider the circuit in Fig. (b).

At node 1, At node 2,

(40 – v1)/10 = 3 + [(v1 – v2)/20] + v1/40, 40 = 7v1 – 2v2 3 + (v1- v2)/20 = 0, or v1 = v2 – 60

Solving (1) and (2),

v1 = 32 V, v2 = 92 V, and VTh = v2 = 92 V

(1) (2)

Chapter 4, Solution 35.

To find RTh, consider the circuit in Fig. (a). RTh = Rab = 6||3 + 12||4 = 2 + 3 =5 ohms To find VTh, consider the circuit shown in Fig. (b). RTh a 6Ω

b

3Ω

12 Ω

4Ω

(a)

2A 6Ω

+ −

v1

v2 4 Ω

+ VTh

+ 12V v1

+ 3Ω

12Ω

v2

−

+ −

19V

−

At node 1,

(b) 2 + (12 – v1)/6 = v1/3, or v1 = 8

At node 2,

(19 – v2)/4 = 2 + v2/12, or v2 = 33/4

But,

-v1 + VTh + v2 = 0, or VTh = v1 – v2 = 8 – 33/4 = -0.25 a +

vo

−

b

10 Ω RTh = 5 Ω

+− VTh = (-1/4)V

vo = VTh/2 = -0.25/2 = -125 mV

Chapter 4, Solution 36.

Remove the 30-V voltage source and the 20-ohm resistor. a

RTh

10Ω a

10Ω

+ + −

40Ω

VTh

40Ω

50V

b

b

(a)

(b)

From Fig. (a),

RTh = 10||40 = 8 ohms

From Fig. (b),

VTh = (40/(10 + 40))50 = 40V 8Ω

+ −

i

a 12 Ω

40V

+ −

30V b

(c)

The equivalent circuit of the original circuit is shown in Fig. (c). Applying KVL, 30 – 40 + (8 + 12)i = 0, which leads to i = 500mA

Chapter 4, Solution 37

RN is found from the circuit below. 20 Ω a 40 Ω

12 Ω

b

R N = 12 //( 20 + 40) = 10Ω IN is found from the circuit below.

2A

20 Ω a 40 Ω

+ 120V -

12 Ω IN b

Applying source transformation to the current source yields the circuit below. 20 Ω

40 Ω

+ 80 V -

+ 120V -

Applying KVL to the loop yields − 120 + 80 + 60 I N = 0  → I N = 40 / 60 = 0.6667 A

IN

Chapter 4, Solution 38

We find Thevenin equivalent at the terminals of the 10-ohm resistor. For RTh, consider the circuit below. 1Ω 4Ω 5Ω RTh

16 Ω

RTh = 1 + 5 //( 4 + 16) = 1 + 4 = 5Ω For VTh, consider the circuit below. V1

4Ω

1Ω

V2 5Ω

3A

+

16 Ω

VTh

+ 12 V At node 1, V V − V2 3= 1 + 1  → 48 = 5V1 − 4V2 16 4 At node 2, V1 − V2 12 − V2 + =0  → 48 = −5V1 + 9V2 4 5 Solving (1) and (2) leads to VTh = V2 = 19.2

-

(1)

(2)

Thus, the given circuit can be replaced as shown below. 5Ω + 19.2V -

+ Vo -

10 Ω

Using voltage division, Vo =

10 (19.2) = 12.8 V 10 + 5

Chapter 4, Solution 39.

To find RTh, consider the circuit in Fig. (a). 10 Ω

3vab10 Ω a +−

+ −

10 Ω +

io

+ 1V v1 − 50V

+ VTh + −

40 Ω4V

−

+ 8 A 2A −

v2

a +

− b

(b)

(a)

(b)

1 – 3 + 10io = 0, or io = 0.4 RTh = 1/io = 2.5 ohms

To find VTh, consider the circuit shown in Fig. (b). [(4 – v)/10] + 2 = 0, or v = 24 But,

vab = VTh

40V

−

b

-

+

3vab 20 Ω v +−

v = VTh + 3vab = 4VTh = 24, which leads to VTh = 6 V

Chapter 4, Solution 40.

To find RTh, consider the circuit in Fig. (a). 10 Ω

RTh a

b

40Ω (a)

20 Ω

RTh = 10||40 + 20 = 28 ohms To get VTh, consider the circuit in Fig. (b). The two loops are independent. From loop 1, v1 = (40/50)50 = 40 V For loop 2, But,

-v2 + 20x8 + 40 = 0, or v2 = 200 VTh + v2 – v1 = 0,

VTh = v1 = v2 = 40 – 200 = -160 volts

This results in the following equivalent circuit. 28 Ω + -160V

+ −

vx

12 Ω

−

vx = [12/(12 + 28)](-160) = -48 V Chapter 4, Solution 41

To find RTh, consider the circuit below 14 Ω a

6Ω

5Ω b

RTh = 5 //(14 + 6) = 4Ω = R N Applying source transformation to the 1-A current source, we obtain the circuit below.

6Ω

14 Ω

- 14V +

VTh

a

+ 6V

5Ω

3A

b At node a, 14 + 6 − VTh V = 3 + Th 6 + 14 5 IN =

 →

VTh = −8 V

VTh = (−8) / 4 = −2 A RTh

Thus, RTh = R N = 4Ω,

VTh = −8V,

I N = −2 A

Chapter 4, Solution 42.

To find RTh, consider the circuit in Fig. (a). 20 Ω 10 Ω

30 Ω a

20 Ω

30 Ω 30 Ω

b

a 10 Ω

10Ω

(a)

10 Ω

b

10 Ω

10 Ω

(b)

20||20 = 10 ohms. Transform the wye sub-network to a delta as shown in Fig. (b). 10||30 = 7.5 ohms. RTh = Rab = 30||(7.5 + 7.5) = 10 ohms. To find VTh, we transform the 20-V and the 5-V sources. We obtain the circuit shown in Fig. (c).

a

10 Ω

+

b

−+

10 Ω

10 Ω i1

30V

10 V

10 Ω

+ −

10 Ω

i2

+ −

50V

(c)

For loop 1,

-30 + 50 + 30i1 – 10i2 = 0, or -2 = 3i1 – i2

(1)

For loop 2,

-50 – 10 + 30i2 – 10i1 = 0, or 6 = -i1 + 3i2

(2)

Solving (1) and (2),

i1 = 0, i2 = 2 A

Applying KVL to the output loop, -vab – 10i1 + 30 – 10i2 = 0, vab = 10 V VTh = vab = 10 volts Chapter 4, Solution 43.

To find RTh, consider the circuit in Fig. (a). RTh a 10Ω

b

10Ω

5Ω (a)

10 Ω

+ −

a +

50V va

b

+ VTh 10 Ω

+ vb

−

− (b)

RTh = 10||10 + 5 = 10 ohms

5Ω

2A

To find VTh, consider the circuit in Fig. (b). vb = 2x5 = 10 V, va = 20/2 = 10 V But,

-va + VTh + vb = 0, or VTh = va – vb = 0 volts

Chapter 4, Solution 44.

(a)

For RTh, consider the circuit in Fig. (a). RTh = 1 + 4||(3 + 2 + 5) = 3.857 ohms

For VTh, consider the circuit in Fig. (b). Applying KVL gives, 10 – 24 + i(3 + 4 + 5 + 2), or i = 1 VTh = 4i = 4 V 3Ω 3Ω

1Ω

a

+ −

+ −

2Ω

2Ω i

b 5Ω

+ VTh

b 10V

5Ω

(b)

(a)

(b)

a

4Ω

24V

RTh

4Ω

1Ω

For RTh, consider the circuit in Fig. (c). 3Ω

1Ω

4Ω

3Ω

b

24V

2Ω RTh

5Ω

1Ω

4Ω vo

+ −

+

2Ω 5Ω

2A

c (c)

b

VTh c

(d)

RTh = 5||(2 + 3 + 4) = 3.214 ohms To get VTh, consider the circuit in Fig. (d). At the node, KCL gives, [(24 – vo)/9] + 2 = vo/5, or vo = 15 VTh = vo = 15 V Chapter 4, Solution 45.

For RN, consider the circuit in Fig. (a). 6Ω

6Ω

6Ω

4Ω

RN

4A

6Ω

(a)

IN

4Ω

(b)

RN = (6 + 6)||4 = 3 ohms For IN, consider the circuit in Fig. (b). The 4-ohm resistor is shorted so that 4-A current is equally divided between the two 6-ohm resistors. Hence, IN = 4/2 = 2 A Chapter 4, Solution 46.

(a)

RN = RTh = 8 ohms. To find IN, consider the circuit in Fig. (a). 10 Ω

20V

+ −

60 Ω

4Ω

Isc

2A

(a)

(b)

IN = Isc = 20/10 = 2 A (b)

To get IN, consider the circuit in Fig. (b).

IN = Isc = 2 + 30/60 = 2.5 A

30 Ω

IN 30V

+ −

Chapter 4, Solution 47

Since VTh = Vab = Vx, we apply KCL at the node a and obtain 30 − VTh VTh = + 2VTh  → VTh = 150 / 126 = 1.19 V 12 60 To find RTh, consider the circuit below.

12 Ω

Vx

a

2Vx

60 Ω

1A

At node a, KCL gives V V 1 = 2V x + x + x  → V x = 60 / 126 = 0.4762 60 12 V V RTh = x = 0.4762Ω, I N = Th = 1.19 / 0.4762 = 2.5 RTh 1 Thus, VTh = 1.19V , RTh = R N = 0.4762Ω, I N = 2 .5 A

Chapter 4, Solution 48.

To get RTh, consider the circuit in Fig. (a). 10Io 10Io

+−

2Ω

+−

−

1A

−

2A

(a)

From Fig. (a),

VTh

4Ω

V

4Ω

+

Io

+

Io

(b)

Io = 1,

2Ω

6 – 10 – V = 0, or V = -4

RN = RTh = V/1 = -4 ohms

To get VTh, consider the circuit in Fig. (b), Io = 2, VTh = -10Io + 4Io = -12 V IN = VTh/RTh = 3A Chapter 4, Solution 49.

RN = RTh = 28 ohms To find IN, consider the circuit below,

3A

10 Ω

40V

At the node,

+ −

vo

20 Ω io Isc = IN

40 Ω

(40 – vo)/10 = 3 + (vo/40) + (vo/20), or vo = 40/7

io = vo/20 = 2/7, but IN = Isc = io + 3 = 3.286 A Chapter 4, Solution 50.

From Fig. (a), RN = 6 + 4 = 10 ohms 6Ω

6Ω Isc = IN

4Ω

(a)

From Fig. (b),

2A

4Ω

+ 12V −

(b)

2 + (12 – v)/6 = v/4, or v = 9.6 V -IN = (12 – v)/6 = 0.4, which leads to IN = -0.4 A

Combining the Norton equivalent with the right-hand side of the original circuit produces the circuit in Fig. (c).

i 10 Ω

0.4A

5Ω

4A

(c)

i = [10/(10 + 5)] (4 – 0.4) = 2.4 A Chapter 4, Solution 51.

(a)

From the circuit in Fig. (a), RN = 4||(2 + 6||3) = 4||4 = 2 ohms RTh 6Ω

VTh

+

4Ω

6Ω

3Ω

2Ω

120V

4Ω

+ −

3Ω

(a)

6A

2Ω

(b)

For IN or VTh, consider the circuit in Fig. (b). After some source transformations, the circuit becomes that shown in Fig. (c). + VTh 2Ω

40V

+ −

4Ω i

2Ω

12V

+ −

(c)

Applying KVL to the circuit in Fig. (c), -40 + 8i + 12 = 0 which gives i = 7/2 VTh = 4i = 14 therefore IN = VTh/RN = 14/2 = 7 A

(b)

To get RN, consider the circuit in Fig. (d). RN = 2||(4 + 6||3) = 2||6 = 1.5 ohms 6Ω

4Ω

2Ω

i

+ 3Ω

RN

2Ω

VTh

(d)

12V

+ −

(e)

To get IN, the circuit in Fig. (c) applies except that it needs slight modification as in Fig. (e). i = 7/2, VTh = 12 + 2i = 19, IN = VTh/RN = 19/1.5 = 12.667 A Chapter 4, Solution 52.

For RTh, consider the circuit in Fig. (a). a 3 kΩ

Io

20Io

2 kΩ

RTh b

(a) 3 kΩ

6V

+ −

a

+ Io

20Io

2 kΩ

VTh b

(b)

For Fig. (a), Io = 0, hence the current source is inactive and RTh = 2 k ohms

For VTh, consider the circuit in Fig. (b). Io = 6/3k = 2 mA VTh = (-20Io)(2k) = -20x2x10-3x2x103 = -80 V Chapter 4, Solution 53.

To get RTh, consider the circuit in Fig. (a). 0.25vo

0.25vo

2Ω

2Ω

a

+ 6Ω

3Ω

+

vo

2Ω

1A

−

1/2

a 1/2

vo

vab

−

− b

b (a)

(b)

From Fig. (b), vo = 2x1 = 2V, -vab + 2x(1/2) +vo = 0 vab = 3V RN = vab/1 = 3 ohms To get IN, consider the circuit in Fig. (c). 0.25vo 6Ω

2Ω

a

+ 18V

+ −

3Ω

vo

Isc = IN

− b (c)

[(18 – vo)/6] + 0.25vo = (vo/2) + (vo/3) or vo = 4V But,

(vo/2) = 0.25vo + IN, which leads to IN = 1 A

+ 1A

Chapter 4, Solution 54 To find VTh =Vx, consider the left loop.

− 3 + 1000io + 2V x = 0  → For the right loop, V x = −50 x 40i o = −2000io Combining (1) and (2), 3 = 1000io − 4000io = −3000io

V x = −2000io = 2

 →

3 = 1000io + 2V x

(1) (2)

 →

io = −1mA

VTh = 2

To find RTh, insert a 1-V source at terminals a-b and remove the 3-V independent source, as shown below. 1 kΩ

ix

. io

+ 2Vx -

40io

+ Vx -

+ 1V -

50 Ω

2V x = −2mA 1000 V 1 i x = 40io + x = −80mA + A = -60mA 50 50

V x = 1,

RTh =

io = −

1 = −1 / 0.060 = − 16.67Ω ix

Chapter 4, Solution 55.

To get RN, apply a 1 mA source at the terminals a and b as shown in Fig. (a). a

I vab/1000 8 kΩ

+ −

80I

+ 50 kΩ

vab

− b (a)

1mA

We assume all resistances are in k ohms, all currents in mA, and all voltages in volts. At node a, (vab/50) + 80I = 1 (1) Also, -8I = (vab/1000), or I = -vab/8000 (2) From (1) and (2),

(vab/50) – (80vab/8000) = 1, or vab = 100 RN = vab/1 = 100 k ohms

To get IN, consider the circuit in Fig. (b). 8 kΩ

2V

+ −

a

I

vab/1000

+

80I

+ −

50 kΩ

IN

vab

− b (b)

Since the 50-k ohm resistor is shorted, IN = -80I, vab = 0 Hence,

8i = 2 which leads to I = (1/4) mA IN = -20 mA

Chapter 4, Solution 56.

We first need RN and IN.

1Ω 20V

2Ω

a

2A

+ −

1Ω

4Ω

4Ω RN

b (a)

2Ω

IN

(b)

− +

16V

To find RN, consider the circuit in Fig. (a). RN = 1 + 2||4 = (7/3) ohms To get IN, short-circuit ab and find Isc from the circuit in Fig. (b). The current source can be transformed to a voltage source as shown in Fig. (c). vo 20V

+ −

1Ω

4Ω a

2V 2Ω

− +

− +

IN

i

16V RN

IN

(c)

(d)

3Ω

b

(20 – vo)/2 = [(vo + 2)/1] + [(vo + 16)/4], or vo = 16/7 IN = (vo + 2)/1 = 30/7 From the Norton equivalent circuit in Fig. (d), i = RN/(RN + 3), IN = [(7/3)/((7/3) + 3)](30/7) = 30/16 = 1.875 A Chapter 4, Solution 57.

To find RTh, remove the 50V source and insert a 1-V source at a – b, as shown in Fig. (a). 2Ω

B

a

A

i

+ 3Ω

vx

6Ω

0.5vx

− (a)

10 Ω

+ −

1V

b

We apply nodal analysis. At node A, At node B,

i + 0.5vx = (1/10) + (1 – vx)/2, or i + vx = 0.6 (1 – vo)/2 = (vx/3) + (vx/6), and vx = 0.5

(1) (2)

From (1) and (2),

i = 0.1 and RTh = 1/i = 10 ohms

To get VTh, consider the circuit in Fig. (b). 3Ω

2Ω

v1

v2

a

+ 50V

+ −

+ 6Ω

vx

10 Ω VTh

0.5vx

−

− b

(b)

At node 1,

(50 – v1)/3 = (v1/6) + (v1 – v2)/2, or 100 = 6v1 – 3v2

(3)

At node 2,

0.5vx + (v1 – v2)/2 = v2/10, vx = v1, and v1 = 0.6v2

(4)

From (3) and (4), v2 = VTh = 166.67 V IN = VTh/RTh = 16.667 A RN = RTh = 10 ohms Chapter 4, Solution 58.

This problem does not have a solution as it was originally stated. The reason for this is that the load resistor is in series with a current source which means that the only equivalent circuit that will work will be a Norton circuit where the value of RN = infinity. IN can be found by solving for Isc. ib

VS

R1

+ −

β ib

vo R2

Writing the node equation at node vo, ib + βib = vo/R2 = (1 + β)ib

Isc

But

ib = (Vs – vo)/R1 vo = Vs – ibR1 Vs – ibR1 = (1 + β)R2ib, or ib = Vs/(R1 + (1 + β)R2)

Isc = IN = -βib = -βVs/(R1 + (1 + β)R2) Chapter 4, Solution 59.

RTh = (10 + 20)||(50 + 40) 30||90 = 22.5 ohms To find VTh, consider the circuit below. i1

i2 10 Ω

20 Ω

+ VTh 8A

50 Ω

40 Ω

i1 = i2 = 8/2 = 4, 10i1 + VTh – 20i2 = 0, or VTh = 20i2 –10i1 = 10i1 = 10x4 VTh = 40V, and IN = VTh/RTh = 40/22.5 = 1.7778 A Chapter 4, Solution 60.

The circuit can be reduced by source transformations. 2A

18 V

12 V

+ −

+ −

10 V + −

10 Ω

5Ω

2A

10 Ω

a

b

3A

5Ω 2A 3A

a

3.333Ω

b

Norton Equivalent Circuit

10 V

3.333Ω

a

+ −

Thevenin Equivalent Circuit

Chapter 4, Solution 61.

To find RTh, consider the circuit in Fig. (a). Let

R = 2||18 = 1.8 ohms,

RTh = 2R||R = (2/3)R = 1.2 ohms.

To get VTh, we apply mesh analysis to the circuit in Fig. (d). 2Ω

a 6Ω

6Ω

2Ω

2Ω 6Ω b (a)

b

2Ω

a 18 Ω 1.8 Ω

2Ω

18 Ω

a

2Ω

18 Ω

1.8 Ω

1.8 Ω

RTh

b

b

(b)

(c)

2Ω

a 6Ω 12V

6Ω

i3

+ −

12V

+ + −

VTh

6Ω 2Ω

i1

− +

i2

2Ω

12V

b

(d)

-12 – 12 + 14i1 – 6i2 – 6i3 = 0, and 7 i1 – 3 i2 – 3i3 = 12

(1)

12 + 12 + 14 i2 – 6 i1 – 6 i3 = 0, and -3 i1 + 7 i2 – 3 i3 = -12

(2)

14 i3 – 6 i1 – 6 i2 = 0, and

(3)

-3 i1 – 3 i2 + 7 i3 = 0

This leads to the following matrix form for (1), (2) and (3),  7 − 3 − 3  i1   12  − 3 7 − 3 i  = − 12   2    − 3 − 3 7  i 3   0 

−3 −3

7

7

∆ = − 3 7 − 3 = 100 , −3 −3 7

−3

12

∆ 2 = − 3 − 12 − 3 = −120 −3 0 7

i2 = ∆/∆2 = -120/100 = -1.2 A VTh = 12 + 2i2 = 9.6 V, and IN = VTh/RTh = 8 A Chapter 4, Solution 62.

Since there are no independent sources, VTh = 0 V To obtain RTh, consider the circuit below. 0.1io

ix

2

+ vo − 1 40 Ω

10 Ω v1 io

2vo

VS

+ −

20 Ω

+ −

At node 2, At node 1,

ix + 0.1io = (1 – v1)/10, or 10ix + io = 1 – v1

(1)

(v1/20) + 0.1io = [(2vo – v1)/40] + [(1 – v1)/10]

(2)

But io = (v1/20) and vo = 1 – v1, then (2) becomes, 1.1v1/20 = [(2 – 3v1)/40] + [(1 – v1)/10] 2.2v1 = 2 – 3v1 + 4 – 4v1 = 6 – 7v1 or

v1 = 6/9.2

(3)

From (1) and (3), 10ix + v1/20 = 1 – v1 10ix = 1 – v1 – v1/20 = 1 – (21/20)v1 = 1 – (21/20)(6/9.2) ix = 31.52 mA, RTh = 1/ix = 31.73 ohms.

Chapter 4, Solution 63.

Because there are no independent sources, IN = Isc = 0 A RN can be found using the circuit below. 10 Ω + vo

20 Ω

− Applying KCL at node 1,

3Ω

v1

0.5vo

io + −

1V

0.5vo + (1 – v1)/3 = v1/30, but vo = (20/30)v1

Hence, 0.5(2/3)(30)v1 + 10 – 10v1 =v1, or v1 = 10 and io = (1 – v1)/3 = -3 RN = 1/io = -1/3 = -333.3 m ohms Chapter 4, Solution 64.

With no independent sources, VTh = 0 V. To obtain RTh, consider the circuit shown below. 4Ω

1Ω

vo

io

ix + –

2Ω

+ −

1V

10ix

ix = [(1 – vo)/1] + [(10ix – vo)/4], or 2vo = 1 + 3ix But ix = vo/2. Hence, 2vo = 1 + 1.5vo, or vo = 2, io = (1 – vo)/1 = -1 Thus, RTh = 1/io = -1 ohm

(1)

Chapter 4, Solution 65

At the terminals of the unknown resistance, we replace the circuit by its Thevenin equivalent. 12 RTh = 2 + 4 // 12 = 2 + 3 = 5Ω, VTh = (32) = 24 V 12 + 4 Thus, the circuit can be replaced by that shown below.

5Ω

Io

+ 24 V -

+ Vo -

Applying KVL to the loop, − 24 + 5I o + Vo = 0

 →

Vo = 24 − 5I o

Chapter 4, Solution 66.

We first find the Thevenin equivalent at terminals a and b. We find RTh using the circuit in Fig. (a). 2Ω 10V − +

3Ω

2Ω

a

b

+ 3Ω

a

VTh

5Ω

b RTh

+ −

5Ω

20V

i 30V

(a)

− +

(b)

RTh = 2||(3 + 5) = 2||8 = 1.6 ohms By performing source transformation on the given circuit, we obatin the circuit in (b).

We now use this to find VTh. 10i + 30 + 20 + 10 = 0, or i = -5 VTh + 10 + 2i = 0, or VTh = 2 V p = VTh2/(4RTh) = (2)2/[4(1.6)] = 625 m watts Chapter 4, Solution 67.

We need to find the Thevenin equivalent at terminals a and b. From Fig. (a), RTh = 4||6 + 8||12 = 2.4 + 4.8 = 7.2 ohms From Fig. (b), 10i1 – 30 = 0, or i1 = 3 +

4Ω

4Ω

6Ω

8Ω

6Ω −

30V + −

RTh 12 Ω

i1

VTh +

12 Ω

+

i2

8Ω

−

−

(a)

(b)

20i2 + 30 = 0, or i2 = 1.5, VTh = 6i1 + 8i2 = 6x3 – 8x1.5 = 6 V For maximum power transfer, p = VTh2/(4RTh) = (6)2/[4(7.2)] = 1.25 watts

Chapter 4, Solution 68.

This is a challenging problem in that the load is already specified. This now becomes a "minimize losses" style problem. When a load is specified and internal losses can be adjusted, then the objective becomes, reduce RThev as much as possible, which will result in maximum power transfer to the load.

+

-

+

-

Removing the 10 ohm resistor and solving for the Thevenin Circuit results in: RTh = (Rx20/(R+20)) and a Voc = VTh = 12x(20/(R +20)) + (-8) As R goes to zero, RTh goes to zero and VTh goes to 4 volts, which produces the maximum power delivered to the 10-ohm resistor. P = vi = v2/R = 4x4/10 = 1.6 watts Notice that if R = 20 ohms which gives an RTh = 10 ohms, then VTh becomes -2 volts and the power delivered to the load becomes 0.1 watts, much less that the 1.6 watts. It is also interesting to note that the internal losses for the first case are 122/20 = 7.2 watts and for the second case are = to 12 watts. This is a significant difference. Chapter 4, Solution 69.

We need the Thevenin equivalent across the resistor R. To find RTh, consider the circuit below. 22 kΩ v1 + 10 kΩ

vo

−

40 kΩ

3vo

30 kΩ

Assume that all resistances are in k ohms and all currents are in mA.

1mA

10||40 = 8, and 8 + 22 = 30 1 + 3vo = (v1/30) + (v1/30) = (v1/15) 15 + 45vo = v1

But vo = (8/30)v1, hence,

15 + 45x(8v1/30) v1, which leads to v1 = 1.3636 RTh = v1/1 = -1.3636 k ohms To find VTh, consider the circuit below. 10 kΩ vo 22 kΩ

v1

+ 100V

+ −

+ 40 kΩ

vo

−

3vo

30 kΩ

VTh

−

(100 – vo)/10 = (vo/40) + (vo – v1)/22

(1)

[(vo – v1)/22] + 3vo = (v1/30)

(2)

Solving (1) and (2), v1 = VTh = -243.6 volts p = VTh2/(4RTh) = (243.6)2/[4(-1363.6)] = -10.882 watts

Chapter 4, Solution 70

We find the Thevenin equivalent across the 10-ohm resistor. To find VTh, consider the circuit below. 3Vx

5Ω

5Ω

+ + 15 Ω

4V -

VTh

6Ω

+

Vx

-

From the figure, 15 (4) = 3V 15 + 5 consider the circuit below:

V x = 0, To find RTh,

VTh =

3Vx

5Ω

5Ω V2

V1 + 4V -

15 Ω

+ At node 1, V V − V2 4 − V1 = 3V x + 1 + 1 , 5 15 5

1A

6Ω

Vx

V x = 6 x1 = 6

-

 →

258 = 3V2 − 7V1

(1)

At node 2, V − V2 1 + 3V x + 1 =0  → V1 = V2 − 95 5 Solving (1) and (2) leads to V2 = 101.75 V 2 VTh V2 9 RTh = = 101.75Ω, p max = = = 22.11 mW 1 4 RTh 4 x101.75

(2)

Chapter 4, Solution 71.

We need RTh and VTh at terminals a and b. To find RTh, we insert a 1-mA source at the terminals a and b as shown below. 10 kΩ a + 3 kΩ

vo

− +

1 kΩ

−

120vo

40 kΩ 1mA b

Assume that all resistances are in k ohms, all currents are in mA, and all voltages are in volts. At node a, 1 = (va/40) + [(va + 120vo)/10], or 40 = 5va + 480vo

(1)

The loop on the left side has no voltage source. Hence, vo = 0. From (1), va = 8 V. RTh = va/1 mA = 8 kohms To get VTh, consider the original circuit. For the left loop, vo = (1/4)8 = 2 V For the right loop,

vR = VTh = (40/50)(-120vo) = -192

The resistance at the required resistor is R = RTh = 8 kohms p = VTh2/(4RTh) = (-192)2/(4x8x103) = 1.152 watts

Chapter 4, Solution 72.

(a) RTh and VTh are calculated using the circuits shown in Fig. (a) and (b) respectively. From Fig. (a),

RTh = 2 + 4 + 6 = 12 ohms

From Fig. (b),

-VTh + 12 + 8 + 20 = 0, or VTh = 40 V

4Ω

6Ω

2Ω

4Ω

12V

6Ω

− +

2Ω

+ −

RTh

VTh

8V 20V

(a)

(b)

+ −

(b)

i = VTh/(RTh + R) = 40/(12 + 8) = 2A

(c)

For maximum power transfer,

(d)

p = VTh2/(4RTh) = (40)2/(4x12) = 33.33 watts.

RL = RTh = 12 ohms

Chapter 4, Solution 73

Find the Thevenin’s equivalent circuit across the terminals of R.

10 Ω

25 Ω RTh

20 Ω

RTh = 10 // 20 + 25 // 5 = 325 / 30 = 10.833Ω

+

5Ω

−

10 Ω + 60 V -

+ VTh -

+ Va

25 Ω

20 Ω

+ 5Ω

20 (60) = 40, 30 − Va + VTh + Vb = 0

Va =

Vb -

5 (60) = 10 30  → VTh = Va − Vb = 40 − 10 = 30 V Vb =

2

p max

V 30 2 = Th = = 20.77 W 4 RTh 4 x10.833

Chapter 4, Solution 74.

When RL is removed and Vs is short-circuited, RTh = R1||R2 + R3||R4 = [R1 R2/( R1 + R2)] + [R3 R4/( R3 + R4)] RL = RTh = (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)/[( R1 + R2)( R3 + R4)] When RL is removed and we apply the voltage division principle, Voc = VTh = vR2 – vR4 = ([R2/(R1 + R2)] – [R4/(R3 + R4)])Vs = {[(R2R3) – (R1R4)]/[(R1 + R2)(R3 + R4)]}Vs pmax = VTh2/(4RTh) = {[(R2R3) – (R1R4)]2/[(R1 + R2)(R3 + R4)]2}Vs2[( R1 + R2)( R3 + R4)]/[4(a)] where a = (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4) pmax = [(R2R3) – (R1R4)]2Vs2/[4(R1 + R2)(R3 + R4) (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)]

Chapter 4, Solution 75.

We need to first find RTh and VTh. R R R R

R

R

RTh

+ 1V

(a)

vo

+ −

2V

+ −

+ −

3V

VTh

− (b)

Consider the circuit in Fig. (a). (1/RTh) = (1/R) + (1/R) + (1/R) = 3/R RTh = R/3 From the circuit in Fig. (b), ((1 – vo)/R) + ((2 – vo)/R) + ((3 – vo)/R) = 0 vo = 2 = VTh For maximum power transfer, RL = RTh = R/3 Pmax = [(VTh)2/(4RTh)] = 3 mW RTh = [(VTh)2/(4Pmax)] = 4/(4xPmax) = 1/Pmax = R/3 R = 3/(3x10-3) = 1 k ohms

Chapter 4, Solution 76.

Follow the steps in Example 4.14. The schematic and the output plots are shown below. From the plot, we obtain,

V = 92 V [i = 0, voltage axis intercept] R = Slope = (120 – 92)/1 = 28 ohms

Chapter 4, Solution 77. (a) The schematic is shown below. We perform a dc sweep on a current source, I1, connected between terminals a and b. We label the top and bottom of source I1 as 2 and 1 respectively. We plot V(2) – V(1) as shown. VTh = 4 V [zero intercept] RTh = (7.8 – 4)/1 = 3.8 ohms

(b)

Everything remains the same as in part (a) except that the current source, I1, is connected between terminals b and c as shown below. We perform a dc sweep on I1 and obtain the plot shown below. From the plot, we obtain, V = 15 V [zero intercept] R = (18.2 – 15)/1 = 3.2 ohms

Chapter 4, Solution 78. The schematic is shown below. We perform a dc sweep on the current source, I1, connected between terminals a and b. The plot is shown. From the plot we obtain, VTh = -80 V [zero intercept] RTh = (1920 – (-80))/1 = 2 k ohms

Chapter 4, Solution 79. After drawing and saving the schematic as shown below, we perform a dc sweep on I1 connected across a and b. The plot is shown. From the plot, we get, V = 167 V [zero intercept] R = (177 – 167)/1 = 10 ohms

Chapter 4, Solution 80. The schematic in shown below. We label nodes a and b as 1 and 2 respectively. We perform dc sweep on I1. In the Trace/Add menu, type v(1) – v(2) which will result in the plot below. From the plot, VTh = 40 V [zero intercept] RTh = (40 – 17.5)/1 = 22.5 ohms [slope]

Chapter 4, Solution 81. The schematic is shown below. We perform a dc sweep on the current source, I2, connected between terminals a and b. The plot of the voltage across I2 is shown below. From the plot, VTh = 10 V [zero intercept] RTh = (10 – 6.4)/1 = 3.4 ohms.

Chapter 4, Solution 82. VTh = Voc = 12 V, Isc = 20 A RTh = Voc/Isc = 12/20 = 0.6 ohm. 0.6 Ω i 12V

i = 12/2.6 ,

+ −

2Ω

p = i2R = (12/2.6)2(2) = 42.6 watts

Chapter 4, Solution 83. VTh = Voc = 12 V, Isc = IN = 1.5 A RTh = VTh/IN = 8 ohms, VTh = 12 V, RTh = 8 ohms

Chapter 4, Solution 84 Let the equivalent circuit of the battery terminated by a load be as shown below. RTh IL +

+

VTh

VL

-

RL

-

For open circuit,

R L = ∞,  → VTh = Voc = VL = 10.8 V When RL = 4 ohm, VL=10.5, IL =

VL = 10.8 / 4 = 2.7 RL

But VTh = VL + I L RTh

 →

RTh =

VTh − V L 12 − 10.8 = = 0.4444Ω IL 2 .7

Chapter 4, Solution 85

(a) Consider the equivalent circuit terminated with R as shown below. RTh a + VTh -

+ Vab -

R

b

Vab

R = VTh R + RTh

 →

10 6= VTh 10 + RTh

or 60 + 6 RTh = 10VTh where RTh is in k-ohm.

(1)

Similarly, 30  → VTh 30 + RTh Solving (1) and (2) leads to 12 =

360 + 12 RTh = 30VTh

(2)

VTh = 24 V, RTh = 30kΩ

(b) Vab =

20 (24) = 9.6 V 20 + 30

Chapter 4, Solution 86.

We replace the box with the Thevenin equivalent. RTh + VTh

+ −

i R

v

− VTh = v + iRTh When i = 1.5, v = 3, which implies that VTh = 3 + 1.5RTh

(1)

When i = 1, v = 8, which implies that VTh = 8 + 1xRTh

(2)

From (1) and (2), RTh = 10 ohms and VTh = 18 V. (a)

When R = 4, i = VTh/(R + RTh) = 18/(4 + 10) = 1.2857 A

(b)

For maximum power, R = RTH Pmax = (VTh)2/4RTh = 182/(4x10) = 8.1 watts

Chapter 4, Solution 87.

(a)

im = 9.975 mA

im = 9.876 mA

+ Is

vm

Rs Rm

Is

Rs

Rs Rm

− (a)

(b)

From Fig. (a), vm = Rmim = 9.975 mA x 20 = 0.1995 V From Fig. (b),

Is = 9.975 mA + (0.1995/Rs)

(1)

vm = Rmim = 20x9.876 = 0.19752 V Is = 9.876 mA + (0.19752/2k) + (0.19752/Rs) = 9.975 mA + (0.19752/Rs)

Solving (1) and (2) gives,

Rs = 8 k ohms,

Is = 10 mA

(b) im’ = 9.876 mA

Is

Rs

Rs Rm

(b)

8k||4k = 2.667 k ohms im’ = [2667/(2667 + 20)](10 mA) = 9.926 mA Chapter 4, Solution 88

To find RTh, consider the circuit below. A

RTh

30k Ω

RTh

5k Ω B 20k Ω

10k Ω = 30 + 10 + 20 // 5 = 44kΩ

(2)

To find VTh , consider the circuit below. 5k Ω A

B io

30k Ω

20k Ω

+

4mA

60 V -

10k Ω V A = 30 x 4 = 120,

VB =

20 (60) = 48, 25

VTh = V A − VB = 72 V

Chapter 4, Solution 89

It is easy to solve this problem using Pspice. (a) The schematic is shown below. We insert IPROBE to measure the desired ammeter reading. We insert a very small resistance in series IPROBE to avoid problem. After the circuit is saved and simulated, the current is displaced on IPROBE as 99.99µA .

(b) By interchanging the ammeter and the 12-V voltage source, the schematic is shown below. We obtain exactly the same result as in part (a).

Chapter 4, Solution 90.

Rx = (R3/R1)R2 = (4/2)R2 = 42.6, R2 = 21.3 which is (21.3ohms/100ohms)% = 21.3% Chapter 4, Solution 91.

Rx = (R3/R1)R2 (a)

Since 0 < R2 < 50 ohms, to make 0 < Rx < 10 ohms requires that when R2 = 50 ohms, Rx = 10 ohms. 10 = (R3/R1)50 or R3 = R1/5 so we select R1 = 100 ohms and R3 = 20 ohms

(b)

For 0 < Rx < 100 ohms 100 = (R3/R1)50, or R3 = 2R1 So we can select R1 = 100 ohms and R3 = 200 ohms

Chapter 4, Solution 92. For a balanced bridge, vab = 0. We can use mesh analysis to find vab. Consider the circuit in Fig. (a), where i1 and i2 are assumed to be in mA. 2 kΩ

3 kΩ 220V

+ −

a

i1

+

6 kΩ

i2 b vab

−

5 kΩ

10 kΩ

0 (a) 220 = 2i1 + 8(i1 – i2) or 220 = 10i1 – 8i2 (1) From (1) and (2),

0 = 24i2 – 8i1 or i2 = (1/3)i1

(2)

i1 = 30 mA and i2 = 10 mA Applying KVL to loop 0ab0 gives 5(i2 – i1) + vab + 10i2 = 0 V Since vab = 0, the bridge is balanced. When the 10 k ohm resistor is replaced by the 18 k ohm resistor, the gridge becomes unbalanced. (1) remains the same but (2) becomes Solving (1) and (3),

0 = 32i2 – 8i1, or i2 = (1/4)i1

(3)

i1 = 27.5 mA, i2 = 6.875 mA vab = 5(i1 – i2) – 18i2 = -20.625 V VTh = vab = -20.625 V

To obtain RTh, we convert the delta connection in Fig. (b) to a wye connection shown in Fig. (c).

2 kΩ

3 kΩ

R2

6 kΩ

a

RTh

R1

b

5 kΩ

6 kΩ

a RTh

R3

18 kΩ

(b)

18 kΩ

(c)

R1 = 3x5/(2 + 3 + 5) = 1.5 k ohms, R2 = 2x3/10 = 600 ohms, R3 = 2x5/10 = 1 k ohm. RTh = R1 + (R2 + 6)||(R3 + 18) = 1.5 + 6.6||9 = 6.398 k ohms RL = RTh = 6.398 k ohms Pmax = (VTh)2/(4RTh) = (20.625)2/(4x6.398) = 16.622 mWatts Chapter 4, Solution 93. ix

VS

+ −

Rs

Ro ix

+ −

βRoix

-Vs + (Rs + Ro)ix + βRoix = 0 ix = Vs/(Rs + (1 + β)Ro)

b

Chapter 4, Solution 94. (a)

Vo/Vg = Rp/(Rg + Rs + Rp)

(1)

Req = Rp||(Rg + Rs) = Rg Rg = Rp(Rg + Rs)/(Rp + Rg + Rs) RgRp + Rg2 + RgRs = RpRg + RpRs RpRs = Rg(Rg + Rs) From (1),

(2)

Rp/α = Rg + Rs + Rp Rg + Rs = Rp((1/α) – 1) = Rp(1 - α)/α

(1a)

Combining (2) and (1a) gives, Rs = [(1 - α)/α]Req = (1 – 0.125)(100)/0.125 = 700 ohms From (3) and (1a), Rp(1 - α)/α = Rg + [(1 - α)/α]Rg = Rg/α Rp = Rg/(1 - α) = 100/(1 – 0.125) = 114.29 ohms (b) RTh I VTh

+ −

RL

VTh = Vs = 0.125Vg = 1.5 V RTh = Rg = 100 ohms I = VTh/(RTh + RL) = 1.5/150 = 10 mA

(3)

Chapter 4, Solution 95. Let 1/sensitivity = 1/(20 k ohms/volt) = 50 µA For the 0 – 10 V scale, Rm = Vfs/Ifs = 10/50 µA = 200 k ohms For the 0 – 50 V scale, Rm = 50(20 k ohms/V) = 1 M ohm RTh

VTh

+ −

Rm

VTh = I(RTh + Rm) (a)

A 4V reading corresponds to I = (4/10)Ifs = 0.4x50 µA = 20 µA VTh = 20 µA RTh + 20 µA 250 k ohms = 4 + 20 µA RTh

(b)

(1)

A 5V reading corresponds to I = (5/50)Ifs = 0.1 x 50 µA = 5 µA VTh = 5 µA x RTh + 5 µA x 1 M ohm

From (1) and (2)

VTh = 5 + 5 µA RTh

(2)

0 = -1 + 15 µA RTh which leads to RTh = 66.67 k ohms From (1),

VTh = 4 + 20x10-6x(1/(15x10-6)) = 5.333 V

Chapter 4, Solution 96. (a)

The resistance network can be redrawn as shown in Fig. (a), 10 Ω

8Ω

10 Ω RTh

9V

+ −

i1

40Ω

+

i2

60 Ω

8Ω

R

VTh

10 Ω

−

+ VTh

+ −

Vo −

(a)

(b)

RTh = 10 + 10 + 60||(8 + 8 + 10||40) = 20 + 60||24 = 37.14 ohms Using mesh analysis, -9 + 50i1 - 40i2 = 0 116i2 – 40i1 = 0 or i1 = 2.9i2 From (1) and (2),

(1) (2)

i2 = 9/105 VTh = 60i2 = 5.143 V

From Fig. (b), Vo = [R/(R + RTh)]VTh = 1.8 R/(R + 37.14) = 1.8/5.143 which leads to R = 20 ohms (b)

R = RTh = 37.14 ohms

Imax = VTh/(2RTh) = 5.143/(2x37.14) = 69.23 mA Chapter 4, Solution 97.

4 kΩ 12V

+ −

B + 4 kΩ

VTh − E

R

RTh = R1||R2 = 6||4 = 2.4 k ohms VTh = [R2/(R1 + R2)]vs = [4/(6 + 4)](12) = 4.8 V Chapter 4, Solution 98. The 20-ohm, 60-ohm, and 14-ohm resistors form a delta connection which needs to be connected to the wye connection as shown in Fig. (b),

20 Ω

30 Ω

30 Ω

R2 R1

14 Ω a

60 Ω

a

b RTh

b

R3

(a)

(b)

R1 = 20x60/(20 + 60 + 14) = 1200/94 = 12.97 ohms R2 = 20x14/94 = 2.98 ohms R3 = 60x14/94 = 8.94 ohms RTh = R3 + R1||(R2 + 30) = 8.94 + 12.77||32.98 = 18.15 ohms To find VTh, consider the circuit in Fig. (c). IT 30 Ω

20 Ω I1

14 Ω b a

60 Ω

+

IT 16 V + −

(c)

VTh

RTh

IT = 16/(30 + 15.74) = 350 mA I1 = [20/(20 + 60 + 14)]IT = 94.5 mA VTh = 14I1 + 30IT = 11.824 V I40 = VTh/(RTh + 40) = 11.824/(18.15 + 40) = 203.3 mA P40 = I402R = 1.654 watts

Chapter 5, Solution 1. (a) (b) (c)

Rin = 1.5 MΩ Rout = 60 Ω A = 8x104 Therefore AdB = 20 log 8x104 = 98.0 dB

Chapter 5, Solution 2. v0 = Avd = A(v2 - v1) = 105 (20-10) x 10-6 = 0.1V Chapter 5, Solution 3. v0 = Avd = A(v2 - v1) = 2 x 105 (30 + 20) x 10-6 = 10V Chapter 5, Solution 4. v0 = Avd = A(v2 - v1) v −4 v2 - v1 = 0 = = −20µV A 2 x10 5 If v1 and v2 are in mV, then v2 - v1 = -20 mV = 0.02 1 - v1 = -0.02 v1 = 1.02 mV Chapter 5, Solution 5.

I

R0

Rin

vd + vi

+

-

+ -

Avd

+ v0

-

-vi + Avd + (Ri - R0) I = 0 But

(1)

vd = RiI, -vi + (Ri + R0 + RiA) I = 0 vd =

vi R i R 0 + (1 + A)R i

(2)

-Avd - R0I + v0 = 0 v0 = Avd + R0I = (R0 + RiA)I =

(R 0 + R i A) v i R 0 + (1 + A)R i

v0 R 0 + RiA 100 + 10 4 x10 5 = ⋅ 10 4 = 5 v i R 0 + (1 + A)R i 100 + (1 + 10 )

≅

100,000 10 9 ⋅ 10 4 = = 0.9999990 5 100,001 1 + 10

(

)

Chapter 5, Solution 6. vi + -

R0 I

vd +

Rin + -

Avd

+ vo

-

(R0 + Ri)R + vi + Avd = 0 But

vd = RiI, vi + (R0 + Ri + RiA)I = 0 I=

− vi R 0 + (1 + A)R i

(1)

-Avd - R0I + vo = 0 vo = Avd + R0I = (R0 + RiA)I Substituting for I in (1),  R 0 + R iA   vi v0 = −  + + R ( 1 A ) R i   0 6 50 + 2 x10 x 2 x10 5 ⋅ 10 −3 = − 50 + 1 + 2x10 5 x 2 x10 6

(

≅

(

)

)

− 200,000 x 2 x10 6 mV 200,001x 2 x10 6

v0 = -0.999995 mV

Chapter 5, Solution 7.

100 kΩ

10 kΩ

VS

+

-

Rout = 100 Ω

1

2

+ Vd

-

Rin

+ -

AVd

+ Vout

-

At node 1,

(VS – V1)/10 k = [V1/100 k] + [(V1 – V0)/100 k] 10 VS – 10 V1 = V1 + V1 – V0 which leads to V1 = (10VS + V0)/12

At node 2,

(V1 – V0)/100 k = (V0 – AVd)/100

But Vd = V1 and A = 100,000, V1 – V0 = 1000 (V0 – 100,000V1) 0= 1001V0 – 100,000,001[(10VS + V0)/12] 0 = -83,333,334.17 VS - 8,332,333.42 V0 which gives us (V0/ VS) = -10 (for all practical purposes) If VS = 1 mV, then V0 = -10 mV Since V0 = A Vd = 100,000 Vd, then Vd = (V0/105) V = -100 nV Chapter 5, Solution 8. (a)

If va and vb are the voltages at the inverting and noninverting terminals of the op amp. va = v b = 0 1mA =

0 − v0 2k

v0 = -2V

(b) 10 kΩ

2V

+

ia

va

2V

+

vb 1V

+

+ vo

-

-

(a)

2 kΩ

+ va

-

10 kΩ

+-

+ ia

(b)

vo

-

Since va = vb = 1V and ia = 0, no current flows through the 10 kΩ resistor. From Fig. (b), -va + 2 + v0 = 0

va = va - 2 = 1 - 2 = -1V

Chapter 5, Solution 9. (a) Let va and vb be respectively the voltages at the inverting and noninverting terminals of the op amp va = vb = 4V At the inverting terminal, 1mA =

4 − v0 2k

v0 = 2V

(b)

1V +-

+

+

vb

vo

-

-

Since va = vb = 3V, -vb + 1 + vo = 0

vo = vb - 1 = 2V

Chapter 5, Solution 10. Since no current enters the op amp, the voltage at the input of the op amp is vs. Hence

 10  v o vs = v o  =  10 + 10  2

vo =2 vs

Chapter 5, Solution 11. 8 kΩ 2 kΩ

3V

vb =

5 kΩ

+ −

a b

io

− +

+

10 kΩ

4 kΩ

vo −

10 (3) = 2V 10 + 5

At node a, 3 − va va − vo = 2 8

12 = 5va – vo

But va = vb = 2V, vo = -2V

12 = 10 – vo –io =

va − vo 0 − vo 2 + 2 2 + = + = 1mA 8 4 8 4

i o = -1mA Chapter 5, Solution 12. 4 kΩ 1 kΩ

1.2V

+ −

a b

− +

4 kΩ

2 kΩ

+ vo −

4 2 2 vo = vo = vo 4+2 3 3

At node b,

vb =

At node a,

1 .2 − v a v a − v o 2 , but va = vb = v o = 1 4 3

4.8 - 4 x

2 2 vo = vo − vo 3 3

va = vb =

2 9.6 vo = 3 7

is =

vo =

3x 4.8 = 2.0570V 7

1 .2 − v a − 1 .2 = 1 7

 − 1.2  p = vsis = 1.2   = -205.7 mW  7  Chapter 5, Solution 13.

10 kΩ a b 1V

+ −

90 kΩ

+ −

io 100 kΩ i2

i1 4 kΩ

50 kΩ

By voltage division, 90 va = (1) = 0.9V 100 v 50 vb = vo = o 3 150 v0 But va = vb = 0 .9 vo = 2.7V 3 v v io = i1 + i2 = o + o = 0.27mA + 0.018mA = 288 µA 10k 150k

+ vo

−

Chapter 5, Solution 14.

Transform the current source as shown below. At node 1, 10 − v1 v1 − v 2 v1 − v o = + 5 20 10

10 kΩ

5 kΩ

10 kΩ

20 kΩ v1

10V

vo

v2

+ −

− +

+ vo

−

But v2 = 0. Hence 40 - 4v1 = v1 + 2v1 - 2vo At node 2,

v1 − v 2 v 2 − v o = , 20 10

40 = 7v1 - 2vo

v 2 = 0 or v1 = -2vo

From (1) and (2), 40 = -14vo - 2vo

(1) (2)

vo = -2.5V

Chapter 5, Solution 15

(a) Let v1 be the voltage at the node where the three resistors meet. Applying KCL at this node gives  1 v −v v 1  vo  − + (1) i s = 1 + 1 o = v1  R2 R3 R R R3 3   2 At the inverting terminal, 0 − v1 (2) is =  → v1 = −i s R1 R1 Combining (1) and (2) leads to   v vo RR  R R   → = − R1 + R3 + 1 3  i s 1 + 1 + 1  = − o R3 is R2    R2 R3 

(b) For this case, vo 20 x 40   = − 20 + 40 +  kΩ = - 92 kΩ is 25  

Chapter 5, Solution 16

10k Ω

5k Ω

ix

va vb

iy

+

vo

2k Ω

+ 0.5V -

8k Ω

Let currents be in mA and resistances be in k Ω . At node a, 0 .5 − v a v a − v o =  → 1 = 3v a − vo 5 10

(1)

But 8 10 vo  → vo = v a (2) 8+2 8 Substituting (2) into (1) gives 10 8 1 = 3v a − v a  → v a = 8 14 Thus, 0 .5 − v a ix = = −1 / 70 mA = − 14.28 µA 5 v − vb v o − v a 10 0 .6 8 iy = o + = 0 .6 ( v o − v a ) = 0 .6 ( v a − v a ) = x mA = 85.71 µA 2 10 8 4 14

v a = vb =

Chapter 5, Solution 17.

(a) (b) (c)

G=

vo R 12 = − 2 = − = -2.4 vi R1 5

vo 80 =− = -16 vi 5 vo 2000 =− = -400 vi 5

Chapter 5, Solution 18.

Converting the voltage source to current source and back to a voltage source, we have the circuit shown below: 10 20 =

20 kΩ 3 1 MΩ

(20/3) kΩ 50 kΩ

− +

+ −

2vi/3

+ vo

−

vo = −

1000 2v i ⋅ 20 3 50 + 3

vo 200 =− = -11.764 v1 17

Chapter 5, Solution 19.

We convert the current source and back to a voltage source. 24=

(4/3) kΩ

(2/3)V

+ −

4 kΩ

0V

4 3 10 kΩ

− +

vo 5 kΩ

10k  2    = -1.25V  4 3  4x k  3 v v −0 = -0.375mA io = o + o 5k 10k vo = −

Chapter 5, Solution 20. 8 kΩ

4 kΩ

9V

a

+ −

4 kΩ

vs

2 kΩ b

− +

+ −

+ vo

−

At node a, 9 − va va − vo va − vb = + 4 8 4

18 = 5va – vo - 2vb

(1)

At node b, va − vb vb − vo = 4 2

va = 3vb - 2vo

But vb = vs = 0; (2) becomes va = –2vo and (1) becomes -18 = -10vo – vo

vo = -18/(11) = -1.6364V

(2)

Chapter 5, Solution 21. Eqs. (1) and (2) remain the same. When vb = vs = 3V, eq. (2) becomes va = 3 x 3 - 2v0 = 9 - 2vo Substituting this into (1), 18 = 5 (9-2vo) – vo – 6 leads to vo = 21/(11) = 1.909V Chapter 5, Solution 22. Av = -Rf/Ri = -15. If Ri = 10kΩ, then Rf = 150 kΩ. Chapter 5, Solution 23 At the inverting terminal, v=0 so that KCL gives vs − 0 0 0 − vo = + R1 R2 Rf

→

vo vs

=−

Rf R1

Chapter 5, Solution 24 v1

Rf R2

R1 - vs +

+ +

R3

R4

vo -

v2

We notice that v1 = v2. Applying KCL at node 1 gives

v1 (v1 − v s ) v1 − vo + + =0 R1 R2 Rf

 →

 1   + 1 + 1 v1 − v s = vo R R R f  R2 R f 2  1

(1)

Applying KCL at node 2 gives R3 v1 v1 − v s + =0  → v1 = vs R3 + R4 R3 R4 Substituting (2) into (1) yields

(2)

 R R R  R3  1   −  v s vo = R f  3 + 3 − 4   R1 R f R2  R3 + R4  R2  i.e.  R R R  R3  1   −  k = R f  3 + 3 − 4   R1 R f R2  R3 + R4  R2 

Chapter 5, Solution 25.

vo = 2 V

+ −

+

+

va

vo

-va + 3 + vo = 0 which leads to va = vo + 3 = 5 V. Chapter 5, Solution 26

+ vb

+ 0.4V -

8k Ω

+ 2k Ω

vo -

vb = 0.4 =

8 vo = 0.8vo 8+ 2

 →

Hence, io =

v o 0 .5 = = 0.1 mA 5k 5k

vo = 0.4 / 0.8 = 0.5 V

io 5k Ω

Chapter 5, Solution 27.

(a)

Let va be the voltage at the noninverting terminal. va = 2/(8+2) vi = 0.2vi  1000  v 0 = 1 +  v a = 10.2v i 20   G = v0/(vi) = 10.2

(b)

vi = v0/(G) = 15/(10.2) cos 120πt = 1.471 cos 120πt V

Chapter 5, Solution 28.

− + + −

At node 1,

0 − v1 v1 − v o = 10k 50k

But v1 = 0.4V, -5v1 = v1 – vo, leads to

vo = 6v1 = 2.4V

Alternatively, viewed as a noninverting amplifier, vo = (1 + (50/10)) (0.4V) = 2.4V io = vo/(20k) = 2.4/(20k) = 120 µA

Chapter 5, Solution 29

R1

va vb

+ vi -

va =

R2

R2 vi , R1 + R2

But v a = vb

vb =

 →

+ -

R1

R1 vo R1 + R2

R2 R1 vi = vo R1 + R2 R1 + R2

Or v o R2 = vi R1

Chapter 5, Solution 30.

The output of the voltage becomes vo = vi = 12 30 20 = 12kΩ By voltage division, vx =

12 (1.2) = 0.2V 12 + 60 ix =

p=

v 2x 0.04 = = 2µW R 20k

vx 0 .2 = = 10µA 20k 20k

+ R2

vo -

Chapter 5, Solution 31. After converting the current source to a voltage source, the circuit is as shown below: 12 kΩ 3 kΩ

1

6 kΩ v o

v1

12 V

+ −

+ −

2

vo

6 kΩ

At node 1, 12 − v1 v1 − v o v1 − v o = + 3 6 12

48 = 7v1 - 3vo

(1)

At node 2, v1 − v o v o − 0 = = ix 6 6

v1 = 2vo

(2)

From (1) and (2),

48 11 vo ix = = 0.7272mA 6k vo =

Chapter 5, Solution 32.

Let vx = the voltage at the output of the op amp. The given circuit is a non-inverting amplifier.  50  v x = 1 +  (4 mV) = 24 mV  10  60 30 = 20kΩ

By voltage division, vo =

v 20 v o = o = 12mV 20 + 20 2

ix =

vx 24mV = = 600nA (20 + 20)k 40k

p=

v o2 144x10 −6 = = 204nW R 60x10 3

Chapter 5, Solution 33.

After transforming the current source, the current is as shown below: 1 kΩ 4 kΩ

4V

vi

+ −

va

+ − 2 kΩ

vo 3 kΩ

This is a noninverting amplifier. 3  1 v o = 1 +  v i = v i 2  2 Since the current entering the op amp is 0, the source resistor has a OV potential drop. Hence vi = 4V. vo =

3 (4) = 6V 2

Power dissipated by the 3kΩ resistor is v o2 36 = = 12mW R 3k ix =

va − vo 4 − 6 = = -2mA R 1k

Chapter 5, Solution 34

v1 − vin v1 − vin + =0 R1 R2

(1)

R3 vo R3 + R 4

(2)

but va =

Combining (1) and (2), v1 − va +

R1 R v 2 − 1 va = 0 R2 R2

 R  R v a 1 + 1  = v1 + 1 v 2 R2  R2  R R  R 3v o  1 + 1  = v1 + 1 v 2 R2 R3 + R 4  R 2 

vo =

vO =

 R3 + R 4  R  v1 + 1 v 2  R2   R  R 3 1 + 1    R2 

R3 + R 4 ( v1R 2 + v 2 ) R 3 ( R1 + R 2 )

Chapter 5, Solution 35. Av =

vo R = 1 + f = 10 Ri vi

If Ri = 10kΩ, Rf = 90kΩ

Rf = 9Ri

Chapter 5, Solution 36 VTh = Vab

But VTh

R1 Vab . Thus, R1 + R2 R R + R2 = Vab = 1 v s = (1 + 2 )v s R1 R1 vs =

To get RTh, apply a current source Io at terminals a-b as shown below. v1

+ -

v2

a + R2

R1

vo

io

b

Since the noninverting terminal is connected to ground, v1 = v2 =0, i.e. no current passes through R1 and consequently R2 . Thus, vo=0 and v RTh = o = 0 io

Chapter 5, Solution 37.  R R R v o = −  f v1 + f v 2 + f v 3  R3  R2  R1

30 30  30  = −  (1) + (2) + (−3) 20 30  10  vo = -3V

Chapter 5, Solution 38.  R R R R v o = −  f v1 + f v 2 + f v 3 + f v 4  R4  R3 R2  R1 50 50 50  50  = −  (10) + (−20) + (50) + (−100) 20 10 50  25 

= -120mV Chapter 5, Solution 39

This is a summing amplifier. Rf Rf   Rf 50 50  50  vo = − v1 + v2 + v3  = − (2) + v 2 + (−1)  = −9 − 2.5v 2 R2 R3  20 50  10   R1 Thus, vo = −16.5 = −9 − 2.5v 2  → v 2 = 3 V

Chapter 5, Solution 40

R1 va

R2 + v1 -

+

R3 + v2 -

vb

-

+ Rf

+ v3 -

R

vo -

Applying KCL at node a, v1 − v a v 2 − v a v3 − v a + + =0 R1 R2 R3

 →

v1 v 2 v3 1 1 1 + + = va ( + + ) (1) R1 R2 R3 R1 R2 R3

But v a = vb =

R vo R + Rf

(2)

Substituting (2) into (1)gives Rvo v1 v 2 v3 1 1 1 + + = ( + + ) R1 R2 R3 R + R f R1 R2 R3 or vo =

R + Rf R

(

v1 v 2 v3 1 1 1 + + ) /( + + ) R1 R2 R3 R1 R2 R3

Chapter 5, Solution 41.

Rf/Ri = 1/(4)

Ri = 4Rf = 40kΩ

The averaging amplifier is as shown below: v1 v2 v3 v4

Chapter 5, Solution 42

1 R f = R1 = 10 kΩ 3

R1 = 40 kΩ

10 kΩ

R2 = 40 kΩ R3 = 40 kΩ R4 = 40 kΩ

− +

vo

Chapter 5, Solution 43. In order for  R R R R v o =  f v1 + f v 2 + f v 3 + f v 4  R2 R3 R4   R1

to become

1 (v 1 + v 2 + v 3 + v 4 ) 4 R Rf 1 12 = Rf = i = = 3kΩ Ri 4 4 4

vo = −

Chapter 5, Solution 44. R4 R3

v1 v2

a R1

− +

b

R2

At node b,

v b − v1 v b − v 2 + =0 R1 R2

At node a,

0 − va va − vo = R3 R4

v1 v 2 + R1 R 2 vb = 1 1 + R1 R 2

(1)

vo 1+ R4 / R3

(2)

va =

But va = vb. We set (1) and (2) equal. vo R v + R 1 v1 = 2 1 1+ R4 / R3 R1 + R 2

or vo =

vo

(R 3 + R 4 ) (R 2 v1 + R 1 v1 ) R 3 (R 1 + R 2 )

Chapter 5, Solution 45.

This can be achieved as follows:

 R (− v1 ) + R v 2  v o = − R/2  R / 3  R R = −  f (− v1 ) + f v 2  R2   R1

i.e. Rf = R, R1 = R/3, and R2 = R/2 Thus we need an inverter to invert v1, and a summer, as shown below (R<100kΩ). R v1

R

R

− +

R/3 -v1 v2

R/2

− +

vo

Chapter 5, Solution 46. v1 1 R R R 1 + ( − v 2 ) + v 3 = f v1 + x ( − v 2 ) + f v 3 3 3 2 R1 R2 R3 i.e. R3 = 2Rf, R1 = R2 = 3Rf. To get -v2, we need an inverter with Rf = Ri. If Rf = 10kΩ, a solution is given below. − vo =

10 kΩ v2

v1

10 kΩ

− +

30 kΩ 30 kΩ

10 kΩ

-v2 v3

20 kΩ

− +

vo

Chapter 5, Solution 47.

If a is the inverting terminal at the op amp and b is the noninverting terminal, then, vb =

10 − v a v a − v o 3 (8) = 6V, v a = v b = 6V and at node a, = 2 4 3 +1

which leads to vo = –2 V and io =

v o (v a − v o ) − = –0.4 – 2 mA = –2.4 mA 5k 4k

Chapter 5, Solution 48.

Since the op amp draws no current from the bridge, the bridge may be treated separately v1 as follows:

i1

+ − i2

v2

For loop 1, (10 + 30) i1 = 5

i1 = 5/(40) = 0.125µA

For loop 2, (40 + 60) i2 = -5

i2 = -0.05µA

But, 10i + v1 - 5 = 0 60i + v2 + 5 = 0

v1 = 5 - 10i = 3.75mV v2 = -5 - 60i = -2mV

As a difference amplifier, R 80 [3.75 − (−2)]mV v o = 2 (v 2 − v 1 ) = 20 R1 = 23mV

Chapter 5, Solution 49.

R1 = R3 = 10kΩ, R2/(R1) = 2 i.e.

R2 = 2R1 = 20kΩ = R4 vo =

Verify:

=2

R 2 1 + R1 / R 2 R v 2 − 2 v1 R1 1 + R 3 / R 4 R1

(1 + 0.5) v 2 − 2v1 = 2(v 2 − v1 ) 1 + 0.5

Thus, R1 = R3 = 10kΩ, R2 = R4 = 20kΩ Chapter 5, Solution 50.

(a)

We use a difference amplifier, as shown below: v1

R1

R2

− + v2

vo =

(b)

R1

vo

R2

R2 (v 2 − v1 ) = 2(v 2 − v1 ), i.e. R2/R1 = 2 R1 If R1 = 10 kΩ then R2 = 20kΩ

We may apply the idea in Prob. 5.35. v 0 = 2 v1 − 2 v 2  R (− v1 ) + R v 2  = − R/2  R / 2 R  R = −  f (− v1 ) + f v 2  R2   R1 i.e. Rf = R, R1 = R/2 = R2

We need an inverter to invert v1 and a summer, as shown below. We may let R = 10kΩ. R v1

R

R

− +

R/2 -v1 v2

R/2

− +

vo

Chapter 5, Solution 51.

We achieve this by cascading an inverting amplifier and two-input inverting summer as shown below: R R R R v2 − va + R vo − v1 +

Verify: But

vo = -va - v1 va = -v2. Hence vo = v2 - v1.

Chapter 5, Solution 52 A summing amplifier shown below will achieve the objective. An inverter is inserted to invert v2. Let R = 10 k Ω . R/2 R v1 R/5 v3 R

+

v4

vo

R

R v2

-

R/4

+

Chapter 5, Solution 53. (a) v1

R2

R1 va vb

v2

− +

R1

vo

R2

At node a,

At node b,

v1 − v a v a − v o = R1 R2 R2 vb = v2 R1 + R 2

va =

But va = vb. Setting (1) and (2) equal gives R v + R 1v o R2 v2 = 2 1 R1 + R 2 R1 + R 2

R 2 v1 + R 1 v o R1 + R 2

(1) (2)

v 2 − v1 =

R1 vo = vi R2

vo R 2 = vi R1

(b)

−

v1

vi + v2

R1/2 v A

R1/2

R2 va

Rg R1/2

R1/2

vb

vB

− +

R2

+ vo

−

At node A,

v1 − v A v B − v A v A − v a + = R1 / 2 Rg R1 / 2

or

v1 − v A +

At node B,

v2 − vB vB − vA vB − vb = + R1 / 2 R1 / 2 Rg

or

v2 − vB −

R1 (v B − v A ) = v A − v a 2R g

R1 (v B − v A ) = v B − v b 2R g

Subtracting (1) from (2), v 2 − v1 − v B + v A −

2R 1 (v B − v A ) = v B − v A − v b + v a 2R g

Since, va = vb,

v 2 − v1  R  v = 1 + 1 (v B − v A ) = i   2 2  2R g 

(1)

(2)

vB − vA =

or

vi ⋅ 2

1 R 1+ 1 2R g

(3)

But for the difference amplifier, R2 (v B − v A ) R1 / 2 R vB − vA = 1 vo 2R 2

vo =

or

Equating (3) and (4),

R1 v vo = i ⋅ 2R 2 2 vo R 2 = ⋅ vi R1

(c)

At node a,

At node b,

(4) 1 R 1+ 1 2R g

1 R 1+ 1 2R g

v1 − v a v a − v A = R1 R2 /2 2R 1 2R 1 vA va − v1 − v a = R2 R2 2R 1 2R 1 vB vb − v2 − vb = R2 R2

(1) (2)

Since va = vb, we subtract (1) from (2), − 2R 1 v (v B − v A ) = i 2 R2 − R2 vi vB − vA = 2R 1 v 2 − v1 =

or

(3)

At node A, va − vA vB − vA vA − vo + = R2 /2 Rg R/2 va − vA +

R2 (v B − v A ) = v A − v o 2R g

(4)

At node B,

vb − vB vB − vA vB − 0 − = R/2 Rg R/2 vb − vB −

R2 (v B − v A ) = v B 2R g

(5)

Subtracting (5) from (4), v B −v A +

R2 (v B − v A ) = v A − v B − v o Rg

 R  2(v B − v A )1 + 2  = − v o  2R  g   Combining (3) and (6), − R 2  R  v i 1 + 2  = −v o  2R  R1 g  

v o R 2  R  = 1+ 2  vi R 1  2R g 

Chapter 5, Solution 54.

(a) (b) But

A0 = A1A2A3 = (-30)(-12.5)(0.8) = 300 A = A1A2A3A4 = A0A4 = 300A4

20Log10 A = 60dB

Log10 A = 3

A = 103 = 1000 A4 = A/(300) = 3.333 Chapter 5, Solution 55.

Let A1 = k, A2 = k, and A3 = k/(4) A = A1A2A3 = k3/(4) 20Log10 A = 42

Thus

Log10 A = 2.1 A = 102 ⋅1 = 125.89 k3 = 4A = 503.57 k = 3 503.57 = 7.956 A1 = A2 = 7.956, A3 = 1.989

(6)

Chapter 5, Solution 56.

There is a cascading system of two inverting amplifiers. − 12  − 12    v s = 6v s 4  6  v i o = s = 3v s mA 2k vo =

(a) (b)

When vs = 12V, io = 36mA When vs = 10 cos 377t V, io = 30 cos 377t mA

Chapter 5, Solution 57

The first stage is a difference amplifier. Since R1/R2 = R3/R4, v o′ =

R2 100 ( v 2 − v1 ) = (1 + 4) = 10 mA R1 50

The second stage is a non-inverter.

R R   v o = 1 +  v o ′ = 1 + 10 mA = 40 mV(given)  40   40  Which leads to, R = 120 kΩ Chapter 5, Solution 58.

By voltage division, the input to the voltage follower is: v1 =

3 (0.6) = 0.45V 3 +1

vo =

10 − 10 v1 − v1 = −7 v1 = −3.15 2 5

io =

0 − vo = 0.7875mA 4k

Thus

Chapter 5, Solution 59.

Let a be the node between the two op amps. va = vo The first stage is a summer va = or

− 10 10 vs − vo = vo 5 20 1.5vs = -2vs vo − 2 = = -1.333 v s 1 .5

Chapter 5, Solution 60.

Transform the current source as shown below: 4 kΩ

10 kΩ

5 kΩ

− + 5is

v1

+ −

io

3 kΩ

+ −

3Ω

2 kΩ

Assume all currents are in mA. The first stage is a summer v1 = By voltage division,

− 10 (5i s ) − 10 v o = −10i s − 2.5v o 5 4

(1)

v1 =

3 1 vo = vo 3+3 2

(2)

Alternatively, we notice that the second stage is a non-inverter.  1  vo =   v1 = 2 v1 3+ 3 From (1) and (2), 0.5v o = −10i s − 2.5v o v o = −2i o = −

10i s 3

3vo = 10is io 5 = = 1.667 is 3

Chapter 5, Solution 61.

Let v01 be the voltage at the left end of R5. The first stage is an inverter, while the second stage is a summer. R2 v1 R1 R R v 0 = − 4 v 01 − 4 v 2 R5 R3

v 01 = −

v1 =

R 2R 4 R v1 − 4 v 2 R 1R 5 R3

Chapter 5, Solution 62.

Let v1 = output of the first op amp v2 = output of the second op amp The first stage is a summer v1 = −

R2 R vi – 2 vo R1 Rf

The second stage is a follower. By voltage division

(1)

vo = v2 =

R4 v1 R3 + R4

v1 =

R3 + R4 vo R4

(2)

From (1) and (2),  R3  R R  v o = − 2 v i − 2 v o 1 + Rf R1  R4   R3 R2  R  v o = − 2 v i 1 + + R1  R4 Rf  vo R 1 =− 2 ⋅ R R vi R1 1+ 3 + 2 R4 R4 − R 2R 4 = R 1 (R 2 + R 3 + R 4 )

Chapter 5, Solution 63.

The two op amps are summer. Let v1 be the output of the first op amp. For the first stage, v1 = −

R2 R vi − 2 vo R1 R3

(1)

For the second stage, vo = −

R4 R v1 − 4 v i R5 Ro

(2)

Combining (1) and (2),  R2  R R  R  v i + 4  2  v o − 4 v i  R5  R3  R6  R1   R R  R R R  v o 1 − 2 4  =  2 4 − 4  v i  R 3 R 5   R 1R 5 R 6  vo =

R4 R5

R 2R 4 R 4 − vo R 1R 3 R 6 = R R vi 1− 2 4 R 3R 5

Chapter 5, Solution 64 G4 G G1 +

G3

1

G +

0V

vs

v

2 0V +

G2

+ vo

-

-

At node 1, v1=0 so that KCL gives G1v s + G4 vo = −Gv

(1)

At node 2, G2 v s + G3 v o = −Gv From (1) and (2), G1v s + G4 v o = G2 v s + G3 vo or vo G1 − G2 = v s G3 − G 4

(2)

→ 

(G1 − G2 )v s = (G3 − G4 )vo

Chapter 5, Solution 65

The output of the first op amp (to the left) is 6 mV. The second op amp is an inverter so that its output is

30 (6mV) = -18 mV 10 The third op amp is a noninverter so that vo ' = −

vo ' =

40 vo 40 + 8

 →

vo =

48 v o ' = − 21.6 mV 40

Chapter 5, Solution 66.

100  40  100 − 110 (6) − (2)  − (4) − 25 20  20  10 = −24 + 40 − 20 = -4V

vo =

Chapter 5, Solution 67.

80  80  80  − (0.5) − (0.2) 40  20  20 = 3.2 − 0.8 = 2.4V

vo = −

Chapter 5, Solution 68.

If Rq = ∞, the first stage is an inverter. Va = −

15 (10) = −30mV 5

when Va is the output of the first op amp. The second stage is a noninverting amplifier.  6 v o = 1 +  v a = (1 + 3)(−30) = -120mV  2 Chapter 5, Solution 69.

In this case, the first stage is a summer va = −

15 15 (10) − v o = −30 − 1.5v o 5 10

For the second stage,

 6 v o = 1 +  v a = 4v a = 4(− 30 − 1.5v o )  2 120 vo = − = -17.143mV 7 v o = −120 7

Chapter 5, Solution 70.

The output of amplifier A is vA = −

30 30 (10) − (2) = −9 10 10

The output of amplifier B is vB = −

20 20 (3) − (4) = −14 10 10 40 kΩ

vA vB

20 kΩ a

60 kΩ

− +

b

vo

10 kΩ

vb =

60 (−14) = −2V 60 + 10

At node a,

vA − va va − vo = 20 40

But va = vb = -2V, 2(-9+2) = -2-vo Therefore,

vo = 12V

Chapter 5, Solution 71 20k Ω

5k Ω

100k Ω 40k Ω

+ + 2V -

v2 10k Ω

80k Ω

+ 20k Ω

+ vo -

+ + 3V -

10k Ω v1

+ -

30k Ω

v3

50k Ω

20 50 (2) = −8, v3 = (1 + )v1 = 8 5 30 100   100 vo = − v2 + v3  = −(−20 + 10) = 10 V 80   40 v1 = 3,

v2 = −

Chapter 5, Solution 72.

Since no current flows into the input terminals of ideal op amp, there is no voltage drop across the 20 kΩ resistor. As a voltage summer, the output of the first op amp is v01 = 0.4 The second stage is an inverter 150 v 01 100 = −2.5(0.4) = -1V

v2 = −

Chapter 5, Solution 73.

The first stage is an inverter. The output is 50 v 01 = − (−1.8) = −9V 10 The second stage is v 2 = v 01 = -9V Chapter 5, Solution 74.

Let v1 = output of the first op amp v2 = input of the second op amp. The two sub-circuits are inverting amplifiers 100 (0.6) = −6V 10 32 v2 = − (0.4) = −8V 1.6 v − v2 −6+8 io = 1 =− = 100 µA 20k 20k v1 = −

Chapter 5, Solution 75.

The schematic is shown below. Pseudo-components VIEWPOINT and IPROBE are involved as shown to measure vo and i respectively. Once the circuit is saved, we click Analysis | Simulate. The values of v and i are displayed on the pseudo-components as:

i = 200 µA (vo/vs) = -4/2 = -2 The results are slightly different than those obtained in Example 5.11.

Chapter 5, Solution 76. The schematic is shown below. IPROBE is inserted to measure io. Upon simulation, the value of io is displayed on IPROBE as io = -374.78 µA

Chapter 5, Solution 77.

The schematic is shown below. IPROBE is inserted to measure io. Upon simulation, the value of io is displayed on IPROBE as io = -374.78 µA

Chapter 5, Solution 78.

The circuit is constructed as shown below. We insert a VIEWPOINT to display vo. Upon simulating the circuit, we obtain, vo = 667.75 mV

Chapter 5, Solution 79.

The schematic is shown below. A pseudo-component VIEWPOINT is inserted to display vo. After saving and simulating the circuit, we obtain, vo = -14.61 V

Chapter 5, Solution 80.

The schematic is shown below. VIEWPOINT is inserted to display vo. After simulation, we obtain, vo = 12 V

Chapter 5, Solution 81.

The schematic is shown below. We insert one VIEWPOINT and one IPROBE to measure vo and io respectively. Upon saving and simulating the circuit, we obtain, vo = 343.37 mV io = 24.51 µA

Chapter 5, Solution 82.

The maximum voltage level corresponds to 11111 = 25 – 1 = 31 Hence, each bit is worth

(7.75/31) = 250 mV

Chapter 5, Solution 83.

The result depends on your design. Hence, let RG = 10 k ohms, R1 = 10 k ohms, R2 = 20 k ohms, R3 = 40 k ohms, R4 = 80 k ohms, R5 = 160 k ohms, R6 = 320 k ohms, then, -vo = (Rf/R1)v1 + --------- + (Rf/R6)v6 = v1 + 0.5v2 + 0.25v3 + 0.125v4 + 0.0625v5 + 0.03125v6 (a)

|vo| = 1.1875 = 1 + 0.125 + 0.0625 = 1 + (1/8) + (1/16) which implies, [v1 v2 v3 v4 v5 v6] = [100110]

(b)

|vo| = 0 + (1/2) + (1/4) + 0 + (1/16) + (1/32) = (27/32) = 843.75 mV

(c)

This corresponds to [1 1 1 1 1 1]. |vo| = 1 + (1/2) + (1/4) + (1/8) + (1/16) + (1/32) = 63/32 = 1.96875 V

Chapter 5, Solution 84. For (a), the process of the proof is time consuming and the results are only approximate, but close enough for the applications where this device is used. (a)

The easiest way to solve this problem is to use superposition and to solve for each term letting all of the corresponding voltages be equal to zero. Also, starting with each current contribution (ik) equal to one amp and working backwards is easiest. 2R

v1

+ −

R

R

R

2R

ik v2

+ −

2R

v3

+ −

2R

v4

+ −

R

For the first case, let v2 = v3 = v4 = 0, and i1 = 1A. Therefore,

v1 = 2R volts or i1 = v1/(2R).

Second case, let v1 = v3 = v4 = 0, and i2 = 1A. Therefore, v2 = 85R/21 volts or i2 = 21v2/(85R). Clearly this is not (1/4th), so where is the difference? (21/85) = 0.247 which is a really good approximation for 0.25. Since this is a practical electronic circuit, the result is good enough for all practical purposes.

Now for the third case, let v1 = v2 = v4 = 0, and i3 = 1A. Therefore, v3 = 8.5R volts or i3 = v3/(8.5R). Clearly this is not (1/8th), so where is the difference? (1/8.5) = 0.11765 which is a really good approximation for 0.125. Since this is a practical electronic circuit, the result is good enough for all practical purposes. Finally, for the fourth case, let v1 = v2 = v4 = 0, and i3 = 1A.

Therefore, v4 = 16.25R volts or i4 = v4/(16.25R). Clearly this is not th (1/16 ), so where is the difference? (1/16.25) = 0.06154 which is a really good approximation for 0.0625. Since this is a practical electronic circuit, the result is good enough for all practical purposes. Please note that a goal of a lot of electronic design is to come up with practical circuits that are economical to design and build yet give the desired results. (b)

If Rf = 12 k ohms and R = 10 k ohms, -vo = (12/20)[v1 + (v2/2) + (v3/4) + (v4/8)] = 0.6[v1 + 0.5v2 + 0.25v3 + 0.125v4] For

[v1 v2 v3 v4] = [1 0 11], |vo| = 0.6[1 + 0.25 + 0.125] = 825 mV

For

[v1 v2 v3 v4] = [0 1 0 1], |vo| = 0.6[0.5 + 0.125] = 375 mV

Chapter 5, Solution 85. Av = 1 + (2R/Rg) = 1 + 20,000/100 = 201 Chapter 5, Solution 86. vo = A(v2 – v1) = 200(v2 – v1) (a) (b)

vo = 200(0.386 – 0.402) = -3.2 V vo = 200(1.011 – 1.002) = 1.8 V

Chapter 5, Solution 87. The output, va, of the first op amp is,

Also,

va = (1 + (R2/R1))v1

(1)

vo = (-R4/R3)va + (1 + (R4/R3))v2

(2)

Substituting (1) into (2), vo = (-R4/R3) (1 + (R2/R1))v1 + (1 + (R4/R3))v2 Or, If

vo = (1 + (R4/R3))v2 – (R4/R3 + (R2R4/R1R3))v1 R4 = R1 and R3 = R2, then, vo = (1 + (R4/R3))(v2 – v1)

which is a subtractor with a gain of (1 + (R4/R3)). Chapter 5, Solution 88. We need to find VTh at terminals a – b, from this, vo = (R2/R1)(1 + 2(R3/R4))VTh = (500/25)(1 + 2(10/2))VTh = 220VTh Now we use Fig. (b) to find VTh in terms of vi. a a 30 kΩ

20 kΩ

20 kΩ vi

vi

30 kΩ

+− 40 kΩ

80 kΩ

40 kΩ

b

80 kΩ

b

(a)

(b) va = (3/5)vi, vb = (2/3)vi VTh = vb – va (1/15)vi (vo/vi) = Av = -220/15 = -14.667

Chapter 5, Solution 89. If we use an inverter, R = 2 k ohms, (vo/vi) = -R2/R1 = -6 R = 6R = 12 k ohms Hence the op amp circuit is as shown below. 12 kΩ 2 kΩ

vi

− +

+ −

+ vo −

Chapter 5, Solution 90. Transforming the current source to a voltage source produces the circuit below, At node b, vb = (2/(2 + 4))vo = vo/3 20 kΩ 5 kΩ a

5is

+ −

b

− + 4 kΩ io 2 kΩ

At node a, But va = vb = vo/3.

+ vo −

(5is – va)/5 = (va – vo)/20 20is – (4/3)vo = (1/3)vo – vo, or is = vo/30 io = [(2/(2 + 4))/2]vo = vo/6 io/is = (vo/6)/(vo/30) = 5

Chapter 5, Solution 91.

− +

vo R2

R1

is

i2 i1

But

io

io = i1 + i2

(1)

i1 = is

(2)

R1 and R2 have the same voltage, vo, across them. R1i1 = R2i2, which leads to i2 = (R1/R2)i1

(3)

Substituting (2) and (3) into (1) gives, io = is(1 + R1/R2) io/is = 1 + (R1/R2) = 1 + 8/1 = 9 Chapter 5, Solution 92 The top op amp circuit is a non-inverter, while the lower one is an inverter. The output at the top op amp is v1 = (1 + 60/30)vi = 3vi while the output of the lower op amp is v2 = -(50/20)vi = -2.5vi Hence,

vo = v1 – v2 = 3vi + 2.5vi = 5.5vi vo/vi = 5.5

Chapter 5, Solution 93. R3 R1 v a vb

− +

io

+

R4

vi

+

−

R2

vL

iL RL

+ vo −

−

At node a,

(vi – va)/R1 = (va – vo)/R3 vi – va = (R1/R2)(va – vo) vi + (R1/R3)vo = (1 + R1/R3)va

(1)

But va = vb = vL. Hence, (1) becomes vi = (1 + R1/R3)vL – (R1/R3)vo

(2)

io = vo/(R4 + R2||RL), iL = (RL/(R2 + RL))io = (R2/(R2 + RL))(vo/( R4 + R2||RL)) Or,

vo = iL[(R2 + RL)( R4 + R2||RL)/R2

(3)

But,

vL = iLRL

(4)

Substituting (3) and (4) into (2), vi = (1 + R1/R3) iLRL – R1[(R2 + RL)/(R2R3)]( R4 + R2||RL)iL = [((R3 + R1)/R3)RL – R1((R2 + RL)/(R2R3)(R4 + (R2RL/(R2 + RL))]iL = (1/A)iL

Thus, A =

1   R + RL R   1 + 1  R L − R 1  2 R3    R 2R 3

 R 2RL  R 4 + R2 + RL 

  

Chapter 6, Solution 1.

i=C

(

)

dv = 5 2e −3t − 6 + e −3 t = 10(1 - 3t)e-3t A dt

p = vi = 10(1-3t)e-3t ⋅ 2t e-3t = 20t(1 - 3t)e-6t W Chapter 6, Solution 2.

1 2 1 Cv1 = (40)(120) 2 2 2 1 1 w2 = Cv12 = (40)(80) 2 2 2 w1 =

∆w = w 1 − w 2 = 20(120 2 − 80 2 ) = 160 kW Chapter 6, Solution 3.

i=C

280 − 160 dv = 40x10 −3 = 480 mA dt 5

Chapter 6, Solution 4.

v= =

1 t idt + v(0) C ∫o

1 6 sin 4 tdt + 1 2∫ = 1 - 0.75 cos 4t

Chapter 6, Solution 5.

1 t idt + v(0) C ∫o For 0 < t < 1, i = 4t, t 1 v= 4t dt + 0 = 100t2 kV − 6 ∫o 20x10 v(1) = 100 kV v=

For 1 < t < 2, i = 8 - 4t, t 1 v= (8 − 4t )dt + v(1) 20x10 −6 ∫1 = 100 (4t - t2 - 3) + 100 kV 100t 2 kV, 0 < t <1 v (t) =  2 100(4t − t − 2)kV, 1 < t < 2

Thus

Chapter 6, Solution 6.

dv = 30x10 −6 x slope of the waveform. dt For example, for 0 < t < 2, i=C

10 dv = dt 2x10 −3 10 dv = 30x10 −6 x = 150mA i= C dt 2x10 −3 Thus the current i is sketched below. i(t) (mA) 150

4

8

2

6

t (msec) 10

-150

Chapter 6, Solution 7.

v=

1 1 idt + v( t o ) = ∫ C 50x10 −3 =

t

∫ 4tx10 o

2t 2 + 10 = 0.04k2 + 10 V 50

−3

dt + 10

12

Chapter 6, Solution 8.

(a) i = C

dv = −100 ACe −100t − 600 BCe −600t dt

i (0) = 2 = −100 AC − 600 BC

 →

(1) 5 = − A − 6B

v (0 + ) = v (0 − )  → 50 = A + B Solving (2) and (3) leads to A=61, B=-11

(b) Energy =

(2) (3)

1 2 1 Cv (0) = x 4 x10 −3 x 2500 = 5 J 2 2

(c ) From (1), i = −100 x61x 4 x10 −3 e −100t − 600 x11x 4 x10 −3 e −600t = − 24.4e −100t − 26.4e −600t A

Chapter 6, Solution 9.

v(t) =

(

)

(

v(2) = 12(2 + e-2) = 25.62 V p = iv = 12 (t + e-t) 6 (1-e-t) = 72(t-e-2t) p(2) = 72(2-e-4) = 142.68 W Chapter 6, Solution 10

i=C

)

1 t 6 1 − e − t dt + 0 = 12 t + e − t V ∫ o 12

dv dv = 2 x10 −3 dt dt

 16t , 0 < t < 1µs  v =  16, 1 < t < 3 µs 64 - 16t, 3 < t < 4 µs 

 16 x10 6 , 0 < t < 1µs dv  =  0, 1 < t < 3 µs dt  6 - 16x10 , 3 < t < 4 µs

 32 kA, 0 < t < 1µs  i (t ) =  0, 1 < t < 3 µs - 32 kA, 3 < t < 4µs 

Chapter 6, Solution 11.

1 t idt + v(0) C ∫o For 0 < t < 1,

v=

t 1 40 x10 −3 dt = 10t kV − 6 ∫o 4x10 v(1) = 10 kV

v=

For 1 < t < 2, v=

1 t vdt + v(1) = 10kV C ∫1

For 2 < t < 3, 1 4x10 −6 = -10t + 30kV v=

t

∫ (−40x10 2

−3

)dt + v(2)

Thus 0 < t <1 10 t ⋅ kV,  v(t) = 10kV, 1< t < 2 − 10 t + 30kV, 2 < t < 3

Chapter 6, Solution 12.

dv = 3x10 −3 x 60(4π)(− sin 4π t) dt = - 0.7e π sin 4πt A

i=C

P = vi = 60(-0.72)π cos 4π t sin 4π t = -21.6π sin 8π t W W=

∫

t

o

1

pdt = − ∫ 8 21.6π sin 8π t dt o

21.6π cos 8π = 8π

1/ 8 o

= -5.4J

Chapter 6, Solution 13. Under dc conditions, the circuit becomes that shown below: i1

10 Ω

50 Ω

i2

20 Ω

+

+

v1

30 Ω

v2

− 60V

−

+ −

i2 = 0, i1 = 60/(30+10+20) = 1A v1 = 30i2 = 30V, v2 = 60-20i1 = 40V Thus, v1 = 30V, v2 = 40V Chapter 6, Solution 14. (a) Ceq = 4C = 120 mF 1 4 4 = = C eq C 30

(b)

Ceq = 7.5 mF

Chapter 6, Solution 15. In parallel, as in Fig. (a), v1 = v2 = 100 +

+ 100V

+ −

v1

C1

−

+ v2

C2

100V

+ −

v1 − C1

v2

− (a)

+ −

(b)

C2

1 2 1 Cv = x 20x10 −6 x100 2 = 0.1J 2 2 1 w30 = x30x10 −6 x100 2 = 0.15J 2

w20 =

(b)

When they are connected in series as in Fig. (b): v1 =

w20 =

C2 30 x100 = 60, v2 = 40 V= 50 C1 + C 2

1 x30x10 −6 x 60 2 = 36 mJ 2

w30 =

1 x30x10 −6 x 40 2 = 24 mJ 2

Chapter 6, Solution 16

C eq = 14 +

Cx80 = 30 C + 80

 →

C = 20 µF

Chapter 6, Solution 17.

(a)

(b) (c)

4F in series with 12F = 4 x 12/(16) = 3F 3F in parallel with 6F and 3F = 3+6+3 = 12F 4F in series with 12F = 3F i.e. Ceq = 3F Ceq = 5 + [6 || (4 + 2)] = 5 + (6 || 6) = 5 + 3 = 8F 3F in series with 6F = (3 x 6)/9 = 6F 1 1 1 1 = + + =1 C eq 2 6 3 Ceq = 1F

Chapter 6, Solution 18.

For the capacitors in parallel C1eq = 15 + 5 + 40 = 60 µF Hence

1 1 1 1 1 = + + = C eq 20 30 60 10

Ceq = 10 µF Chapter 6, Solution 19.

We combine 10-, 20-, and 30- µ F capacitors in parallel to get 60 µ F. The 60 - µ F capacitor in series with another 60- µ F capacitor gives 30 µ F. 30 + 50 = 80 µ F, 80 + 40 = 120 µ F The circuit is reduced to that shown below. 12

120

12

80

120- µ F capacitor in series with 80 µ F gives (80x120)/200 = 48 48 + 12 = 60 60- µ F capacitor in series with 12 µ F gives (60x12)/72 = 10 µ F

Chapter 6, Solution 20.

3 in series with 6 = 6x3/(9) = 2 2 in parallel with 2 = 4 4 in series with 4 = (4x4)/8 = 2 The circuit is reduced to that shown below: 20

1

6

8

2

6 in parallel with 2 = 8 8 in series with 8 = 4 4 in parallel with 1 = 5 5 in series with 20 = (5x20)/25 = 4 Thus Ceq = 4 mF Chapter 6, Solution 21.

4µF in series with 12µF = (4x12)/16 = 3µF 3µF in parallel with 3µF = 6µF 6µF in series with 6µF = 3µF 3µF in parallel with 2µF = 5µF 5µF in series with 5µF = 2.5µF Hence Ceq = 2.5µF Chapter 6, Solution 22.

Combining the capacitors in parallel, we obtain the equivalent circuit shown below: a

b 40 µF

60 µF

30 µF

20 µF

Combining the capacitors in series gives C1eq , where 1 1 1 1 1 = + + = 1 C eq 60 20 30 10 Thus Ceq = 10 + 40 = 50 µF

C1eq = 10µF

Chapter 6, Solution 23. (a)

(b)

3µF is in series with 6µF v4µF = 1/2 x 120 = 60V v2µF = 60V 3 v6µF = (60) = 20V 6+3 v3µF = 60 - 20 = 40V

3x6/(9) = 2µF

Hence w = 1/2 Cv2 w4µF = 1/2 x 4 x 10-6 x 3600 = 7.2mJ w2µF = 1/2 x 2 x 10-6 x 3600 = 3.6mJ w6µF = 1/2 x 6 x 10-6 x 400 = 1.2mJ w3µF = 1/2 x 3 x 10-6 x 1600 = 2.4mJ

Chapter 6, Solution 24.

20µF is series with 80µF = 20x80/(100) = 16µF 14µF is parallel with 16µF = 30µF (a) v30µF = 90V v60µF = 30V v14µF = 60V 80 v20µF = x 60 = 48V 20 + 80 v80µF = 60 - 48 = 12V

(b)

1 2 Cv 2 w30µF = 1/2 x 30 x 10-6 x 8100 = 121.5mJ w60µF = 1/2 x 60 x 10-6 x 900 = 27mJ w14µF = 1/2 x 14 x 10-6 x 3600 = 25.2mJ w20µF = 1/2 x 20 x 10-6 x (48)2 = 23.04mJ w80µF = 1/2 x 80 x 10-6 x 144 = 5.76mJ Since w =

Chapter 6, Solution 25.

(a) For the capacitors in series, Q1 = Q2

C1v1 = C2v2

v1 C 2 = v 2 C1

vs = v1 + v2 =

Similarly, v1 =

C + C2 C2 v2 v2 + v2 = 1 C1 C1

v2 =

C2 vs C1 + C 2

(b) For capacitors in parallel Q1 Q 2 = C1 C 2 C C + C2 Q2 Qs = Q1 + Q2 = 1 Q 2 + Q 2 = 1 C2 C2

v1 = v2 =

or C2 C1 + C 2 C1 Q1 = Qs C1 + C 2

Q2 =

i=

dQ dt

i1 =

C1 is , C1 + C 2

i2 =

C2 is C1 + C 2

Chapter 6, Solution 26.

(a)

Ceq = C1 + C2 + C3 = 35µF

(b)

Q1 = C1v = 5 x 150µC = 0.75mC Q2 = C2v = 10 x 150µC = 1.5mC Q3 = C3v = 20 x 150 = 3mC

(c)

w=

1 1 C eq v 2 = x35x150 2 µJ = 393.8mJ 2 2

C1 vs C1 + C 2

Chapter 6, Solution 27. 1 1 1 1 1 1 1 7 = + + = + + = C eq C1 C 2 C 3 5 10 20 20

(a) Ceq = (b)

(c)

20 µF = 2.857µF 7

Since the capacitors are in series, 20 Q1 = Q2 = Q3 = Q = Ceqv = x 200µV = 0.5714mV 7 1 1 20 w = C eq v 2 = x x 200 2 µJ = 57.143mJ 2 2 7

Chapter 6, Solution 28.

We may treat this like a resistive circuit and apply delta-wye transformation, except that R is replaced by 1/C. Cb

50 µF

Cc

20 µF

Ca

 1  1   1  1   1  1     +    +    1 10 40 10 30 30 40 =          1 Ca 30 3 1 1 2 = + + = 40 10 40 10

Ca = 5µF 1 1 1 + + 2 1 = 400 300 1200 = 1 30 C6 10 Cb = 15µF

1 1 1 + + 1 4 = 400 300 1200 = 1 Cc 15 40 Cc = 3.75µF

Cb in parallel with 50µF = 50 + 15 = 65µF Cc in series with 20µF = 23.75µF 65x 23.75 = 17.39µF 65µF in series with 23.75µF = 88.75 17.39µF in parallel with Ca = 17.39 + 5 = 22.39µF Hence Ceq = 22.39µF Chapter 6, Solution 29.

(a)

C in series with C = C/(2) C/2 in parallel with C = 3C/2 3C in series with C = 2

3

3C 2 = 3C C 5 5 2

Cx

C C in parallel with C = C + 3 = 1.6 C 5 5

(b) 2C

Ceq 2C

1 1 1 1 = + = C eq 2C 2C C

Ceq = C

Chapter 6, Solution 30.

1 t idt + i(0) C ∫o For 0 < t < 1, i = 60t mA, 10 −3 t vo = 60tdt + 0 = 10 t 2 kV 3x10 −6 ∫o vo(1) = 10kV vo =

For 1< t < 2, i = 120 - 60t mA, 10 −3 t vo = (120 − 60t )dt + v o (1) 3x10t −6 ∫1 = [40t – 10t2 ] 1 + 10kV = 40t – 10t2 - 20 10t 2 kV, 0 < t <1 v o (t) =  2 40t − 10 t − 20kV, 1 < t < 2

Chapter 6, Solution 31. 20 tmA, i s ( t ) = 20mA, − 50 + 10 t ,

0 < t <1 1< t < 3 3< t <5

Ceq = 4 + 6 = 10µF 1 t v= idt + v(0) C eq ∫o For 0 < t < 1, v=

10 −3 10x10 −6

For 1 < t < 3, 10 3 v= 10 = 2 t − 1kV For 3 < t < 5, 10 3 v= 10

t

∫ 20t dt + 0 = t o

t

2

kV

∫ 20dt + v(1) = 2(t − 1) + 1kV 1

t

∫ 10(t − 5)dt + v(3) 3

= t 2 − 5 + 3t +5kV = t 2 − 5t + 11kV

 t 2 kV, 0 < t <1  v( t ) = 2t − 1kV, 1< t < 3 2  t − 5t + 11kV, 3 < t < 5  dv dv = 6x10 −6 dt dt 0 < t <1 12 tmA,  = 12mA, 1< t < 3 12 − 30mA, 3 < t < 5

i 1 = C1

dv dv = 4x10 −6 dt dt 0 < t <1 8tmA,  = 8mA, 1< t < 3 8t − 20mA, 3 < t < 5

i1 = C 2

Chapter 6, Solution 32.

(a)

Ceq = (12x60)/72 = 10 µ F t

10 −3 v1 = 30e − 2t dt + v1 (0) = − 1250e − 2t −6 ∫ 12 x10 0 t

10 −3 v2 = 30e − 2t dt + v 2 (0) = 250e − 2t −6 ∫ 60 x10 0

(b)

t 0

t 0

+ 50 = − 1250e − 2t + 1300

+ 20 = 250e − 2t − 230

At t=0.5s, v1 = −1250e −1 + 1300 = 840.15,

w12 µF =

1 x12 x10 −6 x(840.15) 2 = 4.235 J 2

1 x 20 x10 −6 x(−138.03) 2 = 0.1905 J 2 1 = x 40 x10 −6 x(−138.03) 2 = 0.381 J 2

w20 µF = w40 µF

v 2 = 250e −1 − 230 = −138.03

Chapter 6, Solution 33 Because this is a totally capacitive circuit, we can combine all the capacitors using the property that capacitors in parallel can be combined by just adding their values and we combine capacitors in series by adding their reciprocals. 3F + 2F = 5F 1/5 +1/5 = 2/5 or 2.5F The voltage will divide equally across the two 5F capacitors. Therefore, we get: VTh = 7.5 V, CTh = 2.5 F Chapter 6, Solution 34. i = 6e-t/2 di 1 v = L = 10 x10 −3 (6) e − t / 2 dt 2 -t/2 = -30e mV

v(3) = -300e-3/2 mV = -0.9487 mV p = vi = -180e-t mW p(3) = -180e-3 mW = -0.8 mW Chapter 6, Solution 35.

v=L

di dt

L=

V 60 x10 −3 = = 200 mH ∆i / ∆t 0.6 /(2)

Chapter 6, Solution 36.

di 1 = x10 −3 (12)(2)(− sin 2 t )V dt 4 = - 6 sin 2t mV

v=L

p = vi = -72 sin 2t cos 2t mW But 2 sin A cos A = sin 2A p = -36 sin 4t mW Chapter 6, Solution 37.

di = 12 x10 −3 x 4(100) cos100t dt = 4.8 cos 100t V

v=L

p = vi = 4.8 x 4 sin 100t cos 100t = 9.6 sin 200t w=

t

11 / 200

o

o

∫ pdt = ∫

9.6 sin 200 t

9.6 / 200 cos 200t 11 J o 200 = −48(cos π − 1)mJ = 96 mJ =−

Chapter 6, Solution 38.

v=L

di = 40x10 −3 (e − 2 t − 2te − 2 t )dt dt = 40(1 − 2t )e −2 t mV, t > 0

Chapter 6, Solution 39

di 1  → i = ∫ 0t idt + i(0) L dt

v=L

i=

1 200x10

t (3t 2 −3 ∫ 0

= 5( t 3 + t 2 + 4t )

t 0

+ 2t + 4)dt + 1

+1

i(t) = 5t3 + 5t2 + 20t + 1 A Chapter 6, Solution 40

v=L

di di = 20 x10 −3 dt dt

 10t , 0 < t < 1 ms  i =  20 - 10t, 1 < t < 3 ms - 40 + 10t, 3 < t < 4 ms 

 10 x10 3 , di  = - 10x10 3 , dt  3 10x10 ,  200 V,  v = - 200 V,  200 V, 

0 < t < 1 ms 1 < t < 3 ms 3 < t < 4 ms 0 < t < 1 ms 1 < t < 3 ms 3 < t < 4 ms

which is sketched below. v(t) V 200 0 -200

1

2

3

4

t(ms)

Chapter 6, Solution 41.

i=

(

)

1 t 1 t vdt + i(0) =   ∫ 20 1 − 2 − 2 t dt + 0.3 ∫ L 0 2 o

 1  = 10 t + e − 2t  to +0. 3 = 10t + 5e − 2t − 4. 7 A  2  At t = ls, i = 10 - 4.7 + 5e-2 = 5.977 A w=

1 2 L i = 35.72J 2

Chapter 6, Solution 42.

1 t 1 t vdt i ( 0 ) + = v( t )dt − 1 L ∫o 5 ∫o 10 t For 0 < t < 1, i = ∫ dt − 1 = 2t − 1 A 5 0 i=

For 1 < t < 2, i = 0 + i(1) = 1A 1 10dt + i(2) = 2t 2t +1 ∫ 5 = 2t - 3 A

For 2 < t < 3, i =

For 3 < t < 4, i = 0 + i(3) = 3 A 1 t 10dt + i(4) = 2 t 4t +3 5 ∫4 = 2t - 5 A

For 4 < t < 5, i =

 2t − 1A,  1A, Thus, i (t ) =  2t − 3 A,  3 A,  2t − 5,

0 < t <1 1< t < 2 2
Chapter 6, Solution 43. 2

1 1 w = L ∫ idt = Li( t ) − Li (−∞) −∞ 2 2 1 = x80 x10 −3 x 60x10 −3 − 0 2 = 144 µJ t

(

)

Chapter 6, Solution 44.

i=

1 t 1 t vdt + i(t o ) = ∫ (4 + 10 cos 2t )dt − 1 ∫ L to 5 o = 0.8t + sin 2t -1

Chapter 6, Solution 45.

i(t) =

1 t v( t ) + i(0) L ∫o

For 0 < t < 1, v = 5t i=

1 10x10 −3

t

∫ 5t dt + 0 o

= 0.25t2 kA For 1 < t < 2, v = -10 + 5t i=

1 10x10 −3

t

∫ (−10 + 5t )dt + i(1) 1

t

= ∫ (0.5t − 1)dt + 0.25kA 1

= 1 - t + 0.25t2 kA

0.25t 2 kA, 0 < t <1 i( t ) =  2 1 − t + 0.25t kA, 1 < t < 2

Chapter 6, Solution 46.

Under dc conditions, the circuit is as shown below: 2Ω iL + 3A

vC

4Ω

−

By current division, iL =

4 (3) = 2A, vc = 0V 4+2

wL =

1 2 11 2 L i L =  (2) = 1J 2 22

wc =

1 1 C v c2 = (2)( v) = 0J 2 2

Chapter 6, Solution 47.

Under dc conditions, the circuit is equivalent to that shown below: R

+

5A

iL =

2Ω

2 10 10R (5) = , v c = Ri L = R+2 R+2 R+2

vC

−

iL

1 2 100R 2 Cv c = 80x10 −6 x 2 (R + 2) 2 1 100 w L = Li12 = 2x10 −3 x 2 (R + 2) 2 If wc = wL, wc =

100R 2 2x10 −3 x100 80x10 x = (Rx 2) 2 (R + 2) 2

80 x 10-3R2 = 2

−6

R = 5Ω Chapter 6, Solution 48.

Under dc conditions, the circuit is as shown below: 4Ω

iL1 + +

30V

+ −

vC1

−

i L1 = i L 2 =

30 = 3A 4+6

v C1 = 6i L1 = 18V v C 2 = 0V

iL2

vC2

− 6Ω

Chapter 6, Solution 49.

(a)

L eq = 5 + 6 (1 + 4 4) = 5 + 6 3 = 7H

(b)

L eq = 12 (1 + 6 6) = 12 4 = 3H

(c)

L eq = 4 (2 + 3 6) = 4 4 = 2H

Chapter 6, Solution 50.

(

L eq = 10 + 5 4 12 + 3 6

)

= 10 + 5||(3 + 2) = 10 + 2.5 = 12.5 mH Chapter 6, Solution 51.

1 1 1 1 1 = + + = L 60 20 30 10 L eq = 10 (25 + 10) =

L = 10 mH

10x35 45

= 7.778 mH Chapter 6, Solution 52.

3//2//6 = 1H, 4//12 = 3H After the parallel combinations, the circuit becomes that shown below. 3H a 1H

1H Lab = (3+1)//1 = (4x1)/5 = 0.8 H

b

Chapter 6, Solution 53. L eq = 6 + 10 + 8 [5 (8 + 12) + 6 (8 + 4)] = 16 + 8 (4 + 4) = 16 + 4

Leq = 20 mH Chapter 6, Solution 54.

L eq = 4 + (9 + 3) (10 0 + 6 12) = 4 + 12 (0 + 4) = 4 + 3

Leq = 7H Chapter 6, Solution 55.

(a) L//L = 0.5L, L + L = 2L Leq = L + 2 L // 0.5L = L +

2 Lx0.5 L = 1.4 L 2 L + 0.5L

(b) L//L = 0.5L, L//L + L//L = L Leq = L//L = 0.5L Chapter 6, Solution 56.

1 L = 3 3 L Hence the given circuit is equivalent to that shown below: LLL=

L L/3

L/3 L

L eq

5 Lx L 2   3 = 5L = L L + L = 5 8 3   L+ L 3

Chapter 6, Solution 57.

Let v = L eq

di dt

di + v2 dt i2 = i – i1 i = i1 + i2 di di v v 2 = 3 1 or 1 = 2 dt dt 3 and di di − v2 + 2 + 5 2 = 0 dt dt di di v2 = 2 + 5 2 dt dt Incorporating (3) and (4) into (5), di v di di di v2 = 2 + 5 − 5 1 = 7 − 5 2 dt dt dt dt 3 di  5 v 2 1 +  = 7 dt  3 35 di v2 = 8 dt v = v1 + v 2 = 4

Substituting this into (2) gives v=4

di 35 di + dt 8 dt =

67 di 8 dt

Comparing this with (1), L eq =

67 = 8.375H 8

(1)

(2) (3) (4)

(5)

Chapter 6, Solution 58.

v=L

di di = 3 = 3 x slope of i(t). dt dt

Thus v is sketched below: v(t) (V) 6

t (s) 1

2

3

4

5

6

7

-6

Chapter 6, Solution 59.

(a) v s = (L1 + L 2 )

di dt

vs di = dt L1 + L 2 di di v 1 = L1 , v 2 = L 2 dt dt L1 L2 v1 = vs , vL = vs L1 + L 2 L1 + L 2

(b)

v i = v 2 = L1

di1 di = L2 2 dt dt

i s = i1 + i 2 di s di1 di 2 (L + L 2 ) v v = + = + =v 1 L1 L 2 dt dt dt L1 L 2 L1 L 2 di s L2 1 1 dt = is vdt = i1 = ∫ ∫ L1 L1 + L 2 dt L1 + L 2 L1

i2 =

1 1 vdt = ∫ L2 L2

L1 L 2 di s L1 dt = is + L dt L + L 1 2 1 2

∫L

Chapter 6, Solution 60

Leq = 3 // 5 = vo = Leq

(

15 8

)

di 15 d = 4e − 2t = − 15e − 2t dt 8 dt

t

io =

t

I 1 vo (t )dt + io (0) = 2 + ∫ (−15)e − 2t = 2 + 1.5e − 2t ∫ L0 50

t

= 0.5 + 1.5e − 2t A

0

Chapter 6, Solution 61.

(a)

is = i1 + i2 i s (0) = i1 (0) + i 2 (0) 6 = 4 + i 2 ( 0) i2(0) = 2mA (b) Using current division: 20 i1 = i s = 0.4 6e − 2 t = 2.4e-2t mA 30 + 20 i 2 = i s − i1 = 3.6e-2t mA 30 x 20 = 12mH (c) 30 20 = 50 di d v1 = L = 10x10 −3 6e − 2 t x10 −3 = -120e-2t µV dt dt di d v 2 = L = 12x10 −3 6e −2 t x10 −3 = -144e-2t µV dt dt

(

(d)

w 10 mH = = 0.8e − 4 t

)

(

)

(

)

(

1 x30x10 −3 36e − 4 t x10 −6 2 t=

1 2

)

µJ

= 24.36nJ 1 w 30 mH = x30 x10 −3 5.76e − 4 t x10 −6 t =1 / 2 2 = 11.693nJ 1 w 20 mH = x 20x10 −3 12.96e − 4 t x10 −6 t =1 / 2 2 = 17.54 nJ

(

(

)

)

Chapter 6, Solution 62.

(a)

Leq = 25 + 20 // 60 = 25 +

v = Leq

di dt

20 x60 = 40 mH 80 t

 →

i=

1 10 −3 ( ) ( 0 ) v t dt + i = 12e −3t dt + i (0) = −0.1(e −3t − 1) + i (0) Leq ∫ 40 x10 −3 ∫0

Using current division, 60 3 1 i1 = i = i, i 2 = i 80 4 4 3 i1 (0) = i (0)  → 0.75i (0) = −0.01 4

 →

1 (−0.1e −3t + 0.08667) A = - 25e -3t + 21.67 mA 4 i2 (0) = −25 + 21.67 = − 3.33 mA i2 =

3 (−0.1e −3t + 0.08667) A = - 75e -3t + 65 mA 4 i2 = - 25e -3t + 21.67 mA (b) i1 =

Chapter 6, Solution 63.

We apply superposition principle and let vo = v1 + v 2

where v1 and v2 are due to i1 and i2 respectively. v1 = L

di1 di  2, =2 1 = dt dt − 2,

 4, di2 di2  v2 = L =2 =  0, dt dt  − 4,

0
i (0) = −0.01333

v1

v2

2

4

0 3

6

t

0

-2

2

4

6

-4

Adding v1 and v2 gives vo, which is shown below. vo(t) V 6

2 0 2 3

4

6

t (s)

-2

-6

Chapter 6, Solution 64.

(a) When the switch is in position A, i=-6 =i(0) When the switch is in position B, i (∞) = 12 / 4 = 3, τ = L / R = 1/ 8 i (t ) = i (∞) + [i (0) − i (∞)]e − t / ι = 3 − 9e −8t A

(b) -12 + 4i(0) + v=0, i.e. v=12 – 4i(0) = 36 V (c) At steady state, the inductor becomes a short circuit so that v= 0 V

t

Chapter 6, Solution 65.

1 1 L1i12 = x5x (4) 2 = 40 W 2 2 1 w 20 = (20)(−2) 2 = 40 W 2 (b) w = w5 + w20 = 80 W dv (c) i1 = L1 = 5(− 200)(50e − 200 t x10 −3 ) dt = -50e-200tA (a)

w5 =

i2 = L2

dv = 20(−200)(50e − 200 t x10 −3 ) dt = -200e-200tA

(

dv = 20(−200) 50e − 200 t x10 −3 dt = -200e-200t A i2 = L2

(d)

)

i = i1 + i2 = -250e-200t A

Chapter 6, Solution 66. L eq = 20 + 16 + 60 40 = 36 + 24 = 60mH

v=L

di dt

1 t vdt + i(0) L ∫o t 1 = 12 sin 4t dt + 0 mA −3 ∫o 60x10 i = −50 cos 4t ot = 50(1 - cos 4t) mA i=

60 40 = 24mH

d di = 24x10 −3 (50)(1 − cos 4t )mV dt dt = 4.8 sin 4t mV v=L

Chapter 6, Solution 67.

1 vi dt, RC = 50 x 103 x 0.04 x 10-6 = 2 x 10-3 RC ∫ − 10 3 vo = 10 sin 50t dt 2 ∫ vo = 100 cos 50t mV vo = −

Chapter 6, Solution 68.

1 vi dt + v(0), RC = 50 x 103 x 100 x 10-6 = 5 ∫ RC 1 t vo = − ∫ 10dt + 0 = −2t 5 o The op amp will saturate at vo = ± 12 vo = −

-12 = -2t

t = 6s

Chapter 6, Solution 69.

RC = 4 x 106 x 1 x 10-6 = 4 vo = −

1 1 v i dt = − ∫ v i dt ∫ RC 4

For 0 < t < 1, vi = 20, v o = −

1 t 20dt = -5t mV 4 ∫o

1 t 10dt + v(1) = −2.5( t − 1) − 5 4 ∫1 = -2.5t - 2.5mV

For 1 < t < 2, vi = 10, v o = −

1 t 20dt + v(2) = 5( t − 2) − 7.5 4 ∫2 = 5t - 17.5 mV

For 2 < t < 4, vi = - 20, v o = +

1 t 10dt + v(4) = 2.5( t − 4) + 2.5 4 ∫4 = 2.5t - 7.5 mV

For 4 < t < 5m, vi = -10, v o =

1 t 20dt + v(5) = −5( t − 5) + 5 4 ∫5 = - 5t + 30 mV

For 5 < t < 6, vi = 20, v o = −

Thus vo(t) is as shown below:

5 25 0

1

2

3

4

5

6

7

5

Chapter 6, Solution 70.

One possibility is as follows: 1 = 50 RC 1 Let R = 100 kΩ, C = = 0.2µF 50 x100 x10 3 Chapter 6, Solution 71.

By combining a summer with an integrator, we have the circuit below:

− + 1 1 1 v1dt − v 2 dt − v 2 dt ∫ ∫ R 1C R 2C R 2C ∫ For the given problem, C = 2µF, vo = −

R1C = 1 R2C = 1/(4) R3C = 1/(10)

R1 = 1/(C) = 1006/(2) = 500 kΩ R2 = 1/(4C) = 500kΩ/(4) = 125 kΩ R3 = 1/(10C) = 50 kΩ

Chapter 6, Solution 72. The output of the first op amp is

v1 = −

1 1 v i dt = − 3 ∫ RC 10x10 x 2 x10 −6

t

∫ idt = − o

100 t 2

= - 50t

vo = −

1 1 v i dt = − 3 ∫ RC 20x10 x 0.5x10 −6

t

∫ (−50t )dt o

= 2500t2 At t = 1.5ms, v o = 2500(1.5) 2 x10 −6 = 5.625 mV Chapter 6, Solution 73.

Consider the op amp as shown below: Let va = vb = v At node a,

0 − v v − vo = R R

2v - vo = 0

(1) R

R a R

v

− + R

v

vo

b vi

At node b,

+ −

vi − v v − vo dv = +C R R dt

+

C

−

v i = 2v − v o + RC

dv dt

(2)

Combining (1) and (2), v i =v o −v o +

RC dv o 2 dt

or vo =

2 v i dt RC ∫

showing that the circuit is a noninverting integrator. Chapter 6, Solution 74.

RC = 0.01 x 20 x 10-3 sec v o = − RC

dv i dv = −0.2 m sec dt dt

 − 2V, v o = 2V, − 2V,

0 < t <1 1< t < 3 3< t < 4

Thus vo(t) is as sketched below: vo(t) (V) 2

t (ms) 1

-2

2

3

Chapter 6, Solution 75.

v 0 = − RC

dv i , RC = 250 x10 3 x10x10 −6 = 2.5 dt

v o = −2.5

d (12t ) = -30 mV dt

Chapter 6, Solution 76. dv i , RC = 50 x 103 x 10 x 10-6 = 0.5 dt − 10, 0 < t < 5 dv v o = 0.5 i =  5
The input is sketched in Fig. (a), while the output is sketched in Fig. (b). vo(t) (V)

vi(t) (V) 5

5

t (ms) 0

5

10

t (ms)

15

0

5

10

(a)

-10 (b) Chapter 6, Solution 77.

i = iR + i C vi − 0 0 − v0 d = + C (0 − v o ) dt R RF

R F C = 10 6 x10 −6 = 1

15

dv   Hence v i = − v o + o  dt   Thus vi is obtained from vo as shown below:

–dvo(t)/dt

– vo(t) (V)

4

4

t (ms)

t (ms) 0

1

2

3

0

4

1

2

3

4

-4

-4

vi(t) (V) 8

t (ms) -4

1

2

3

4

-8

Chapter 6, Solution 78.

d 2 vo 2dv o = 10 sin 2 t − − vo dt dt Thus, by combining integrators with a summer, we obtain the appropriate analog computer as shown below:

2vo t=0

− +

C

C

R

R

− +

d2vo/dt 2

R R

− +

-dvo/dt

− +

vo R

d2vo/dt 2

R R/2

− +

dvo/dt

R R + −

sin2t

R/10

− +

-sin2t

Chapter 6, Solution 79. We can write the equation as dy = f (t ) − 4 y (t ) dt which is implemented by the circuit below.

1V

t=0 C

R

R

R R/4 dy/dt

+

+

-y R f(t)

R + dy/dt

Chapter 6, Solution 80.

From the given circuit, d 2 vo 1000kΩ 1000kΩ dv o = f (t) − vo − 2 5000kΩ 200kΩ dt dt or d 2 vo dv + 5 o + 2v o = f ( t ) 2 dt dt

Chapter 6, Solution 81

We can write the equation as d 2v = −5v − 2 f (t ) dt 2

which is implemented by the circuit below. C

C R

R 2

2

d v/dt

+

R

R/5 -

-dv/dt

+

v

+ R/2

f(t)

d2v/dt2

Chapter 6, Solution 82

The circuit consists of a summer, an inverter, and an integrator. Such circuit is shown below. 10R R

R R

+

+

vo

R C=1/(2R) R +

+ vs -

Chapter 6, Solution 83.

Since two 10µF capacitors in series gives 5µF, rated at 600V, it requires 8 groups in parallel with each group consisting of two capacitors in series, as shown below:

+ 600

− Answer: 8 groups in parallel with each group made up of 2 capacitors in series.

Chapter 6, Solution 84.

∆I =

∆q ∆t

∆I x ∆t = ∆q

∆q = 0.6 x 4 x 10-6 = 2.4µC ∆q 2.4 x10−6 = = 150nF C= ∆v (36 − 20)

Chapter 6, Solution 85. It is evident that differentiating i will give a waveform similar to v. Hence, di v=L dt 4 t ,0 < t < 1 i= 8 − 4 t ,1 < t < 2 v=L

But,

di 4L,0 < t < 1 = dt − 4L,1 < t < 2

5mV,0 < t < 1 v= − 5mV,1 < t < 2

Thus, 4L = 5 x 10-3

L = 1.25 mH in a 1.25 mH inductor

Chapter 6, Solution 86. (a) For the series-connected capacitor Cs =

1 1 1 1 + + .... + C C C

=

C 8

For the parallel-connected strings, C eq = 10C s =

10C s 1000 = 10 x µF = 1250µF 3 8

(b)

vT = 8 x 100V = 800V w=

(

)

1 1 C eq v T2 = 1250 x10 −6 (800) 2 2 2

= 400J

Chapter 7, Solution 1. Applying KVL to Fig. 7.1. 1 t ∫ i dt + Ri = 0 C -∞ Taking the derivative of each term, i di +R =0 C dt di dt or =− i RC

Integrating,  i( t )  - t = ln  I 0  RC i( t ) = I 0 e - t RC v( t ) = Ri( t ) = RI 0 e - t RC or

v(t ) = V0e- t RC

Chapter 7, Solution 2. τ = R th C where R th is the Thevenin equivalent at the capacitor terminals.

R th = 120 || 80 + 12 = 60 Ω

τ = 60 × 0.5 × 10 -3 = 30 ms Chapter 7, Solution 3.

(a) RTh = 10 // 10 = 5kΩ,

τ = RTh C = 5 x10 3 x 2 x10 −6 = 10 ms

(b) RTh = 20 //(5 + 25) + 8 = 20Ω,

τ = RTh C = 20 x0.3 = 6s

Chapter 7, Solution 4.

τ = R eq C eq where C eq =

C1C 2 , C1 + C 2

τ=

R eq =

R 1R 2 R1 + R 2

R 1 R 2 C1C 2 ( R 1 + R 2 )(C 1 + C 2 )

Chapter 7, Solution 5.

v( t ) = v(4) e -(t -4) τ where v(4) = 24 , τ = RC = (20)(0.1) = 2 -(t - 4) 2 v( t ) = 24 e v(10) = 24 e -6 2 = 1.195 V

Chapter 7, Solution 6.

v o = v ( 0) =

2 (24) = 4 V 10 + 2

v( t ) = voe − t / τ , τ = RC = 40 x10−6 x 2 x103 =

2 25

v( t ) = 4e −12.5t V

Chapter 7, Solution 7.

v( t ) = v(0) e - t τ , τ = R th C where R th is the Thevenin resistance across the capacitor. To determine R th , we insert a 1-V voltage source in place of the capacitor as shown below. 8Ω

i2

i i1

0.5 V

+ −

10 Ω

+ v=1

− i1 =

1 = 0.1 , 10

i = i1 + i 2 = 0.1 +

i2 = 1 13 = 16 80

1 80 R th = = i 13 80 8 τ = R th C = × 0.1 = 13 13 -13t 8 v( t ) = 20 e V

1 − 0.5 1 = 8 16

+ −

1V

Chapter 7, Solution 8.

(a)

τ = RC =

1 4

dv dt -4t - 0.2 e = C (10)(-4) e-4t -i = C

 → C = 5 mF

1 = 50 Ω 4C 1 τ = RC = = 0.25 s 4 1 1 w C (0) = CV02 = (5 × 10 -3 )(100) = 250 mJ 2 2 1 1 1 w R = × CV02 = CV02 (1 − e -2t 0 τ ) 2 2 2 1 → e -8t 0 = 0.5 = 1 − e -8t 0  2 8t 0 or e =2 1 t 0 = ln (2) = 86.6 ms 8 R=

(b) (c) (d)

Chapter 7, Solution 9.

v( t ) = v(0) e- t τ ,

τ = R eq C

R eq = 2 + 8 || 8 + 6 || 3 = 2 + 4 + 2 = 8 Ω τ = R eq C = (0.25)(8) = 2 v( t ) = 20 e -t 2 V

Chapter 7, Solution 10. io

15 Ω

i

10 Ω iT

+ 10 mF

4Ω

v

− (10)(3) =2A 15 i.e. if i(0) = 3 A , then i o (0) = 2 A i T (0) = i(0) + i o (0) = 5 A v(0) = 10i(0) + 4i T (0) = 30 + 20 = 50 V across the capacitor terminals. 15 i o = 10 i  → i o =

R th = 4 + 10 || 15 = 4 + 6 = 10 Ω τ = R th C = (10)(10 × 10 -3 ) = 0.1 v( t ) = v(0) e - t τ = 50 e -10t dv iC = C = (10 × 10 -3 )(-500 e -10t ) dt i C = - 5 e -10t A By applying the current division principle, 15 i( t ) = ( - i ) = -0.6 i C = 3 e -10t A 10 + 15 C Chapter 7, Solution 11.

Applying KCL to the RL circuit, 1 v v dt + = 0 ∫ L R Differentiating both sides, v 1 dv + =0  → L R dt v = A e -Rt L

dv R + v=0 dt L

If the initial current is I 0 , then v(0) = I 0 R = A v = I 0 R e -t τ ,

τ=

L R

1 t ∫ v(t ) dt L -∞ - τ I 0 R -t τ t i= e -∞ L i = - I 0 R e -t τ i=

i( t ) = I 0 e - t τ

Chapter 7, Solution 12.

When t < 0, the switch is closed and the inductor acts like a short circuit to dc. The 4 Ω resistor is short-circuited so that the resulting circuit is as shown in Fig. (a). 3Ω

12 V

i(0-)

+ −

4Ω

(a)

2H

(b)

12 =4A 3 Since the current through an inductor cannot change abruptly, i(0) = i(0 − ) = i(0 + ) = 4 A i (0 − ) =

When t > 0, the voltage source is cut off and we have the RL circuit in Fig. (b). L 2 τ = = = 0.5 R 4 Hence, i( t ) = i(0) e - t τ = 4 e -2t A

Chapter 7, Solution 13.

L R th where R th is the Thevenin resistance at the terminals of the inductor. τ=

R th = 70 || 30 + 80 || 20 = 21 + 16 = 37 Ω 2 × 10 -3 τ= = 81.08 µs 37

Chapter 7, Solution 14

Converting the wye-subnetwork to delta gives 16 Ω

R2 80mH

R1 R3

30 Ω

R1 =

10 x 20 + 20 x50 + 50 x10 = 1700 / 20 = 85Ω, 20

R2 =

1700 = 34Ω , 50

R3 =

1700 = 170Ω 10

30//170 = (30x170)/200 = 25.5 Ω , 34//16=(34x16)/50 =10.88 Ω RTh = 85 //( 25.5 + 10.88) =

85 x36.38 = 25.476Ω, 121.38

τ=

80 x10 −3 L = = 3.14 ms RTh 25.476

Chapter 7, Solution 15

(a) RTh = 12 + 10 // 40 = 20Ω, (b) RTh = 40 // 160 + 8 = 40Ω,

L = 5 / 20 = 0.25s RTh L τ= = (20 x10 −3 ) / 40 = 0.5 ms RTh

τ=

Chapter 7, Solution 16.

τ= (a)

L eq R eq

L eq = L and R eq = R 2 +

τ=

(b)

R 1R 3 R 2 (R 1 + R 3 ) + R 1 R 3 = R1 + R 3 R1 + R 3

L( R 1 + R 3 ) R 2 (R 1 + R 3 ) + R 1 R 3 R 3 (R 1 + R 2 ) + R 1 R 2 L1 L 2 R 1R 2 = and R eq = R 3 + L1 + L 2 R1 + R 2 R1 + R 2 L1L 2 (R 1 + R 2 ) τ= (L 1 + L 2 ) ( R 3 ( R 1 + R 2 ) + R 1 R 2 )

where L eq =

Chapter 7, Solution 17.

i( t ) = i(0) e - t τ ,

τ=

14 1 L = = R eq 4 16

i( t ) = 2 e -16t v o ( t ) = 3i + L

di = 6 e-16t + (1 4)(-16) 2 e-16t dt

v o ( t ) = - 2 e -16t V

Chapter 7, Solution 18.

If v( t ) = 0 , the circuit can be redrawn as shown below. +

0.4 H

Req

vo(t)

−

i(t)

6 L 2 5 1 τ= = × = , 5 R 5 6 3 -t τ -3t i( t ) = i(0) e = e di - 2 v o ( t ) = -L = (-3) e -3t = 1.2 e -3t V dt 5 R eq = 2 || 3 =

Chapter 7, Solution 19. i

1V − +

10 Ω

i1

i1

i2

i/2

i2

40 Ω

To find R th we replace the inductor by a 1-V voltage source as shown above. 10 i1 − 1 + 40 i 2 = 0 i = i2 + i 2 and i = i1 But i1 = 2 i 2 = 2 i i.e. 1 10 i − 1 + 20 i = 0  → i = 30 1 R th = = 30 Ω i L 6 τ= = = 0.2 s R th 30 i( t ) = 2 e -5t A

Chapter 7, Solution 20.

(a).

L 1 =  → R = 50L R 50 di -v= L dt -50t - 150 e = L(30)(-50) e -50t  → L = 0.1 H τ=

R = 50L = 5 Ω

(b). (c). (d).

i.e.

L 1 = = 20 ms R 50 1 1 w = L i 2 (0) = (0.1)(30) 2 = 45 J 2 2 Let p be the fraction 1 1 L I 0 ⋅ p = L I 0 ( 1 − e -2t 0 τ ) 2 2 -(2)(10) 50 p = 1− e = 1 − e -0.4 = 0.3296 p = 33% τ=

Chapter 7, Solution 21.

The circuit can be replaced by its Thevenin equivalent shown below. Rth

Vth

Vth =

+ −

2H

80 (60) = 40 V 80 + 40 80 +R 3 Vth 40 I = i(0) = i(∞) = = R th 80 3 + R

R th = 40 || 80 + R =

2

1 1  40   =1 w = L I 2 = (2) 2 2  R + 80 3  40 40 =1  → R = R + 80 3 3 R = 13.33 Ω

Chapter 7, Solution 22. i( t ) = i(0) e - t τ ,

τ=

L R eq

R eq = 5 || 20 + 1 = 5 Ω ,

τ=

2 5

i( t ) = 10 e -2.5t A Using current division, the current through the 20 ohm resistor is 5 -i io = (-i) = = -2 e -2.5t 5 + 20 5 v( t ) = 20 i o = - 40 e -2.5t V

Chapter 7, Solution 23.

Since the 2 Ω resistor, 1/3 H inductor, and the (3+1) Ω resistor are in parallel, they always have the same voltage. 2 2 + = 1.5  → i(0) = -1.5 2 3 +1 The Thevenin resistance R th at the inductor’s terminals is 13 1 L 4 R th = 2 || (3 + 1) = , τ= = = 3 R th 4 3 4 -i =

i( t ) = i(0) e - t τ = -1.5 e -4t , t > 0 di v L = v o = L = -1.5(-4)(1/3) e -4t dt -4t v o = 2 e V, t > 0

vx =

1 v = 0.5 e -4t V , t > 0 3 +1 L

Chapter 7, Solution 24.

(a) v( t ) = - 5 u(t) (b) i( t ) = -10 [ u ( t ) − u ( t − 3)] + 10[ u ( t − 3) − u ( t − 5)] = - 10 u(t ) + 20 u(t − 3) − 10 u(t − 5)

(c) x ( t ) = ( t − 1) [ u ( t − 1) − u ( t − 2)] + [ u ( t − 2) − u ( t − 3)] + (4 − t ) [ u ( t − 3) − u ( t − 4)] = ( t − 1) u ( t − 1) − ( t − 2) u ( t − 2) − ( t − 3) u ( t − 3) + ( t − 4) u ( t − 4) = r(t − 1) − r(t − 2) − r(t − 3) + r(t − 4) (d) y( t ) = 2 u (-t ) − 5 [ u ( t ) − u ( t − 1)] = 2 u(-t ) − 5 u(t ) + 5 u(t − 1)

Chapter 7, Solution 25. v(t) = [u(t) + r(t – 1) – r(t – 2) – 2u(t – 2)] V Chapter 7, Solution 26. v1 ( t ) = u ( t + 1) − u ( t ) + [ u ( t − 1) − u ( t )] v1 ( t ) = u(t + 1) − 2 u(t ) + u(t − 1)

(a)

v 2 ( t ) = ( 4 − t ) [ u ( t − 2) − u ( t − 4) ] v 2 ( t ) = -( t − 4) u ( t − 2) + ( t − 4) u ( t − 4) v 2 ( t ) = 2 u(t − 2) − r(t − 2) + r(t − 4)

(b)

v 3 ( t ) = 2 [ u(t − 2) − u(t − 4)] + 4 [ u(t − 4) − u(t − 6)] v 3 ( t ) = 2 u(t − 2) + 2 u(t − 4) − 4 u(t − 6)

(c)

v 4 ( t ) = -t [ u ( t − 1) − u ( t − 2)] = -t u(t − 1) + t u ( t − 2) v 4 ( t ) = (-t + 1 − 1) u ( t − 1) + ( t − 2 + 2) u ( t − 2) v 4 ( t ) = - r(t − 1) − u(t − 1) + r(t − 2) + 2 u(t − 2)

(d)

Chapter 7, Solution 27. v(t) is sketched below.

v(t) 2 1 0 -1

1

2

3

4

t

Chapter 7, Solution 28. i(t) is sketched below.

i(t) 1

0

1

3

2

4

t

-1

Chapter 7, Solution 29 x(t) (a)

3.679

0 (b)

1

t

y(t) 27.18

0

t

(c)

z (t ) = cos 4tδ (t − 1) = cos 4δ (t − 1) = −0.6536δ (t − 1) , which is sketched below. z(t)

0

1

t

-0.653 δ (t )

Chapter 7, Solution 30.

(a)

∫

4 t 2 δ( t − 1) dt = 4 t 2

(b)

∫

cos(2πt ) δ( t − 0.5) dt = cos(2πt )

10

0 ∞

-∞

t =1

=4 t = 0.5

= cos π = - 1

Chapter 7, Solution 31.

(a) (b)

= e = 112 × 10 ∫ [ e δ(t − 2)] dt = e ∫ [ 5 δ(t ) + e δ(t ) + cos 2πt δ(t )] dt = ( 5 + e + cos(2πt )) ∞

- 4t 2

- 4t 2

-∞ ∞

-t

-t

-∞

Chapter 7, Solution 32.

(a) (b)

t

t

t

1 4

1

1

∫ u (λ )dλ = ∫ 1dλ = λ

5

(c )

1

4

0

1

= t −1

∫ r (t − 1)dt = ∫ 0dt + ∫ (t − 1)dt = 0

∫ (t − 6) 1

2

-9

-16

t=2

δ (t − 2)dt = (t − 6) 2

t2 − t 14 = 4.5 2

t =2

= 16

t =0

= 5 +1+1 = 7

Chapter 7, Solution 33.

i( t ) =

1 t ∫ v(t ) dt + i(0) L 0

i( t ) =

10 -3 10 × 10 -3

∫ 20 δ(t − 2) dt + 0 t

0

i ( t ) = 2 u( t − 2 ) A

Chapter 7, Solution 34.

(a)

d [u ( t − 1) u ( t + 1)] = δ( t − 1)u ( t + 1) + dt u ( t − 1)δ( t + 1) = δ( t − 1) • 1 + 0 • δ( t + 1) = δ( t − 1)

(b)

d [r ( t − 6) u ( t − 2)] = u ( t − 6)u ( t − 2) + dt r ( t − 6)δ( t − 2) = u ( t − 6) • 1 + 0 • δ( t − 2) = u ( t − 6)

d [sin 4t u (t − 3)] = 4 cos 4t u ( t − 3) + sin 4tδ( t − 3) dt = 4 cos 4t u ( t − 3) + sin 4x3δ( t − 3)

(c)

= 4 cos 4t u ( t − 3) − 0.5366δ( t − 3)

Chapter 7, Solution 35.

(a)

v( t ) = A e -5t 3 , v(0) = A = -2 v( t ) = - 2 e -5t 3 V

(b)

v( t ) = A e 2t 3 , v(0) = A = 5 v( t ) = 5 e 2t 3 V

Chapter 7, Solution 36.

(a)

(b)

v( t ) = A + B e-t , t > 0 A = 1, v(0) = 0 = 1 + B v( t ) = 1 − e -t V , t > 0

or

B = -1

v( t ) = A + B e t 2 , t > 0 A = -3 , v(0) = -6 = -3 + B v( t ) = - 3 ( 1 + e t 2 ) V , t > 0

or

B = -3

Chapter 7, Solution 37.

Let v = vh + vp, vp =10. •

1

vh + 4 v

h

=0

v h = Ae −t / 4

 →

v = 10 + Ae −0.25t v(0) = 2 = 10 + A v = 10 − 8e −0.25t

 →

A = −8

(a) τ = 4 s (b) v(∞) = 10 V (c ) v = 10 − 8e −0.25t

Chapter 7, Solution 38

Let i = ip +ih •

i h + 3ih = 0

Let i p = ku (t ),

•

ip = 0,

 →

3ku (t ) = 2u (t )

ih = Ae −3t u (t )

 →

k=

2 3

ip =

2 u (t ) 3

2 i = ( Ae −3t + )u (t ) 3 If i(0) =0, then A + 2/3 = 0, i.e. A=-2/3. Thus i=

2 (1 − e −3t )u (t ) 3

Chapter 7, Solution 39.

(a)

Before t = 0, v( t ) =

1 (20) = 4 V 4 +1

After t = 0, v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ τ = RC = (4)(2) = 8 , v(0) = 4 , v( t ) = 20 + (8 − 20) e - t 8 v( t ) = 20 − 12 e -t 8 V

v(∞) = 20

Before t = 0, v = v1 + v 2 , where v1 is due to the 12-V source and v 2 is due to the 2-A source. v1 = 12 V To get v 2 , transform the current source as shown in Fig. (a). v 2 = -8 V Thus, v = 12 − 8 = 4 V

(b)

After t = 0, the circuit becomes that shown in Fig. (b). 2F +

v2

2F

4Ω − + −

8V

12 V

+ −

3Ω

3Ω

(a)

(b)

v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v(∞) = 12 , v(0) = 4 , τ = RC = (2)(3) = 6 -t 6 v( t ) = 12 + (4 − 12) e v( t ) = 12 − 8 e -t 6 V Chapter 7, Solution 40.

(a)

(b)

Before t = 0, v = 12 V .

After t = 0, v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v(∞) = 4 , v(0) = 12 , τ = RC = (2)(3) = 6 -t 6 v( t ) = 4 + (12 − 4) e v( t ) = 4 + 8 e - t 6 V Before t = 0, v = 12 V .

After t = 0, v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ After transforming the current source, the circuit is shown below. t=0 2Ω + −

12 V

v(∞) = 12 ,

v(0) = 12 , v = 12 V Chapter 7, Solution 41.

v(0) = 0 ,

v(∞) =

R eq C = (6 || 30)(1) =

30 (12) = 10 16

(6)(30) =5 36

v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v( t ) = 10 + (0 − 10) e - t 5 v( t ) = 10 ( 1 − e -0.2t ) V

4Ω

5F

τ = RC = (2)(5) = 10

Chapter 7, Solution 42. (a)

v o ( t ) = v o (∞) + [ v o (0) − v o (∞)] e - t τ 4 v o (∞) = (12) = 8 v o (0) = 0 , 4+2 4 τ = R eq C eq , R eq = 2 || 4 = 3 4 τ = (3) = 4 3 v o ( t ) = 8 − 8 e -t 4 v o ( t ) = 8 ( 1 − e -0.25t ) V

(b)

For this case, v o (∞) = 0 so that v o ( t ) = v o (0) e -t τ 4 v o (0) = (12) = 8 , 4+2 v o ( t ) = 8 e -t 12 V

τ = RC = (4)(3) = 12

Chapter 7, Solution 43.

Before t = 0, the circuit has reached steady state so that the capacitor acts like an open circuit. The circuit is equivalent to that shown in Fig. (a) after transforming the voltage source. vo vo , 0.5i = 2 − i= 40 80 vo 1 vo 320 = 2−  → v o = = 64 Hence, 2 80 40 5 vo i= = 0.8 A 80 After t = 0, the circuit is as shown in Fig. (b). v C ( t ) = v C (0) e - t τ ,

τ = R th C

To find R th , we replace the capacitor with a 1-V voltage source as shown in Fig. (c). 0.5i

vC i

1V

+ −

0.5i (c)

80 Ω

vC 1 0.5 = , i o = 0.5 i = 80 80 80 1 80 R th = = = 160 Ω , τ = R th C = 480 i o 0.5 v C (0) = 64 V

i=

v C ( t ) = 64 e - t 480 dv C  1   64 e - t 480 0.5 i = -i C = -C = -3   dt 480  i( t ) = 0.8 e -t 480 A Chapter 7, Solution 44. R eq = 6 || 3 = 2 Ω ,

τ = RC = 4

v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ

Using voltage division, 3 3 v(0) = (30) = 10 V , v(∞) = (12) = 4 V 3+ 6 3+ 6 Thus, v( t ) = 4 + (10 − 4) e - t 4 = 4 + 6 e - t 4  - 1 dv i( t ) = C = (2)(6)   e - t 4 = - 3 e -0.25t A 4 dt

Chapter 7, Solution 45. For t < 0, v s = 5 u ( t ) = 0  → v(0) = 0 For t > 0, v s = 5 ,

v(∞) =

4 5 (5) = 4 + 12 4

R eq = 7 + 4 || 12 = 10 ,

τ = R eq C = (10)(1 2) = 5

v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v( t ) = 1.25 ( 1 − e -t 5 ) V i( t ) = C

dv  1  - 5  - 1  - t 5 =     e dt  2  4  5 

i( t ) = 0.125 e -t 5 A

Chapter 7, Solution 46.

τ = RTh C = (2 + 6) x0.25 = 2s,

v(0) = 0,

v(∞) = 6i s = 6 x5 = 30

v(t ) = v(∞) + [v(0) − v(∞)]e − t / τ = 30(1 − e − t / 2 ) V

Chapter 7, Solution 47. For t < 0, u ( t ) = 0 ,

u ( t − 1) = 0 ,

v(0) = 0

For 0 < t < 1, τ = RC = (2 + 8)(0.1) = 1 v(0) = 0 , v(∞) = (8)(3) = 24 v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v( t ) = 24( 1 − e - t ) For t > 1,

v(1) = 24( 1 − e -1 ) = 15.17

- 6 + v(∞) - 24 = 0  → v(∞) = 30 v( t ) = 30 + (15.17 − 30) e -(t-1) v( t ) = 30 − 14.83 e -(t-1) Thus,

(

)

 24 1 − e - t V , 01  30 − 14.83 e

Chapter 7, Solution 48. For t < 0,

u (-t) = 1 ,

For t > 0, u (-t) = 0 , R th = 20 + 10 = 30 ,

v(0) = 10 V v(∞) = 0 τ = R th C = (30)(0.1) = 3

v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v( t ) = 10 e -t 3 V

 - 1 dv = (0.1)  10 e - t 3 3 dt - 1 -t 3 e A i( t ) = 3 i( t ) = C

Chapter 7, Solution 49. For 0 < t < 1, v(0) = 0 , R eq = 4 + 6 = 10 ,

v(∞) = (2)(4) = 8 τ = R eq C = (10)(0.5) = 5

v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v( t ) = 8 ( 1 − e - t 5 ) V

For t > 1, v(1) = 8 ( 1 − e -0.2 ) = 1.45 , v( t ) = v(∞) + [ v(1) − v(∞)] e -( t −1) τ v( t ) = 1.45 e -( t −1) 5 V Thus,

(

v(∞) = 0

)

 8 1 − e -t 5 V , 0 < t < 1 v( t ) =  - ( t −1 ) 5 V, t >1  1.45 e

Chapter 7, Solution 50. For the capacitor voltage, v( t ) = v(∞) + [ v(0) − v(∞)] e- t τ v(0) = 0 For t < 0, we transform the current source to a voltage source as shown in Fig. (a).

1 kΩ

1 kΩ +

30 V

+ −

v −

(a) 2 (30) = 15 V 2 +1+1 R th = (1 + 1) || 2 = 1 kΩ 1 1 τ = R th C = 10 3 × × 10 -3 = 4 4 -4t v( t ) = 15 ( 1 − e ) , t > 0 v(∞) =

2 kΩ

We now obtain i x from v(t). Consider Fig. (b).

iT 1 kΩ

v

ix 1 kΩ

30 mA

1/4 mF (b)

But

i x = 30 mA − i T v dv iT = +C R3 dt

i T ( t ) = 7.5 ( 1 − e -4t ) mA + i T ( t ) = 7.5 ( 1 + e -4t ) mA

1 × 10 -3 (-15)(-4) e -4t A 4

Thus, i x ( t ) = 30 − 7.5 − 7.5 e -4t mA i x ( t ) = 7.5 ( 3 − e -4t ) mA , t > 0

Chapter 7, Solution 51. Consider the circuit below.

t=0

R

+ VS

+ −

i

L

v −

After the switch is closed, applying KVL gives di VS = Ri + L dt  VS  di  or L = -R  i −  dt R di -R = dt i − VS R L Integrating both sides,

2 kΩ

 V  i(t ) - R ln  i − S  I 0 = t  R L  i − VS R  - t = ln   I0 − VS R  τ or

i − VS R = e- t τ I0 − VS R

i( t ) =

VS  VS  -t τ e +  I0 − R  R which is the same as Eq. (7.60).

Chapter 7, Solution 52. 20 = 2 A, i(∞) = 2 A 10 i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ

i(0) =

i( t ) = 2 A

Chapter 7, Solution 53. 25 =5A 3+ 2 After t = 0, i( t ) = i(0) e - t τ L 4 τ = = = 2, i(0) = 5 R 2 i( t ) = 5 e - t 2 A i=

(a)

Before t = 0,

(b)

Before t = 0, the inductor acts as a short circuit so that the 2 Ω and 4 Ω resistors are short-circuited. i( t ) = 6 A After t = 0, we have an RL circuit. i( t ) = i(0) e - t τ ,

i( t ) = 6 e - 2 t 3 A

τ=

L 3 = R 2

Chapter 7, Solution 54. (a)

Before t = 0, i is obtained by current division or 4 i( t ) = (2) = 1 A 4+4 After t = 0, i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ L τ= , R eq = 4 + 4 || 12 = 7 Ω R eq τ=

3.5 1 = 7 2

i(0) = 1 ,

i(∞) =

6 3 4 || 12 (2) = (2) = 7 4+3 4 + 4 || 12

6  6 +  1 −  e -2 t 7  7 1 i( t ) = ( 6 − e - 2t ) A 7 10 =2A Before t = 0, i( t ) = 2+3 After t = 0, R eq = 3 + 6 || 2 = 4.5 i( t ) =

(b)

L 2 4 = = R eq 4.5 9 i(0) = 2 To find i(∞) , consider the circuit below, at t = when the inductor becomes a short circuit, v τ=

i 10 V 2Ω

+ −

24 V

+ −

6Ω

10 − v 24 − v v + =  → v = 9 2 6 3 v i(∞) = = 3 A 3 i( t ) = 3 + (2 − 3) e -9 t 4 i( t ) = 3 − e - 9 t 4 A

2H 3Ω

Chapter 7, Solution 55. For t < 0, consider the circuit shown in Fig. (a).

0.5 H

io 3Ω 24 V

io + −

+ −

0.5 H i

+ 4io

2Ω

v

− (a)

8Ω 20 V

+ v

+ −

− (b)

3i o + 24 − 4i o = 0  → i o = 24 v v( t ) = 4i o = 96 V i = = 48 A 2

For t > 0, consider the circuit in Fig. (b). i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ 20 i(∞) = =2A i(0) = 48 , 8+ 2 L 0.5 1 R th = 2 + 8 = 10 Ω , τ = = = R th 10 20

i( t ) = 2 + (48 − 2) e -20t = 2 + 46 e -20t v( t ) = 2 i( t ) = 4 + 92 e -20t V

Chapter 7, Solution 56. R eq = 6 + 20 || 5 = 10 Ω ,

τ=

L = 0.05 R

i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ

i(0) is found by applying nodal analysis to the following circuit.

2Ω

5Ω i

vx

12 Ω

2A

6Ω +

20 Ω

0.5 H

+ −

20 V

v

− 20 − v x v x v x v x = + + 5 12 20 6 vx i ( 0) = =2A 6 2+

 → v x = 12

Since 20 || 5 = 4 , 4 i(∞) = (4) = 1.6 4+6 i( t ) = 1.6 + (2 − 1.6) e- t 0.05 = 1.6 + 0.4 e-20t di 1 v( t ) = L = (0.4) (-20) e -20t dt 2 v( t ) = - 4 e -20t V

Chapter 7, Solution 57.

At t = 0 − , the circuit has reached steady state so that the inductors act like short circuits. 6Ω

30 V

+ −

i i1

i2

5Ω

20 Ω

20 30 30 = = 3, i1 = (3) = 2.4 , 6 + 5 || 20 10 25 i 1 ( 0 ) = 2 .4 A , i 2 ( 0 ) = 0 .6 A

i=

i 2 = 0 .6

For t > 0, the switch is closed so that the energies in L1 and L 2 flow through the closed switch and become dissipated in the 5 Ω and 20 Ω resistors. L 2.5 1 i1 ( t ) = i1 (0) e - t τ1 , τ1 = 1 = = R1 5 2 i1 ( t ) = 2.4 e -2t A

i 2 ( t ) = i 2 (0) e - t τ 2 ,

τ2 =

L2 4 1 = = R 2 20 5

i 2 ( t ) = 0.6 e -5t A

Chapter 7, Solution 58. For t < 0,

v o (t) = 0

For t > 0,

i(0) = 10 ,

R th = 1 + 3 = 4 Ω ,

20 =5 1+ 3 L 14 1 τ= = = R th 4 16

i(∞) =

i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ i( t ) = 5 ( 1 + e-16t ) A

di 1 = 15 ( 1 + e -16t ) + (-16)(5) e-16t dt 4 -16t v o ( t ) = 15 − 5 e V vo (t ) = 3i + L

Chapter 7, Solution 59. Let I be the current through the inductor. i(0) = 0 For t < 0, vs = 0 , For t > 0,

R eq = 4 + 6 || 3 = 6 ,

2 (3) = 1 2+ 4 i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ i( t ) = 1 − e-4t i(∞) =

di = (1.5)(-4)(-e- 4t ) dt v o ( t ) = 6 e -4t V vo (t ) = L

τ=

L 1 .5 = = 0.25 R eq 6

Chapter 7, Solution 60. Let I be the inductor current. For t < 0, u(t) = 0  → i(0) = 0 For t > 0,

R eq = 5 || 20 = 4 Ω ,

τ=

L 8 = =2 R eq 4

i(∞) = 4 i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ i( t ) = 4 ( 1 − e - t 2 )  - 1 di = (8)(-4)  e - t 2 2 dt v( t ) = 16 e -0.5t V v( t ) = L

Chapter 7, Solution 61.

The current source is transformed as shown below. 4Ω

20u(-t) + 40u(t)

+ −

0.5 H

L 12 1 = = , i(0) = 5 , R 4 8 i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ i( t ) = 10 − 5 e -8t A τ=

di  1  =  (-5)(-8) e -8t dt  2  v( t ) = 20 e -8t V v( t ) = L

Chapter 7, Solution 62. L 2 = =1 R eq 3 || 6 For 0 < t < 1, u ( t − 1) = 0 so that τ=

i(∞) = 10

i(0) = 0 ,

i( t ) =

i(∞) =

1 6

1 ( 1 − e -t ) 6

1 ( 1 − e -1 ) = 0.1054 6 1 1 1 i(∞) = + = 3 6 2 i( t ) = 0.5 + (0.1054 − 0.5) e-(t -1) i( t ) = 0.5 − 0.3946 e-(t -1) i(1) =

For t > 1,

Thus,

 1 ( 1 − e -t ) A 01

Chapter 7, Solution 63.

10 =2 5

For t < 0,

u (- t ) = 1 ,

i(0) =

For t > 0,

u (-t) = 0 ,

i(∞) = 0 L 0.5 1 τ= = = R th 4 8

R th = 5 || 20 = 4 Ω ,

i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ i( t ) = 2 e -8t A di  1  =  (-8)(2) e-8t dt  2  v( t ) = - 8 e -8t V

v( t ) = L

Chapter 7, Solution 64.

Let i be the inductor current. For t < 0, the inductor acts like a short circuit and the 3 Ω resistor is shortcircuited so that the equivalent circuit is shown in Fig. (a). 6Ω

10 Ω

+ −

3Ω

(a)

6Ω

i

10 Ω

+ −

io

v

i

3Ω 2Ω (b)

i = i(0) = For t > 0,

10 = 1.667 A 6

R th = 2 + 3 || 6 = 4 Ω ,

τ=

L 4 = =1 R th 4

To find i(∞) , consider the circuit in Fig. (b). 10 − v v v 10 = +  → v = 6 3 2 6 v 5 i = i(∞) = = 2 6 i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ 5  10 5  5 i( t ) = +  −  e - t = ( 1 − e - t ) A 6  6 6 6 v o is the voltage across the 4 H inductor and the 2 Ω resistor 5 di 10 10 - t 10 10 - t − e = + e + (4) (-1) e - t = 6 dt 6 6 6 6 v o ( t ) = 1.667 ( 1 − e -t ) V v o (t) = 2 i + L

Chapter 7, Solution 65.

Since v s = 10 [ u ( t ) − u ( t − 1)] , this is the same as saying that a 10 V source is turned on at t = 0 and a -10 V source is turned on later at t = 1. This is shown in the figure below. vs 10 1 t -10 For 0 < t < 1, i(0) = 0 , R th = 5 || 20 = 4 ,

10 =2 5 L 2 1 τ= = = R th 4 2

i(∞) =

i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ

i( t ) = 2 ( 1 − e -2t ) A i(1) = 2 ( 1 − e-2 ) = 1.729 For t > 1,

i(∞) = 0

since vs = 0

i( t ) = i(1) e- ( t −1) τ i( t ) = 1.729 e-2( t −1) A Thus,

 2 ( 1 − e - 2t ) A 0 < t < 1 i( t ) =  t>1  1.729 e - 2( t −1) A

Chapter 7, Solution 66.

Following Practice Problem 7.14, v( t ) = VT e - t τ τ = R f C = (10 × 103 )(2 × 10- 6 ) =

VT = v(0) = -4 ,

1 50

v( t ) = -4 e -50t v o ( t ) = -v( t ) = 4 e -50t , t > 0 i o (t) =

v o (t) 4 = e -50t = 0.4 e -50t mA , t > 0 Ro 10 × 10 3

Chapter 7, Solution 67.

The op amp is a voltage follower so that v o = v as shown below. R R

− +

vo

v1 + R

vo

−

C

vo

At node 1, v o − v1 v1 − 0 v1 − v o = + R R R

 → v1 =

2 v 3 o

At the noninverting terminal, dv v − v1 C o + o =0 dt R dv 1 2 − RC o = v o − v1 = v o − v o = v o 3 3 dt dv o v =− o dt 3RC v o ( t ) = VT e - t 3RC VT = vo (0) = 5 V ,

τ = 3RC = (3)(10 × 103 )(1 × 10- 6 ) =

3 100

v o ( t ) = 5 e -100t 3 V

Chapter 7, Solution 68.

This is a very interesting problem and has both an important ideal solution as well as an important practical solution. Let us look at the ideal solution first. Just before the switch closes, the value of the voltage across the capacitor is zero which means that the voltage at both terminals input of the op amp are each zero. As soon as the switch closes, the output tries to go to a voltage such that the input to the op amp both go to 4 volts. The ideal op amp puts out whatever current is necessary to reach this condition. An infinite (impulse) current is necessary if the voltage across the capacitor is to go to 8 volts in zero time (8 volts across the capacitor will result in 4 volts appearing at the negative terminal of the op amp). So vo will be equal to 8 volts for all t > 0. What happens in a real circuit? Essentially, the output of the amplifier portion of the op amp goes to whatever its maximum value can be. Then this maximum voltage appears across the output resistance of the op amp and the capacitor that is in series with it. This results in an exponential rise in the capacitor voltage to the steady-state value of 8 volts. vC(t) = Vop amp max(1 – e-t/(RoutC)) volts, for all values of vC less than 8 V, = 8 V when t is large enough so that the 8 V is reached. Chapter 7, Solution 69.

Let v x be the capacitor voltage. v x ( 0) = 0 For t < 0,

For t > 0, the 20 kΩ and 100 kΩ resistors are in series since no current enters the op amp terminals. As t → ∞ , the capacitor acts like an open circuit so that 20 + 100 48 v x (∞) = (4) = 20 + 100 + 10 13 R th = 20 + 100 = 120 kΩ , τ = R th C = (120 × 103 )(25 × 10-3 ) = 3000

v x ( t ) = v x (∞) + [ v x (0) − v x (∞)] e- t τ 48 v x ( t ) = ( 1 − e - t 3000 ) 13 vo (t ) =

40 100 ( 1 − e -t 3000 ) V vx (t) = 120 13

Chapter 7, Solution 70.

Let v = capacitor voltage. For t < 0, the switch is open and v(0) = 0 . For t > 0, the switch is closed and the circuit becomes as shown below. 1

+ −

2 vS

+

+ −

vo v

−

C R

v1 = v 2 = v s 0 − vs dv =C R dt where v = v s − v o  → v o = v s − v From (1), dv v s = =0 dt RC - t vs -1 v= v s dt + v(0) = ∫ RC RC Since v is constant,

(1) (2) (3)

RC = (20 × 10 3 )(5 × 10 -6 ) = 0.1 - 20 t mV = -200 t mV v= 0.1 From (3), v o = v s − v = 20 + 200 t v o = 20 ( 1 + 10t ) mV

Chapter 7, Solution 71.

Let v = voltage across the capacitor. Let v o = voltage across the 8 kΩ resistor. For t < 2, v = 0 so that v(2) = 0 . For t > 2, we have the circuit shown below. 10 kΩ 10 kΩ

20 kΩ − +

+ 4V

+ −

100 mF

v

−

+

io 8 kΩ

vo

−

Since no current enters the op amp, the input circuit forms an RC circuit. τ = RC = (10 × 10 3 )(100 × 10 -3 ) = 1000 v( t ) = v(∞) + [ v(2) − v(∞)] e -( t − 2 ) τ v( t ) = 4 ( 1 − e -( t − 2 ) 1000 ) As an inverter, - 10k v = 2 ( e -( t − 2 ) 1000 − 1 ) vo = 20k vo io = = 0.25 ( e -( t − 2 ) 1000 − 1 ) A 8

Chapter 7, Solution 72. The op amp acts as an emitter follower so that the Thevenin equivalent circuit is shown below. C +

3u(t)

Hence,

v

−

io

+ −

R

v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v(0) = -2 V , v(∞) = 3 V , τ = RC = (10 × 10 3 )(10 × 10 -6 ) = 0.1 v( t ) = 3 + (-2 - 3) e -10t = 3 − 5 e -10t

dv = (10 × 10 -6 )(-5)(-10) e -10t dt i o = 0.5 e -10t mA , t > 0

io = C

Chapter 7, Solution 73.

Consider the circuit below. Rf v1

R1

v2 +

v1

+ −

C v

v3 −

− +

+ vo

− At node 2, v1 − v 2 dv =C dt R1 At node 3,

(1)

C

dv v 3 − v o = dt Rf

(2)

But v 3 = 0 and v = v 2 − v 3 = v 2 . Hence, (1) becomes v1 − v dv =C R1 dt dv v1 − v = R 1C dt v1 dv v or + = dt R 1C R 1C which is similar to Eq. (7.42). Hence,  vT t<0 v( t ) =  -t τ t>0  v1 + ( v T − v1 ) e

where v T = v(0) = 1 and v1 = 4 τ = R 1C = (10 × 10 3 )(20 × 10 -6 ) = 0.2  1 t<0 v( t ) =  -5t t>0 4 − 3 e From (2), dv = (20 × 10 3 )(20 × 10 -6 )(15 e -5t ) dt v o = -6 e -5t , t > 0

v o = -R f C

v o = - 6 e -5t u(t ) V

Chapter 7, Solution 74.

Let v = capacitor voltage. Rf v1

R1

v2 +

v1

+ −

C v

v3 −

− +

+ vo

−

v(0) = 0 i s = 10 µA . Consider the circuit below.

For t < 0, For t > 0,

dv v + dt R v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ

is = C

(1) (2)

It is evident from the circuit that τ = RC = (2 × 10 −6 )(50 × 10 3 ) = 0.1 Rf

C is R

− +

is

+ vo

− At steady state, the capacitor acts like an open circuit so that i s passes through R. Hence, v(∞) = i s R = (10 × 10 −6 )(50 × 10 3 ) = 0.5 V

Then,

But

v( t ) = 0.5 ( 1 − e -10t ) V is =

0 − vo Rf

 → v o = -i s R f

Combining (1), (3), and (4), we obtain - Rf dv vo = v − RfC R dt -1 dv v o = v − (10 × 10 3 )(2 × 10 -6 ) 5 dt -10t -2 v o = -0.1 + 0.1e − (2 × 10 )(0.5)( - 10 e -10t ) v o = 0.2 e -10t − 0.1

v o = 0.1 ( 2 e -10t − 1) V

(3) (4)

Chapter 7, Solution 75.

Let v1 = voltage at the noninverting terminal. Let v 2 = voltage at the inverting terminal. For t > 0, v1 = v 2 = v s = 4 0 − vs = i o , R 1 = 20 kΩ R1 vo = -ioR Also, i o = i.e.

v dv +C , R2 dt

(1)

R 2 = 10 kΩ , C = 2 µF

- vs dv v = +C dt R1 R 2

(2)

This is a step response. v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ , where τ = R 2 C = (10 × 10 3 )(2 × 10 -6 ) =

v(0) = 1

1 50

At steady state, the capacitor acts like an open circuit so that i o passes through R 2 . Hence, as t → ∞ - vs v(∞) = io = R2 R1 - R2 - 10 (4) = -2 v(∞) = vs = i.e. 20 R1

v( t ) = -2 + (1 + 2) e -50t v( t ) = -2 + 3 e -50t But

v = vs − vo

or

v o = v s − v = 4 + 2 − 3 e -50 t v o = 6 − 3 e -50 t V - vs -4 = = -0.2 mA R 1 20k dv v +C = - 0.2 mA io = dt R2 io =

or

Chapter 7, Solution 76.

The schematic is shown below. For the pulse, we use IPWL and enter the corresponding values as attributes as shown. By selecting Analysis/Setup/Transient, we let Print Step = 25 ms and Final Step = 2 s since the width of the input pulse is 1 s. After saving and simulating the circuit, we select Trace/Add and display –V(C1:2). The plot of V(t) is shown below.

Chapter 7, Solution 77.

The schematic is shown below. We click Marker and insert Mark Voltage Differential at the terminals of the capacitor to display V after simulation. The plot of V is shown below. Note from the plot that V(0) = 12 V and V(∞) = -24 V which are correct.

Chapter 7, Solution 78.

(a)

When the switch is in position (a), the schematic is shown below. We insert IPROBE to display i. After simulation, we obtain, i(0) = 7.714 A

from the display of IPROBE.

(b) When the switch is in position (b), the schematic is as shown below. For inductor I1, we let IC = 7.714. By clicking Analysis/Setup/Transient, we let Print Step = 25 ms and Final Step = 2 s. After Simulation, we click Trace/Add in the probe menu and display I(L1) as shown below. Note that i(∞) = 12A, which is correct.

Chapter 7, Solution 79.

When the switch is in position 1, io(0) = 12/3 = 4A. When the switch is in position 2, R 4 i o (∞ ) = − = −0.5 A, RTh = (3 + 5) // 4 = 8 / 3, τ = Th = 80 / 3 5+3 L

io (t ) = io (∞) + [io (0) − io (∞)]e −t / τ = − 0.5 + 4.5e −3t / 80 A Chapter 7, Solution 80.

(a) When the switch is in position A, the 5-ohm and 6-ohm resistors are shortcircuited so that i1 (0) = i2 (0) = vo (0) = 0

but the current through the 4-H inductor is iL(0) =30/10 = 3A. (b) When the switch is in position B, RTh = 3 // 6 = 2Ω,

τ=

RTh = 2 / 4 = 0 .5 L

i L (t ) = i L (∞) + [i L (0) − i L (∞)]e −t / τ = 0 + 3e −t / 0.5 = 3e −2t A

(c) i1 (∞) =

30 = 2 A, 10 + 5 vo (t ) = L

3 i 2 (∞ ) = − i L (∞ ) = 0 A 9 di L dt

 →

v o (∞ ) = 0 V

Chapter 7, Solution 81.

The schematic is shown below. We use VPWL for the pulse and specify the attributes as shown. In the Analysis/Setup/Transient menu, we select Print Step = 25 ms and final Step = 3 S. By inserting a current marker at one termial of LI, we automatically obtain the plot of i after simulation as shown below.

Chapter 7, Solution 82. τ = RC  → R =

3 × 10 -3 τ = = 30 Ω C 100 × 10 -6

Chapter 7, Solution 83. v(∞) = 120,

v(0) = 0,

τ = RC = 34 x10 6 x15 x10 −6 = 510s

v(t ) = v(∞) + [v(0) − v(∞)]e − t / τ

85.6 = 120(1 − e − t / 510 )

 →

Solving for t gives t = 510 ln 3.488 = 637.16 s speed = 4000m/637.16s = 6.278m/s Chapter 7, Solution 84. Let Io be the final value of the current. Then

i (t ) = I o (1 − e − t / τ ), 0.6 I o = I o (1 − e −50t )

τ = R / L = 0.16 / 8 = 1 / 50  →

t=

1 1 ln = 18.33 ms. 50 0.4

Chapter 7, Solution 85.

(a)

τ = RC = (4 × 106 )(6 × 10-6 ) = 24 s Since v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v( t 1 ) − v(∞) = [ v(0) − v(∞)] e - t1 τ v( t 2 ) − v(∞) = [ v(0) − v(∞)] e- t 2 τ Dividing (1) by (2), v( t1 ) − v(∞) = e( t 2 − t1 ) τ v( t 2 ) − v(∞)  v( t ) − v(∞)   t 0 = t 2 − t1 = τ ln  1  v( t 2 ) − v(∞) 

(b)

 75 − 120   = 24 ln (2) = 16.63 s t 0 = 24 ln   30 − 120  Since t 0 < t , the light flashes repeatedly every τ = RC = 24 s

(1) (2)

Chapter 7, Solution 86. v( t ) = v(∞) + [ v(0) − v(∞)] e- t τ v(∞) = 12 , v(0) = 0 -t τ v( t ) = 12 ( 1 − e ) v( t 0 ) = 8 = 12 ( 1 − e- t 0 τ ) 8 1 = 1 − e- t 0 τ  → e- t 0 τ = 12 3 t 0 = τ ln (3)

For R = 100 kΩ , τ = RC = (100 × 103 )(2 × 10-6 ) = 0.2 s t 0 = 0.2 ln (3) = 0.2197 s For R = 1 MΩ , τ = RC = (1 × 106 )(2 × 10-6 ) = 2 s t 0 = 2 ln (3) = 2.197 s Thus, 0.2197 s < t 0 < 2.197 s

Chapter 7, Solution 87.

Let i be the inductor current. For t < 0,

i (0 − ) =

120 = 1.2 A 100

For t > 0, we have an RL circuit L 50 τ= = = 0.1 , i(∞) = 0 R 100 + 400 i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ i( t ) = 1.2 e -10t At t = 100 ms = 0.1 s, i(0.1) = 1.2 e -1 = 0.441 A which is the same as the current through the resistor.

Chapter 7, Solution 88.

(a)

τ = RC = (300 × 10 3 )(200 × 10 -12 ) = 60 µs As a differentiator, T > 10 τ = 600 µs = 0.6 ms Tmin = 0.6 ms i.e.

(b)

τ = RC = 60 µs As an integrator, T < 0.1τ = 6 µs Tmax = 6 µs i.e.

Chapter 7, Solution 89.

Since τ < 0.1 T = 1 µs L < 1 µs R L < R × 10 -6 = (200 × 10 3 )(1 × 10 -6 )

L < 200 mH

Chapter 7, Solution 90.

We determine the Thevenin equivalent circuit for the capacitor C s . Rs v th = v, R th = R s || R p Rs + Rp i Rth

Vth

+ −

Cs

The Thevenin equivalent is an RC circuit. Since Rs 1 1 v th = v i  → = 10 10 R s + R p Rs = Also,

1 6 2 R p = = MΩ 9 9 3

τ = R th C s = 15 µs 6 (2 3) where R th = R p || R s = = 0.6 MΩ 6+2 3 τ 15 × 10 -6 = = 25 pF Cs = R th 0.6 × 10 6

Chapter 7, Solution 91.

12 = 240 mA , i(∞) = 0 50 i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ i( t ) = 240 e - t τ L 2 τ= = R R i( t 0 ) = 10 = 240 e - t 0 τ i o (0) =

e t 0 τ = 24  → t 0 = τ ln (24) t0 2 5 τ= = = 1.573 = R ln (24) ln (24) 2 R= = 1.271 Ω 1.573

Chapter 7, Solution 92.  10  10 -3 dv = 4 × 10 -9 ⋅  2 ×- 10 i=C dt   5 × 10 -6

0 < t < tR tR < t < tD

 20 µA 0 < t < 2 ms i( t ) =  - 8 mA 2 ms < t < 2 ms + 5 µs which is sketched below. i(t)

5 µs

20 µA t

2 ms -8 mA (not to scale)

Chapter 8, Solution 1. (a) At t = 0-, the circuit has reached steady state so that the equivalent circuit is shown in Figure (a).

6Ω VS

+ −

6Ω

6Ω +

+ vL

10 H

−

(a)

v

10 µF −

(b) i(0-) = 12/6 = 2A, v(0-) = 12V At t = 0+, i(0+) = i(0-) = 2A, v(0+) = v(0-) = 12V (b)

For t > 0, we have the equivalent circuit shown in Figure (b). vL = Ldi/dt or di/dt = vL/L

Applying KVL at t = 0+, we obtain, vL(0+) – v(0+) + 10i(0+) = 0 vL(0+) – 12 + 20 = 0, or vL(0+) = -8 Hence,

di(0+)/dt = -8/2 = -4 A/s

Similarly,

iC = Cdv/dt, or dv/dt = iC/C iC(0+) = -i(0+) = -2 dv(0+)/dt = -2/0.4 = -5 V/s

(c)

As t approaches infinity, the circuit reaches steady state. i(∞) = 0 A, v(∞) = 0 V

Chapter 8, Solution 2. (a)

At t = 0-, the equivalent circuit is shown in Figure (a). 25 kΩ

20 kΩ iR

+ −

80V

iL

+ 60 kΩ v − (a)

25 kΩ

20 kΩ iL

iR 80V

+ −

(b) 60||20 = 15 kohms, iR(0-) = 80/(25 + 15) = 2mA. By the current division principle, iL(0-) = 60(2mA)/(60 + 20) = 1.5 mA At t = 0+,

vC(0-) = 0 vC(0+) = vC(0-) = 0 iL(0+) = iL(0-) = 1.5 mA 80 = iR(0+)(25 + 20) + vC(0-) iR(0+) = 80/45k = 1.778 mA

But,

iR = i C + iL 1.778 = iC(0+) + 1.5 or iC(0+) = 0.278 mA

(b)

vL(0+) = vC(0+) = 0 But,

vL = LdiL/dt and diL(0+)/dt = vL(0+)/L = 0 diL(0+)/dt = 0 Again, 80 = 45iR + vC 0

But,

= 45diR/dt + dvC/dt

dvC(0+)/dt = iC(0+)/C = 0.278 mohms/1 µF = 278 V/s Hence,

diR(0+)/dt = (-1/45)dvC(0+)/dt = -278/45 diR(0+)/dt = -6.1778 A/s Also, iR = iC + iL diR(0+)/dt = diC(0+)/dt + diL(0+)/dt

-6.1788 = diC(0+)/dt + 0, or diC(0+)/dt = -6.1788 A/s (c)

As t approaches infinity, we have the equivalent circuit in Figure (b). iR(∞) = iL(∞) = 80/45k = 1.778 mA iC(∞) = Cdv(∞)/dt = 0.

Chapter 8, Solution 3. At t = 0-, u(t) = 0. Consider the circuit shown in Figure (a). iL(0-) = 0, and vR(0-) = 0. But, -vR(0-) + vC(0-) + 10 = 0, or vC(0-) = -10V. (a)

At t = 0+, since the inductor current and capacitor voltage cannot change abruptly, the inductor current must still be equal to 0A, the capacitor has a voltage equal to –10V. Since it is in series with the +10V source, together they represent a direct short at t = 0+. This means that the entire 2A from the current source flows through the capacitor and not the resistor. Therefore, vR(0+) = 0 V.

(b)

At t = 0+, vL(0+) = 0, therefore LdiL(0+)/dt = vL(0+) = 0, thus, diL/dt = 0A/s, iC(0+) = 2 A, this means that dvC(0+)/dt = 2/C = 8 V/s. Now for the value of dvR(0+)/dt. Since vR = vC + 10, then dvR(0+)/dt = dvC(0+)/dt + 0 = 8 V/s.

40 Ω

40 Ω

+ vC

+ vR

+ +

−

10 Ω

−

vR + −

−

10V

2A

iL

vC −

10 Ω

+ −

(a)

10V

(b)

(c) As t approaches infinity, we end up with the equivalent circuit shown in Figure (b). iL(∞) = 10(2)/(40 + 10) = 400 mA vC(∞) = 2[10||40] –10 = 16 – 10 = 6V vR(∞) = 2[10||40] = 16 V Chapter 8, Solution 4. (a) At t = 0-, u(-t) = 1 and u(t) = 0 so that the equivalent circuit is shown in Figure (a). i(0-) = 40/(3 + 5) = 5A, and v(0-) = 5i(0-) = 25V. i(0+) = i(0-) = 5A

Hence,

v(0+) = v(0-) = 25V 3Ω i 40V

+ −

+ v

5Ω −

(a)

3Ω

0.25 H + vL − iC

i + −

40V

iR

0.1F

4A 5Ω

(b) (b)

iC = Cdv/dt or dv(0+)/dt = iC(0+)/C

For t = 0+, 4u(t) = 4 and 4u(-t) = 0. The equivalent circuit is shown in Figure (b). Since i and v cannot change abruptly, iR = v/5 = 25/5 = 5A, i(0+) + 4 =iC(0+) + iR(0+) 5 + 4 = iC(0+) + 5 which leads to iC(0+) = 4 dv(0+)/dt = 4/0.1 = 40 V/s Chapter 8, Solution 5. (a)

For t < 0, 4u(t) = 0 so that the circuit is not active (all initial conditions = 0). iL(0-) = 0 and vC(0-) = 0. For t = 0+, 4u(t) = 4. Consider the circuit below. iL

A i 4A

+ 4 Ω vC

1H

iC + 0.25F

vL −

+

6Ω

−

v −

Since the 4-ohm resistor is in parallel with the capacitor, i(0+) = vC(0+)/4 = 0/4 = 0 A Also, since the 6-ohm resistor is in series with the inductor, v(0+) = 6iL(0+) = 0V.

(b)

di(0+)/dt = d(vR(0+)/R)/dt = (1/R)dvR(0+)/dt = (1/R)dvC(0+)/dt = (1/4)4/0.25 A/s = 4 A/s v = 6iL or dv/dt = 6diL/dt and dv(0+)/dt = 6diL(0+)/dt = 6vL(0+)/L = 0 Therefore dv(0+)/dt = 0 V/s

(c)

As t approaches infinity, the circuit is in steady-state. i(∞) = 6(4)/10 = 2.4 A v(∞) = 6(4 – 2.4) = 9.6 V

Chapter 8, Solution 6. (a)

Let i = the inductor current. For t < 0, u(t) = 0 so that i(0) = 0 and v(0) = 0. For t > 0, u(t) = 1. Since, v(0+) = v(0-) = 0, and i(0+) = i(0-) = 0. vR(0+) = Ri(0+) = 0 V Also, since v(0+) = vR(0+) + vL(0+) = 0 = 0 + vL(0+) or vL(0+) = 0 V. (1)

(b)

Since i(0+) = 0,

iC(0+) = VS/RS

But,

iC = Cdv/dt which leads to dv(0+)/dt = VS/(CRS)

(2)

From (1),

dv(0+)/dt = dvR(0+)/dt + dvL(0+)/dt vR = iR or dvR/dt = Rdi/dt

(3) (4)

But,

vL = Ldi/dt, vL(0+) = 0 = Ldi(0+)/dt and di(0+)/dt = 0

From (4) and (5),

dvR(0+)/dt = 0 V/s

From (2) and (3),

dvL(0+)/dt = dv(0+)/dt = Vs/(CRs)

(5)

(c) As t approaches infinity, the capacitor acts like an open circuit, while the inductor acts like a short circuit. vR(∞) = [R/(R + Rs)]Vs vL(∞) = 0 V

Chapter 8, Solution 7. s2 + 4s + 4 = 0, thus s1,2 =

− 4 ± 4 2 − 4x 4 = -2, repeated roots. 2

v(t) = [(A + Bt)e-2t], v(0) = 1 = A dv/dt = [Be-2t] + [-2(A + Bt)e-2t] dv(0)/dt = -1 = B – 2A = B – 2 or B = 1. Therefore, v(t) = [(1 + t)e-2t] V Chapter 8, Solution 8. s2 + 6s + 9 = 0, thus s1,2 =

− 6 ± 6 2 − 36 = -3, repeated roots. 2

i(t) = [(A + Bt)e-3t], i(0) = 0 = A di/dt = [Be-3t] + [-3(Bt)e-3t] di(0)/dt = 4 = B. Therefore, i(t) = [4te-3t] A Chapter 8, Solution 9. s2 + 10s + 25 = 0, thus s1,2 =

− 10 ± 10 − 10 = -5, repeated roots. 2

i(t) = [(A + Bt)e-5t], i(0) = 10 = A di/dt = [Be-5t] + [-5(A + Bt)e-5t] di(0)/dt = 0 = B – 5A = B – 50 or B = 50. Therefore, i(t) = [(10 + 50t)e-5t] A

Chapter 8, Solution 10.

s2 + 5s + 4 = 0, thus s1,2 =

− 5 ± 25 − 16 = -4, -1. 2

v(t) = (Ae-4t + Be-t), v(0) = 0 = A + B, or B = -A dv/dt = (-4Ae-4t - Be-t) dv(0)/dt = 10 = – 4A – B = –3A or A = –10/3 and B = 10/3. Therefore, v(t) = (–(10/3)e-4t + (10/3)e-t) V Chapter 8, Solution 11.

s2 + 2s + 1 = 0, thus s1,2 =

−2± 4−4 = -1, repeated roots. 2

v(t) = [(A + Bt)e-t], v(0) = 10 = A dv/dt = [Be-t] + [-(A + Bt)e-t] dv(0)/dt = 0 = B – A = B – 10 or B = 10. Therefore, v(t) = [(10 + 10t)e-t] V Chapter 8, Solution 12.

(a)

Overdamped when C > 4L/(R2) = 4x0.6/400 = 6x10-3, or C > 6 mF

(b)

Critically damped when C = 6 mF

(c)

Underdamped when C < 6mF

Chapter 8, Solution 13.

Let R||60 = Ro. For a series RLC circuit, ωo =

1 LC

=

1 0.01x 4

= 5

For critical damping, ωo = α = Ro/(2L) = 5 or Ro = 10L = 40 = 60R/(60 + R) which leads to R = 120 ohms Chapter 8, Solution 14.

This is a series, source-free circuit. 60||30 = 20 ohms α = R/(2L) = 20/(2x2) = 5 and ωo =

1 LC

=

1 0.04

= 5

ωo = α leads to critical damping i(t) = [(A + Bt)e-5t], i(0) = 2 = A v = Ldi/dt = 2{[Be-5t] + [-5(A + Bt)e-5t]} v(0) = 6 = 2B – 10A = 2B – 20 or B = 13. Therefore, i(t) = [(2 + 13t)e-5t] A Chapter 8, Solution 15.

This is a series, source-free circuit. 60||30 = 20 ohms α = R/(2L) = 20/(2x2) = 5 and ωo =

1

LC ωo = α leads to critical damping

=

1 0.04

i(t) = [(A + Bt)e-5t], i(0) = 2 = A v = Ldi/dt = 2{[Be-5t] + [-5(A + Bt)e-5t]} v(0) = 6 = 2B – 10A = 2B – 20 or B = 13. Therefore, i(t) = [(2 + 13t)e-5t] A

= 5

Chapter 8, Solution 16.

At t = 0, i(0) = 0, vC(0) = 40x30/50 = 24V For t > 0, we have a source-free RLC circuit. α = R/(2L) = (40 + 60)/5 = 20 and ωo =

1 LC

=

1 −3

10 x 2.5

ωo = α leads to critical damping i(t) = [(A + Bt)e-20t], i(0) = 0 = A di/dt = {[Be-20t] + [-20(Bt)e-20t]}, but di(0)/dt = -(1/L)[Ri(0) + vC(0)] = -(1/2.5)[0 + 24] Hence,

B = -9.6 or i(t) = [-9.6te-20t] A

Chapter 8, Solution 17.

i(0) = I0 = 0, v(0) = V0 = 4 x15 = 60 di(0) 1 = − (RI0 + V0 ) = −4(0 + 60) = −240 dt L 1 1 ωo = = = 10 LC 1 1 4 25 R 10 α= = = 20, which is > ωo . 2L 2 1 4 s = −α ± α 2 − ωo2 = −20 ± 300 = −20 ± 10 3 = −2.68, − 37.32 i( t ) = A1e − 2.68t + A 2e −37.32 t di(0) i(0) = 0 = A1 + A 2 , = −2.68A1 − 37.32A 2 = −240 dt This leads to A1 = −6.928 = −A 2

(

i( t ) = 6.928 e −37.32 t − e − 268t

Since, v( t ) =

)

1 t ∫ i( t )dt + 60, we get C 0

v(t) = (60 + 64.53e-2.68t – 4.6412e-37.32t) V

= 20

Chapter 8, Solution 18.

When the switch is off, we have a source-free parallel RLC circuit.

ωo = α < ωo

1 LC

=

1 0.25 x1

 →

α=

= 2,

1 = 0.5 2 RC

underdamped case ω d = ω o − α 2 = 4 − 0.25 = 1.936 2

Io(0) = i(0) = initial inductor current = 20/5 = 4A Vo(0) = v(0) = initial capacitor voltage = 0 V v(t ) = e −αt ( A1 cos ω d t + A2 sin ω d t ) = e −0.5αt ( A1 cos1.936t + A2 sin 1.936t ) v(0) =0 = A1

dv = e −0.5αt (−0.5)( A1 cos1.936t + A2 sin 1.936t ) + e −0.5αt (−1.936 A1 sin 1.936t + 1.936 A2 cos1.936t ) dt (V + RI o ) dv(0) ( 0 + 4) =− o =− = −4 = −0.5 A1 + 1.936 A2 dt RC 1 Thus,

 →

A2 = −2.066

v(t ) = −2.066e −0.5t sin 1.936t

Chapter 8, Solution 19.

For t < 0, the equivalent circuit is shown in Figure (a). 10 Ω

i +

120V

+ −

i

+ v

L

− (a)

i(0) = 120/10 = 12, v(0) = 0

v

C

− (b)

For t > 0, we have a series RLC circuit as shown in Figure (b) with R = 0 = α. 1

ωo =

LC

=

1 4

= 0.5 = ωd

i(t) = [Acos0.5t + Bsin0.5t], i(0) = 12 = A v = -Ldi/dt, and -v/L = di/dt = 0.5[-12sin0.5t + Bcos0.5t], which leads to -v(0)/L = 0 = B Hence,

i(t) = 12cos0.5t A and v = 0.5

However, v = -Ldi/dt = -4(0.5)[-12sin0.5t] = 24sin0.5t V Chapter 8, Solution 20.

For t < 0, the equivalent circuit is as shown below. 2Ω

i

12 + −

−

vC

+

v(0) = -12V and i(0) = 12/2 = 6A For t > 0, we have a series RLC circuit. α = R/(2L) = 2/(2x0.5) = 2 ωo = 1/ LC = 1 / 0.5x 1 4 = 2 2 Since α is less than ωo, we have an under-damped response. ωd = ωo2 − α 2 = 8 − 4 = 2

i(t) = (Acos2t + Bsin2t)e-2t i(0) = 6 = A

di/dt = -2(6cos2t + Bsin2t)e-2t + (-2x6sin2t + 2Bcos2t)e-αt di(0)/dt = -12 + 2B = -(1/L)[Ri(0) + vC(0)] = -2[12 – 12] = 0 Thus, B = 6 and i(t) = (6cos2t + 6sin2t)e-2t A Chapter 8, Solution 21.

By combining some resistors, the circuit is equivalent to that shown below. 60||(15 + 25) = 24 ohms. 12 Ω

24V

6Ω

t=0

i

3H

+ −

24 Ω

+ (1/27)F

v

− At t = 0-,

i(0) = 0, v(0) = 24x24/36 = 16V

For t > 0, we have a series RLC circuit.

R = 30 ohms, L = 3 H, C = (1/27) F

α = R/(2L) = 30/6 = 5 ωo = 1 / LC = 1 / 3x1 / 27 = 3, clearly α > ωo (overdamped response)

s1,2 = − α ± α 2 − ωo2 = −5 ± 5 2 − 3 2 = -9, -1 v(t) = [Ae-t + Be-9t], v(0) = 16 = A + B

(1)

i = Cdv/dt = C[-Ae-t - 9Be-9t] i(0) = 0 = C[-A – 9B] or A = -9B From (1) and (2),

B = -2 and A = 18. Hence,

v(t) = (18e-t – 2e-9t) V

(2)

Chapter 8, Solution 22. α = 20 = 1/(2RC) or RC = 1/40

(1)

ωd = 50 = ωo2 − α 2 which leads to 2500 + 400 = ωo2 = 1/(LC)

Thus, LC 1/2900

(2)

In a parallel circuit, vC = vL = vR But,

iC = CdvC/dt or iC/C = dvC/dt = -80e-20tcos50t – 200e-20tsin50t + 200e-20tsin50t – 500e-20tcos50t = -580e-20tcos50t iC(0)/C = -580 which leads to C = -6.5x10-3/(-580) = 11.21 µF R = 1/(40C) = 106/(2900x11.21) = 2.23 kohms L = 1/(2900x11.21) = 30.76 H

Chapter 8, Solution 23. Let Co = C + 0.01. For a parallel RLC circuit, α = 1/(2RCo), ωo = 1/ LC o α = 1 = 1/(2RCo), we then have Co = 1/(2R) = 1/20 = 50 mF ωo = 1/ 0.5x 0.5 = 6.32 > α (underdamped) Co = C + 10 mF = 50 mF or 40 mF Chapter 8, Solution 24. For t < 0, u(-t) 1, namely, the switch is on. v(0) = 0, i(0) = 25/5 = 5A For t > 0, the voltage source is off and we have a source-free parallel RLC circuit. α = 1/(2RC) = 1/(2x5x10-3) = 100

ωo = 1/ LC = 1 / 0.1x10 −3 = 100 ωo = α (critically damped) v(t) = [(A1 + A2t)e-100t] v(0) = 0 = A1 dv(0)/dt = -[v(0) + Ri(0)]/(RC) = -[0 + 5x5]/(5x10-3) = -5000 But,

dv/dt = [(A2 + (-100)A2t)e-100t]

Therefore, dv(0)/dt = -5000 = A2 – 0 v(t) = -5000te-100t V Chapter 8, Solution 25. In the circuit in Fig. 8.76, calculate io(t) and vo(t) for t>0. 1H

2Ω

30V

+ −

io(t) +

t=0, note this is a make before break switch so the inductor current is not interrupted.

Figure 8.78

8Ω

For Problem 8.25.

At t = 0-, vo(0) = (8/(2 + 8)(30) = 24 For t > 0, we have a source-free parallel RLC circuit. α = 1/(2RC) = ¼ ωo = 1/ LC = 1 / 1x 1 4 = 2 Since α is less than ωo, we have an under-damped response. ωd = ωo2 − α 2 = 4 − (1 / 16) = 1.9843

vo(t) = (A1cosωdt + A2sinωdt)e-αt

(1/4)F

vo(t) −

vo(0) = 24 = A1 and io(t) = C(dvo/dt) = 0 when t = 0. dvo/dt = -α(A1cosωdt + A2sinωdt)e-αt + (-ωdA1sinωdt + ωdA2cosωdt)e-αt at t = 0, we get dvo(0)/dt = 0 = -αA1 + ωdA2 Thus, A2 = (α/ωd)A1 = (1/4)(24)/1.9843 = 3.024 vo(t) = (24cosωdt + 3.024sinωdt)e-t/4 volts Chapter 8, Solution 26. s2 + 2s + 5 = 0, which leads to s1,2 =

− 2 ± 4 − 20 = -1±j4 2

i(t) = Is + [(A1cos4t + A2sin4t)e-t], Is = 10/5 = 2 i(0) = 2 = = 2 + A1, or A1 = 0 di/dt = [(A2cos4t)e-t] + [(-A2sin4t)e-t] = 4 = 4A2, or A2 = 1 i(t) = 2 + sin4te-t A Chapter 8, Solution 27.

s2 + 4s + 8 = 0 leads to s =

− 4 ± 16 − 32 = −2 ± j2 2

v(t) = Vs + (A1cos2t + A2sin2t)e-2t 8Vs = 24 means that Vs = 3 v(0) = 0 = 3 + A1 leads to A1 = -3 dv/dt = -2(A1cos2t + A2sin2t)e-2t + (-2A1sin2t + 2A2cos2t)e-2t 0 = dv(0)/dt = -2A1 +2A2 or A2 = A1 = -3 v(t) = [3 – 3(cos2t + sin2t)e-2t] volts

Chapter 8, Solution 28.

The characteristic equation is s2 + 6s + 8 with roots − 6 ± 36 − 32 s1, 2 = = −4,−2 2 Hence,

i (t ) = I s + Ae −2t + Be −4t 8I s = 12

 →

i (0) = 0

 →

I s = 1.5 0 = 1.5 + A + B

(1)

di = −2 Ae − 2t − 4 Be − 4t dt di(0) = 2 = −2 A − 4 B  → 0 = 1 + A + 2 B dt Solving (1) and (2) leads to A=-2 and B=0.5.

(2)

i (t ) = 1.5 − 2e −2t + 0.5e −4t A

Chapter 8, Solution 29.

(a)

s2 + 4 = 0 which leads to s1,2 = ±j2 (an undamped circuit) v(t) = Vs + Acos2t + Bsin2t 4Vs = 12 or Vs = 3 v(0) = 0 = 3 + A or A = -3 dv/dt = -2Asin2t + 2Bcos2t dv(0)/dt = 2 = 2B or B = 1, therefore v(t) = (3 – 3cos2t + sin2t) V

(b)

s2 + 5s + 4 = 0 which leads to s1,2 = -1, -4 i(t) = (Is + Ae-t + Be-4t) 4Is = 8 or Is = 2 i(0) = -1 = 2 + A + B, or A + B = -3

(1)

di/dt = -Ae-t - 4Be-4t di(0)/dt = 0 = -A – 4B, or B = -A/4 From (1) and (2) we get A = -4 and B = 1 i(t) = (2 – 4e-t + e-4t) A (c)

s2 + 2s + 1 = 0, s1,2 = -1, -1 v(t) = [Vs + (A + Bt)e-t], Vs = 3. v(0) = 5 = 3 + A or A = 2 dv/dt = [-(A + Bt)e-t] + [Be-t] dv(0)/dt = -A + B = 1 or B = 2 + 1 = 3 v(t) = [3 + (2 + 3t)e-t] V

Chapter 8, Solution 30.

s1 = −500 = −α + α 2 − ω o ,

s 2 = −800 = −α − α 2 − ω o

2

s1 + s 2 = −1300 = −2α

 →

α = 650 =

2

R 2L

Hence, L= s1 − s 2 = 300 = 2 α 2 − ω o

R 200 = = 153.8 mH 2α 2 x650 2

C=

 →

ω o = 623.45 =

1 = 16.25µF (632.45) 2 L

1 LC

(2)

Chapter 8, Solution 31.

For t = 0-, we have the equivalent circuit in Figure (a). For t = 0+, the equivalent circuit is shown in Figure (b). By KVL, v(0+) = v(0-) = 40, i(0+) = i(0-) = 1 By KCL, 2 = i(0+) + i1 = 1 + i1 which leads to i1 = 1. By KVL, -vL + 40i1 + v(0+) = 0 which leads to vL(0+) = 40x1 + 40 = 80 vL(0+) = 80 V, 40 Ω i

vC(0+) = 40 V

10 Ω

i1 40 Ω

+

+

+

v

50V

−

+ −

v

vL

−

10 Ω

−

0.5H

(a)

50V

(b)

Chapter 8, Solution 32.

For t = 0-, the equivalent circuit is shown below. 2A

i +

v

−

6Ω

i(0-) = 0, v(0-) = -2x6 = -12V For t > 0, we have a series RLC circuit with a step input. α = R/(2L) = 6/2 = 3, ωo = 1/ LC = 1 / 0.04 s = − 3 ± 9 − 25 = −3 ± j4 Thus, v(t) = Vf + [(Acos4t + Bsin4t)e-3t]

+ −

where Vf = final capacitor voltage = 50 V v(t) = 50 + [(Acos4t + Bsin4t)e-3t] v(0) = -12 = 50 + A which gives A = -62 i(0) = 0 = Cdv(0)/dt dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t] 0 = dv(0)/dt = -3A + 4B or B = (3/4)A = -46.5 v(t) = {50 + [(-62cos4t – 46.5sin4t)e-3t]} V Chapter 8, Solution 33.

We may transform the current sources to voltage sources. For t = 0-, the equivalent circuit is shown in Figure (a). 10 Ω

i

i

+ 30V

+ −

1H

+

v

5Ω

−

v

10 Ω

30V

4F

+ −

−

(a)

(b)

i(0) = 30/15 = 2 A, v(0) = 5x30/15 = 10 V For t > 0, we have a series RLC circuit. α = R/(2L) = 5/2 = 2.5 ω o = 1 / LC = 1 / 4 = 0.25, clearly α > ωo (overdamped response)

s1,2 = − α ± α 2 − ω 2o = −2.5 ± 6.25 − 0.25 = -4.95, -0.05 v(t) = Vs + [A1e-4.95t + A2e-0.05t], v = 20. v(0) = 10 = 20 + A1 + A2

(1)

i(0) = Cdv(0)/dt or dv(0)/dt = 2/4 = 1/2 Hence,

½ = -4.95A1 – 0.05A2

From (1) and (2),

A1 = 0, A2 = -10.

(2)

v(t) = {20 – 10e-0.05t} V Chapter 8, Solution 34.

Before t = 0, the capacitor acts like an open circuit while the inductor behaves like a short circuit. i(0) = 0, v(0) = 20 V For t > 0, the LC circuit is disconnected from the voltage source as shown below. Vx + − i

(1/16)F (¼) H

This is a lossless, source-free, series RLC circuit. α = R/(2L) = 0, ωo = 1/ LC = 1/

1 1 + = 8, s = ±j8 16 4

Since α is less than ωo, we have an underdamped response. Therefore, i(t) = A1cos8t + A2sin8t where i(0) = 0 = A1 di(0)/dt = (1/L)vL(0) = -(1/L)v(0) = -4x20 = -80 However, di/dt = 8A2cos8t, thus, di(0)/dt = -80 = 8A2 which leads to A2 = -10 Now we have

i(t) = -10sin8t A

Chapter 8, Solution 35. At t = 0-, iL(0) = 0, v(0) = vC(0) = 8 V For t > 0, we have a series RLC circuit with a step input. α = R/(2L) = 2/2 = 1, ωo = 1/ LC = 1/ 1 / 5 =

5

s1,2 = − α ± α 2 − ω 2o = −1 ± j2 v(t) = Vs + [(Acos2t + Bsin2t)e-t], Vs = 12. v(0) = 8 = 12 + A or A = -4, i(0) = Cdv(0)/dt = 0. But dv/dt = [-(Acos2t + Bsin2t)e-t] + [2(-Asin2t + Bcos2t)e-t] 0

= dv(0)/dt = -A + 2B or 2B = A = -4 and B = -2 v(t) = {12 – (4cos2t + 2sin2t)e-t V.

Chapter 8, Solution 36. For t = 0-, 3u(t) = 0. Thus, i(0) = 0, and v(0) = 20 V. For t > 0, we have the series RLC circuit shown below. 10 Ω

i

10 Ω

5H

+ 15V

+ −

2Ω

20 V

0.2 F

+ −

v −

α = R/(2L) = (2 + 5 + 1)/(2x5) = 0.8 ωo = 1/ LC = 1/ 5x 0.2 = 1

s1,2 = − α ± α 2 − ω2o = −0.8 ± j0.6 v(t) = Vs + [(Acos0.6t + Bsin0.6t)e-0.8t] Vs = 15 + 20 = 35V and v(0) = 20 = 35 + A or A = -15 i(0) = Cdv(0)/dt = 0 But dv/dt = [-0.8(Acos0.6t + Bsin0.6t)e-0.8t] + [0.6(-Asin0.6t + Bcos0.6t)e-0.8t] 0

= dv(0)/dt = -0.8A + 0.6B which leads to B = 0.8x(-15)/0.6 = -20 v(t) = {35 – [(15cos0.6t + 20sin0.6t)e-0.8t]} V

i = Cdv/dt = 0.2{[0.8(15cos0.6t + 20sin0.6t)e-0.8t] + [0.6(15sin0.6t – 20cos0.6t)e-0.8t]} i(t) = [(5sin0.6t)e-0.8t] A Chapter 8, Solution 37. For t = 0-, the equivalent circuit is shown below.

+

i2

6Ω

6Ω

6Ω v(0) 30V

+ −

i1

10V

+ −

−

18i2 – 6i1 = 0 or i1 = 3i2

(1)

-30 + 6(i1 – i2) + 10 = 0 or i1 – i2 = 10/3

(2)

From (1) and (2).

i1 = 5, i2 = 5/3 i(0) = i1 = 5A -10 – 6i2 + v(0) = 0

v(0) = 10 + 6x5/3 = 20 For t > 0, we have a series RLC circuit. R = 6||12 = 4 ωo = 1/ LC = 1/ (1 / 2)(1 / 8) = 4 α = R/(2L) = (4)/(2x(1/2)) = 4 α = ωo, therefore the circuit is critically damped v(t) = Vs +[(A + Bt)e-4t], and Vs = 10

v(0) = 20 = 10 + A, or A = 10 i = Cdv/dt = -4C[(A + Bt)e-4t] + C[(B)e-4t] i(0) = 5 = C(-4A + B) which leads to 40 = -40 + B or B = 80 i(t) = [-(1/2)(10 + 80t)e-4t] + [(10)e-4t] i(t) = [(5 – 40t)e-4t] A Chapter 8, Solution 38. At t = 0-, the equivalent circuit is as shown. 2A + i 10 Ω v

i1 5Ω

−

10 Ω

i(0) = 2A, i1(0) = 10(2)/(10 + 15) = 0.8 A v(0) = 5i1(0) = 4V For t > 0, we have a source-free series RLC circuit. R = 5||(10 + 10) = 4 ohms ωo = 1/ LC = 1/ (1 / 3)(3 / 4) = 2 α = R/(2L) = (4)/(2x(3/4)) = 8/3 s1,2 = − α ± α 2 − ω 2o = -4.431, -0.903 i(t) = [Ae-4.431t + Be-0.903t] i(0) = A + B = 2

(1)

di(0)/dt = (1/L)[-Ri(0) + v(0)] = (4/3)(-4x2 + 4) = -16/3 = -5.333 Hence, -5.333 = -4.431A – 0.903B

(2)

From (1) and (2), A = 1 and B = 1. i(t) = [e-4.431t + e-0.903t] A Chapter 8, Solution 39. For t = 0-, the equivalent circuit is shown in Figure (a). Where 60u(-t) = 60 and 30u(t) = 0. 30 Ω

60V

+ −

+ v − 20 Ω

(a)

30 Ω

0.5F

0.25H

20 Ω 30V

(b)

v(0) = (20/50)(60) = 24 and i(0) = 0

+ −

For t > 0, the circuit is shown in Figure (b). R = 20||30 = 12 ohms ωo = 1/ LC = 1/ (1 / 2)(1 / 4) =

8

α = R/(2L) = (12)/(0.5) = 24 Since α > ωo, we have an overdamped response. s1,2 = − α ± α 2 − ω 2o = -47.833, -0.167 v(t) = Vs + [Ae-47.833t + Be-0.167t], Vs = 30

Thus,

v(0) = 24 = 30 + A + B or -6 = A + B

(1)

i(0) = Cdv(0)/dt = 0 But,

dv(0)/dt = -47.833A – 0.167B = 0 B = -286.43A

From (1) and (2),

(2)

A = 0.021 and B = -6.021

v(t) = 30 + [0.021e-47.833t – 6.021e-0.167t] V Chapter 8, Solution 40. At t = 0-, vC(0) = 0 and iL(0) = i(0) = (6/(6 + 2))4 = 3A For t > 0, we have a series RLC circuit with a step input as shown below. i

0.02 F

2H +

6Ω

v

14 Ω −

24V

12V

+ −

ωo = 1/ LC = 1/ 2 x 0.02 = 5 α = R/(2L) = (6 + 14)/(2x2) = 5

+ −

Since α = ωo, we have a critically damped response. v(t) = Vs + [(A + Bt)e-5t], Vs = 24 – 12 = 12V v(0) = 0 = 12 + A or A = -12 i = Cdv/dt = C{[Be-5t] + [-5(A + Bt)e-5t]} i(0) = 3 = C[-5A + B] = 0.02[60 + B] or B = 90 Thus, i(t) = 0.02{[90e-5t] + [-5(-12 + 90t)e-5t]} i(t) = {(3 – 9t)e-5t} A Chapter 8, Solution 41. At t = 0-, the switch is open. i(0) = 0, and v(0) = 5x100/(20 + 5 + 5) = 50/3 For t > 0, we have a series RLC circuit shown in Figure (a). After source transformation, it becomes that shown in Figure (b). 10 H 4Ω 5A

20 Ω

5Ω

1H

i +

10 µF 20V

+ −

0.04F

v −

(a)

(b) ωo = 1/ LC = 1/ 1x1 / 25 = 5 α = R/(2L) = (4)/(2x1) = 2

s1,2 = − α ± α 2 − ω 2o = -2 ± j4.583 Thus,

v(t) = Vs + [(Acosωdt + Bsinωdt)e-2t], where ωd = 4.583 and Vs = 20 v(0) = 50/3 = 20 + A or A = -10/3

i(t) = Cdv/dt = C(-2) [(Acosωdt + Bsinωdt)e-2t] + Cωd[(-Asinωdt + Bcosωdt)e-2t] i(0) = 0 = -2A + ωdB B = 2A/ωd = -20/(3x4.583) = -1.455 i(t) = C{[(0cosωdt + (-2B - ωdA)sinωdt)]e-2t} = (1/25){[(2.91 + 15.2767) sinωdt)]e-2t} i(t) = {0.7275sin(4.583t)e-2t} A Chapter 8, Solution 42. For t = 0-, we have the equivalent circuit as shown in Figure (a). i(0) = i(0) = 0, and v(0) = 4 – 12 = -8V 4V − +

1Ω

5Ω

6Ω

12V + −

i

1H

+ v(0)

+ − +

12V

−

v

0.04F

−

(a) (b) For t > 0, the circuit becomes that shown in Figure (b) after source transformation. ωo = 1/ LC = 1/ 1x1 / 25 = 5 α = R/(2L) = (6)/(2) = 3 s1,2 = − α ± α 2 − ω 2o = -3 ± j4 Thus,

v(t) = Vs + [(Acos4t + Bsin4t)e-3t], Vs = -12 v(0) = -8 = -12 + A or A = 4

i = Cdv/dt, or i/C = dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t] i(0) = -3A + 4B or B = 3 v(t) = {-12 + [(4cos4t + 3sin4t)e-3t]} A

Chapter 8, Solution 43. For t>0, we have a source-free series RLC circuit.

α=

R 2L

R = 2αL = 2 x8 x0.5 = 8Ω

 →

ω d = ω o 2 − α 2 = 30 ωo =

1 LC

ω o = 900 − 64 = 836

 →

 →

C=

1

ω oL 2

=

1 = 2.392 mF 836 x0.5

Chapter 8, Solution 44.

α=

R 1000 = = 500, 2L 2 x1

ωo > α

 →

ωo =

1 LC

=

1 100 x10

−9

= 10 4

underdamped.

Chapter 8, Solution 45. ωo = 1/ LC = 1/ 1x 0.5 =

2

α = R/(2L) = (1)/(2x2x0.5) = 0.5

Since α < ωo, we have an underdamped response. s1,2 = − α ± α 2 − ω 2o = -0.5 ± j1.323 Thus,

i(t) = Is + [(Acos1.323t + Bsin1.323t)e-0.5t], Is = 4 i(0) = 1 = 4 + A or A = -3 v = vC = vL = Ldi(0)/dt = 0

di/dt = [1.323(-Asin1.323t + Bcos1.323t)e-0.5t] + [-0.5(Acos1.323t + Bsin1.323t)e-0.5t] di(0)/dt = 0 = 1.323B – 0.5A or B = 0.5(-3)/1.323 = -1.134 Thus,

i(t) = {4 – [(3cos1.323t + 1.134sin1.323t)e-0.5t]} A

Chapter 8, Solution 46.

For t = 0-, u(t) = 0, so that v(0) = 0 and i(0) = 0. For t > 0, we have a parallel RLC circuit with a step input, as shown below. +

i 8mH

5µF

v

2 kΩ

−

6mA

α = 1/(2RC) = (1)/(2x2x103 x5x10-6) = 50 ωo = 1/ LC = 1/ 8x10 3 x 5x10 −6 = 5,000

Since α < ωo, we have an underdamped response. s1,2 = − α ± α 2 − ωo2 ≅ -50 ± j5,000 Thus,

i(t) = Is + [(Acos5,000t + Bsin5,000t)e-50t], Is = 6mA i(0) = 0 = 6 + A or A = -6mA v(0) = 0 = Ldi(0)/dt

di/dt = [5,000(-Asin5,000t + Bcos5,000t)e-50t] + [-50(Acos5,000t + Bsin5,000t)e-50t] di(0)/dt = 0 = 5,000B – 50A or B = 0.01(-6) = -0.06mA Thus,

i(t) = {6 – [(6cos5,000t + 0.06sin5,000t)e-50t]} mA

Chapter 8, Solution 47.

At t = 0-, we obtain,

iL(0) = 3x5/(10 + 5) = 1A

and vo(0) = 0. For t > 0, the 20-ohm resistor is short-circuited and we have a parallel RLC circuit with a step input. α = 1/(2RC) = (1)/(2x5x0.01) = 10 ωo = 1/ LC = 1/ 1x 0.01 = 10

Since α = ωo, we have a critically damped response. s1,2 = -10 i(t) = Is + [(A + Bt)e-10t], Is = 3

Thus,

i(0) = 1 = 3 + A or A = -2 vo = Ldi/dt = [Be-10t] + [-10(A + Bt)e-10t] vo(0) = 0 = B – 10A or B = -20 Thus, vo(t) = (200te-10t) V Chapter 8, Solution 48.

For t = 0-, we obtain i(0) = -6/(1 + 2) = -2 and v(0) = 2x1 = 2. For t > 0, the voltage is short-circuited and we have a source-free parallel RLC circuit. α = 1/(2RC) = (1)/(2x1x0.25) = 2 ωo = 1/ LC = 1/ 1x 0.25 = 2

Since α = ωo, we have a critically damped response. s1,2 = -2 Thus,

i(t) = [(A + Bt)e-2t], i(0) = -2 = A v = Ldi/dt = [Be-2t] + [-2(-2 + Bt)e-2t] vo(0) = 2 = B + 4 or B = -2 Thus, i(t) = [(-2 - 2t)e-2t] A and v(t) = [(2 + 4t)e-2t] V

Chapter 8, Solution 49.

For t = 0-, i(0) = 3 + 12/4 = 6 and v(0) = 0. For t > 0, we have a parallel RLC circuit with a step input. α = 1/(2RC) = (1)/(2x5x0.05) = 2 ωo = 1/ LC = 1/ 5x 0.05 = 2

Since α = ωo, we have a critically damped response. s1,2 = -2 i(t) = Is + [(A + Bt)e-2t], Is = 3

Thus,

i(0) = 6 = 3 + A or A = 3 v = Ldi/dt or v/L = di/dt = [Be-2t] + [-2(A + Bt)e-2t] v(0)/L = 0 = di(0)/dt = B – 2x3 or B = 6 Thus, i(t) = {3 + [(3 + 6t)e-2t]} A Chapter 8, Solution 50.

For t = 0-, 4u(t) = 0, v(0) = 0, and i(0) = 30/10 = 3A. For t > 0, we have a parallel RLC circuit. i

+ 3A

10 Ω

10 mF 6A

40 Ω

v −

Is = 3 + 6 = 9A and R = 10||40 = 8 ohms α = 1/(2RC) = (1)/(2x8x0.01) = 25/4 = 6.25 ωo = 1/ LC = 1/ 4x 0.01 = 5

Since α > ωo, we have a overdamped response. s1,2 = − α ± α 2 − ω o2 = -10, -2.5

10 H

i(t) = Is + [Ae-10t] + [Be-2.5t], Is = 9

Thus,

i(0) = 3 = 9 + A + B or A + B = -6 di/dt = [-10Ae-10t] + [-2.5Be-2.5t], v(0) = 0 = Ldi(0)/dt or di(0)/dt = 0 = -10A – 2.5B or B = -4A Thus, A = 2 and B = -8 Clearly, i(t) = { 9 + [2e-10t] + [-8e-2.5t]} A Chapter 8, Solution 51.

Let i = inductor current and v = capacitor voltage. At t = 0, v(0) = 0 and i(0) = io. For t > 0, we have a parallel, source-free LC circuit (R = ∞). α = 1/(2RC) = 0 and ωo = 1/ LC which leads to s1,2 = ± jωo

v = Acosωot + Bsinωot, v(0) = 0 A iC = Cdv/dt = -i dv/dt = ωoBsinωot = -i/C dv(0)/dt = ωoB = -io/C therefore B = io/(ωoC) v(t) = -(io/(ωoC))sinωot V where ωo =

LC

Chapter 8, Solution 52.

α = 300 =

1 2 RC

ω d = ω o 2 − α 2 = 400

(1)  →

ω o = 400 2 − 300 2 = 264.575 =

From (2), C=

1 = 285.71µF (264.575) 2 x50 x10 −3

From (1), R=

1 1 = (3500) = 5.833Ω 2αC 2 x300

1 LC

(2)

Chapter 8, Solution 53. C1 + + −

vS

v1

R2 −

i1

+ R1

C2

i2

vo −

i2 = C2dvo/dt

(1)

i1 = C1dv1/dt

(2)

0 = R2i2 + R1(i2 – i1) +vo

(3)

Substituting (1) and (2) into (3) we get, 0 = R2C2dvo/dt + R1(C2dvo/dt – C1dv1/dt)

(4)

Applying KVL to the outer loop produces, vs = v1 + i2R2 + vo = v1 + R2C2dvo/dt + vo, which leads to v1 = vs – vo – R2C2dvo/dt

(5)

Substituting (5) into (4) leads to, 0 = R1C2dvo/dt + R1C2dvo/dt – R1C1(dvs/dt – dvo/dt – R2C2d2vo/dt2) Hence, (R1C1R2C2)(d2vo/dt2) + (R1C1 + R2C2 +R1C2)(dvo/dt) = R1C1(dvs/dt) Chapter 8, Solution 54. Let i be the inductor current.

v dv + 0.5 4 dt di v = 2i + dt Substituting (1) into (2) gives −i =

(1)

(2)

−v =

v dv 1 dv 1 d 2 v + + + 2 dt 4 dt 2 dt 2

s 2 + 2.5s + 3 = 0

 →

d 2v dv + 2.5 + 3v = 0 2 dt dt

 →

s = −1.25 ± j1.199

v = Ae −1.25t cos1.199t + Be −1.25t sin 1.199t v(0) = 2=A. Let w=1.199 dv = −1.25( Ae −1.25t cos wt + Be −1.25t sin wt ) + w(− Ae −1.25t sin wt + Be −1.25t cos wt ) dt dv(0) = 0 = −1.25 A + Bw dt

 →

B=

1.25 X 2 = 2.085 1.199

v = 2e −1.25t cos1.199t + 2.085e −1.25t sin 1.199t V

Chapter 8, Solution 55.

At the top node, writing a KCL equation produces, i/4 +i = C1dv/dt, C1 = 0.1 5i/4 = C1dv/dt = 0.1dv/dt i = 0.08dv/dt But,

(1)

v = − (2i + (1 / C 2 ) ∫ idt ) , C2 = 0.5 or

-dv/dt = 2di/dt + 2i

(2)

Substituting (1) into (2) gives, -dv/dt = 0.16d2v/dt2 + 0.16dv/dt 0.16d2v/dt2 + 0.16dv/dt + dv/dt = 0, or d2v/dt2 + 7.25dv/dt = 0 Which leads to s2 + 7.25s = 0 = s(s + 7.25) or s1,2 = 0, -7.25 v(t) = A + Be-7.25t

(3)

v(0) = 4 = A + B

(4)

From (1),

i(0) = 2 = 0.08dv(0+)/dt or dv(0+)/dt = 25

But,

dv/dt = -7.25Be-7.25t, which leads to,

dv(0)/dt = -7.25B = 25 or B = -3.448 and A = 4 – B = 4 + 3.448 = 7.448 Thus, v(t) = {7.45 – 3.45e-7.25t} V Chapter 8, Solution 56.

For t < 0, i(0) = 0 and v(0) = 0. For t > 0, the circuit is as shown below. 4Ω i

6Ω

i

0.04F + −

20

io

0.25H

Applying KVL to the larger loop, -20 +6io +0.25dio/dt + 25 ∫ (i o + i)dt = 0 Taking the derivative, 6dio/dt + 0.25d2io/dt2 + 25(io + i) = 0

For the smaller loop,

4 + 25 ∫ (i + i o )dt = 0

Taking the derivative,

25(i + io) = 0 or i = -io

From (1) and (2)

6dio/dt + 0.25d2io/dt2 = 0 This leads to, 0.25s2 + 6s = 0 or s1,2 = 0, -24

(1)

(2)

io(t) = (A + Be-24t) and io(0) = 0 = A + B or B = -A As t approaches infinity, io(∞) = 20/10 = 2 = A, therefore B = -2 Thus, io(t) = (2 - 2e-24t) = -i(t) or i(t) = (-2 + 2e-24t) A Chapter 8, Solution 57.

(a) Let v = capacitor voltage and i = inductor current. At t = 0-, the switch is closed and the circuit has reached steady-state. v(0-) = 16V and i(0-) = 16/8 = 2A At t = 0+, the switch is open but, v(0+) = 16 and i(0+) = 2. We now have a source-free RLC circuit. R 8 + 12 = 20 ohms, L = 1H, C = 4mF. α = R/(2L) = (20)/(2x1) = 10 ωo = 1/ LC = 1/ 1x (1 / 36) = 6 Since α > ωo, we have a overdamped response. s1,2 = − α ± α 2 − ωo2 = -18, -2 Thus, the characteristic equation is (s + 2)(s + 18) = 0 or s2 + 20s +36 = 0. (b)

i(t) = [Ae-2t + Be-18t] and i(0) = 2 = A + B To get di(0)/dt, consider the circuit below at t = 0+. i 12 Ω +

+ (1/36)F

v

−

8Ω vL

−

1H

-v(0) + 20i(0) + vL(0) = 0, which leads to,

(1)

-16 + 20x2 + vL(0) = 0 or vL(0) = -24 Ldi(0)/dt = vL(0) which gives di(0)/dt = vL(0)/L = -24/1 = -24 A/s Hence -24 = -2A – 18B or 12 = A + 9B From (1) and (2),

(2)

B = 1.25 and A = 0.75

i(t) = [0.75e-2t + 1.25e-18t] = -ix(t) or ix(t) = [-0.75e-2t - 1.25e-18t] A v(t) = 8i(t) = [6e-2t + 10e-18t] A Chapter 8, Solution 58.

(a) Let i =inductor current, v = capacitor voltage i(0) =0, v(0) = 4 [v(0) + Ri(0)] (4 + 0) dv(0) =− =− = − 8 V/s dt RC 0.5 (b) For t ≥ 0 , the circuit is a source-free RLC parallel circuit.

α=

1 1 = = 1, 2 RC 2 x0.5 x1

ωo =

1 LC

=

1 0.25 x1

=2

ω d = ω 2 o − α 2 = 4 − 1 = 1.732 Thus, v(t ) = e − t ( A1 cos1.732t + A2 sin 1.732t ) v(0) = 4 = A1 dv = −e −t A1 cos1.732t − 1.732e −t A1 sin 1.732t − e −t A2 sin 1.732t + 1.732e −t A2 cos1.732t dt dv(0) = −8 = − A1 + 1.732 A2  → A2 = −2.309 dt v(t ) = e − t (4 cos 1.732t − 2.309 sin 1.732t ) V

Chapter 8, Solution 59.

Let i = inductor current and v = capacitor voltage v(0) = 0, i(0) = 40/(4+16) = 2A For t>0, the circuit becomes a source-free series RLC with R 16 1 1 = = 2, ω o = = = 2,  → α = ω o = 2 2L 2 x4 4 x1 / 16 LC i (t ) = Ae −2t + Bte −2t i(0) = 2 = A di = −2 Ae − 2t + Be − 2t − 2 Bte − 2t dt 1 1 di (0) = −2 A + B = − [ Ri(0) + v(0)]  → − 2 A + B = − (32 + 0), dt L 4

α=

B = −4

i (t ) = 2e −2t − 4te −2t t

v=

t

t

1 idt + v(0) = 32 ∫ e − 2t dt − 64 ∫ te − 2t dt = −16e − 2t ∫ C0 0 0

t

−

0

64 − 2t e (−2t − 1) 4

t 0

v = 32te −2t V

Chapter 8, Solution 60.

At t = 0-, 4u(t) = 0 so that i1(0) = 0 = i2(0)

(1)

Applying nodal analysis, 4 = 0.5di1/dt + i1 + i2 Also,

i2 = [1di1/dt – 1di2/dt]/3 or 3i2 = di1/dt – di2/dt

Taking the derivative of (2), 0 = d2i1/dt2 + 2di1/dt + 2di2/dt From (2) and (3),

(2) (3) (4)

di2/dt = di1/dt – 3i2 = di1/dt – 3(4 – i1 – 0.5di1/dt) = di1/dt – 12 + 3i1 + 1.5di1/dt

Substituting this into (4), d2i1/dt2 + 7di1/dt + 6i1 = 24 which gives s2 + 7s + 6 = 0 = (s + 1)(s + 6)

Thus, i1(t) = Is + [Ae-t + Be-6t], 6Is = 24 or Is = 4 i1(t) = 4 + [Ae-t + Be-6t] and i1(0) = 4 + [A + B]

(5)

i2 = 4 – i1 – 0.5di1/dt = i1(t) = 4 + -4 - [Ae-t + Be-6t] – [-Ae-t - 6Be-6t] = [-0.5Ae-t + 2Be-6t] and i2(0) = 0 = -0.5A + 2B From (5) and (6),

(6)

A = -3.2 and B = -0.8 i1(t) = {4 + [-3.2e-t – 0.8e-6t]} A i2(t) = [1.6e-t – 1.6e-6t] A

Chapter 8, Solution 61.

For t > 0, we obtain the natural response by considering the circuit below. 1H

a

iL

+ 4Ω

vC

0.25F

6Ω

− At node a,

vC/4 + 0.25dvC/dt + iL = 0

(1)

But,

vC = 1diL/dt + 6iL

(2)

Combining (1) and (2), (1/4)diL/dt + (6/4)iL + 0.25d2iL/dt2 + (6/4)diL/dt + iL = 0 d2iL/dt2 + 7diL/dt + 10iL = 0 s2 + 7s + 10 = 0 = (s + 2)(s + 5) or s1,2 = -2, -5 Thus, iL(t) = iL(∞) + [Ae-2t + Be-5t], where iL(∞) represents the final inductor current = 4(4)/(4 + 6) = 1.6 iL(t) = 1.6 + [Ae-2t + Be-5t] and iL(0) = 1.6 + [A+B] or -1.6 = A+B diL/dt = [-2Ae-2t - 5Be-5t]

(3)

and diL(0)/dt = 0 = -2A – 5B or A = -2.5B

(4)

From (3) and (4), A = -8/3 and B = 16/15 iL(t) = 1.6 + [-(8/3)e-2t + (16/15)e-5t] v(t) = 6iL(t) = {9.6 + [-16e-2t + 6.4e-5t]} V vC = 1diL/dt + 6iL = [ (16/3)e-2t - (16/3)e-5t] + {9.6 + [-16e-2t + 6.4e-5t]} vC = {9.6 + [-(32/3)e-2t + 1.0667e-5t]} i(t) = vC/4 = {2.4 + [-2.667e-2t + 0.2667e-5t]} A Chapter 8, Solution 62.

This is a parallel RLC circuit as evident when the voltage source is turned off. α = 1/(2RC) = (1)/(2x3x(1/18)) = 3 ωo = 1/ LC = 1/ 2x1 / 18 = 3 Since α = ωo, we have a critically damped response. s1,2 = -3 Let v(t) = capacitor voltage Thus, v(t) = Vs + [(A + Bt)e-3t] where Vs = 0 But -10 + vR + v = 0 or vR = 10 – v Therefore vR = 10 – [(A + Bt)e-3t] where A and B are determined from initial conditions.

Chapter 8, Solution 63.

R

v1

vs

+

R vo

v2 C

C

At node 1, v s − v1 dv =C 1 R dt

(1)

At node 2, dv v2 − vo =C o (2) dt R As a voltage follower, v1 = v 2 = v . Hence (2) becomes dv (3) v = v o + RC o dt and (1) becomes dv v s = v + RC (4) dt Substituting (3) into (4) gives

v s = vo + RC

dvo dv d 2 vo + RC o + R 2 C 2 dt dt dt 2

or R 2C 2

d 2 vo dv + 2 RC o + vo = v s 2 dt dt

Chapter 8, Solution 64. C2 R2 R1 vs

1 v1

C1

2

− +

vo

At node 1,

(vs – v1)/R1 = C1 d(v1 – 0)/dt or vs = v1 + R1C1dv1/dt

At node 2,

C1dv1/dt = (0 – vo)/R2 + C2d(0 – vo)/dt or

From (1) and (2), or

(1)

–R2C1dv1/dt = vo + C2dvo/dt

(2)

(vs – v1)/R1 = C1 dv1/dt = -(1/R2)(vo + C2dvo/dt) v1 = vs + (R1/R2)(vo + C2dvo/dt)

(3)

Substituting (3) into (1) produces, vs = vs + (R1/R2)(vo + C2dvo/dt) + R1C1d{vs + (R1/R2)(vo + C2dvo/dt)}/dt = vs + (R1/R2)(vo)+ (R1C2/R2) dvo/dt) + R1C1dvs/dt + (R1R1C1/R2)dvo/dt + (R12 C1C2/R2)[d2vo/dt2] Simplifying we get, d2vo/dt2 + [(1/ R1C1) + (1/ C2)]dvo/dt + [1/(R1C1C2)](vo) = - [R2/(R1C2)]dvs/dt Chapter 8, Solution 65. At the input of the first op amp, (vo – 0)/R = Cd(v1 – 0)

(1)

At the input of the second op amp, (-v1 – 0)/R = Cdv2/dt

(2)

Let us now examine our constraints. Since the input terminals are essentially at ground, then we have the following, vo = -v2 or v2 = -vo Combining (1), (2), and (3), eliminating v1 and v2 we get, d 2 vo  1  d 2vo − 100 v o = 0 − v o = dt 2  R 2 C 2  dt 2

(3)

Which leads to s2 – 100 = 0 Clearly this produces roots of –10 and +10. And, we obtain, vo(t) = (Ae+10t + Be-10t)V At t = 0, vo(0+) = – v2(0+) = 0 = A + B, thus B = –A This leads to vo(t) = (Ae+10t – Ae-10t)V. Now we can use v1(0+) = 2V. From (2), v1 = –RCdv2/dt = 0.1dvo/dt = 0.1(10Ae+10t + 10Ae-10t) v1(0+) = 2 = 0.1(20A) = 2A or A = 1 Thus, vo(t) = (e+10t – e-10t)V It should be noted that this circuit is unstable (clearly one of the poles lies in the righthalf-plane). Chapter 8, Solution 66. C2

vS

R2

R1

2

+ –

1

vo R4

C1

Note that the voltage across C1 is

R3

v2 = [R3/(R3 + R4)]vo

This is the only difference between this problem and Example 8.11, i.e. v = kv, where k = [R3/(R3 + R4)].

At node 1, (vs – v1)/R1 = C2[d(v1 – vo)/dt] + (v1 – v2)/R2 vs/R1 = (v1/R1) + C2[d(v1)/dt] – C2[d(vo)/dt] + (v1 – kvo)/R2

(1)

At node 2, (v1 – kvo)/R2 = C1[d(kvo)/dt] or

v1 = kvo + kR2C1[d(vo)/dt]

(2)

Substituting (2) into (1), vs/R1 = (kvo/R1) + (kR2C1/R1)[d(vo)/dt] + kC2[d(vo)/dt] + kR2C1C2[d2(vo)/dt2] – (kvo/R2) + kC1[d(vo)/dt] – (kvo/R2) + C2[d(vo)/dt] We now rearrange the terms. [d2(vo)/dt2] + [(1/C2R1) + (1/ R2C2) + (1/R2C1) – (1/ kR2C1)][d(vo)/dt] + [vo/(R1R2C1C2)] = vs/(kR1R2C1C2) If R1 = R2 10 kohms, C1 = C2 = 100 µF, R3 = 20 kohms, and R4 = 60 kohms, k = [R3/(R3 + R4)] = 1/3 R1R2C1C2 = 104 x104 x10-4 x10-4 = 1 (1/C2R1) + (1/ R2C2) + (1/R2C1) – (1/ kR2C1) = 1 + 1 + 1 – 3 = 3 – 3 = 0 Hence,

[d2(vo)/dt2] + vo = 3vs = 6, t > 0, and s2 + 1 = 0, or s1,2 = ±j vo(t) = Vs + [Acost + B sint], Vs = 6 vo(0) = 0 = 6 + A or A = –6 dvo/dt = –Asint + Bcost, but dvo(0)/dt = 0 = B

Hence,

vo(t) = 6(1 – cost)u(t) volts.

Chapter 8, Solution 67. At node 1, d ( v1 − v o ) d ( v 1 − 0) v in − v1 = C1 + C2 dt dt R1

At node 2,

C2

(1)

− vo d ( v 1 − 0) 0 − v o dv1 , or = = dt R2 dt C2R 2

(2)

From (1) and (2), v in − v1 = −

v1 = v in +

v dv R 1C1 dv o − R 1 C1 o − R 1 o R2 dt C 2 R 2 dt

v dv R 1C1 dv o + R 1 C1 o + R 1 o R2 dt C 2 R 2 dt

(3)

C1 R2 R1 vin

1

v1

C2 2 0V

− +

vo

From (2) and (3), −

vo d 2 v o R 1 dv o dv dv R C dv o = 1 = in + 1 1 + R 1 C1 + C2R 2 dt dt C 2 R 2 dt R 2 dt dt 2 d 2 vo vo 1  1 1  dv o 1 dv in   + =− + + 2 R 2  C1 C 2  dt C1 C 2 R 2 R 1 R 1C1 dt dt

But C1C2R1R2 = 10-4 x10-4 x104 x104 = 1 1 R2

 1 1  2 2  =  = 4 =2 + −4  C1 C 2  R 2 C1 10 x10

d 2 vo dv dv + 2 o + v o = − in 2 dt dt dt

Which leads to s2 + 2s + 1 = 0 or (s + 1)2 = 0 and s = –1, –1 Therefore,

vo(t) = [(A + Bt)e-t] + Vf

As t approaches infinity, the capacitor acts like an open circuit so that Vf = vo(∞) = 0 vin = 10u(t) mV and the fact that the initial voltages across each capacitor is 0 means that vo(0) = 0 which leads to A = 0. vo(t) = [Bte-t] dv o = [(B – Bt)e-t] dt dv o (0+ ) v (0+ ) =− o =0 dt C2R 2

From (2), From (1) at t = 0+,

dv (0+) dv o (0+ ) 1 1− 0 = −C1 o which leads to =− = −1 dt dt C1 R 1 R1

Substituting this into (4) gives B = –1 Thus,

v(t) = –te-tu(t) V

(4)

Chapter 8, Solution 68. The schematic is as shown below. The unit step is modeled by VPWL as shown. We insert a voltage marker to display V after simulation. We set Print Step = 25 ms and final step = 6s in the transient box. The output plot is shown below.

Chapter 8, Solution 69. The schematic is shown below. The initial values are set as attributes of L1 and C1. We set Print Step to 25 ms and the Final Time to 20s in the transient box. A current marker is inserted at the terminal of L1 to automatically display i(t) after simulation. The result is shown below.

Chapter 8, Solution 70. The schematic is shown below.

After the circuit is saved and simulated, we obtain the capacitor voltage v(t) as shown below.

Chapter 8, Solution 71. The schematic is shown below. We use VPWL and IPWL to model the 39 u(t) V and 13 u(t) A respectively. We set Print Step to 25 ms and Final Step to 4s in the Transient box. A voltage marker is inserted at the terminal of R2 to automatically produce the plot of v(t) after simulation. The result is shown below.

Chapter 8, Solution 72. When the switch is in position 1, we obtain IC=10 for the capacitor and IC=0 for the inductor. When the switch is in position 2, the schematic of the circuit is shown below.

When the circuit is simulated, we obtain i(t) as shown below.

Chapter 8, Solution 73. (a)

For t < 0, we have the schematic below. When this is saved and simulated, we obtain the initial inductor current and capacitor voltage as iL(0) = 3 A and vc(0) = 24 V.

(b) For t > 0, we have the schematic shown below. To display i(t) and v(t), we insert current and voltage markers as shown. The initial inductor current and capacitor voltage are also incorporated. In the Transient box, we set Print Step = 25 ms and the Final Time to 4s. After simulation, we automatically have io(t) and vo(t) displayed as shown below.

Chapter 8, Solution 74. 10 Ω

5Ω

+ 20 V -

2F

4H

Hence the dual circuit is shown below. 2H

20A

0.1 Ω

4F

0.2 Ω

Chapter 8, Solution 75. The dual circuit is connected as shown in Figure (a). It is redrawn in Figure (b). 0.1 Ω 12V

+ −

10 Ω

12A

24A

0.5 F 24V 0.25 Ω

4Ω

10 H

10 H 10 µF

(a) 0.1 Ω 2F 0.5 H

24A

12A 0.25 Ω

(b)

+ −

Chapter 8, Solution 76. The dual is obtained from the original circuit as shown in Figure (a). It is redrawn in Figure (b). 0.1 Ω

0.05 Ω

1/3 Ω

10 Ω

20 Ω

60 A

30 Ω 120 A

+ −

– +

60 V

2V

120 V + −

4H

1F

1H

2A

4F

(a) 0.05 Ω

60 A

120 A

1H

0.1 Ω

1/30 Ω 1/4 F 2V (b)

+ −

Chapter 8, Solution 77. The dual is constructed in Figure (a) and redrawn in Figure (b). – +

5A

5V

2Ω

1/3 Ω

1/2 Ω 1F

1Ω

1/4 H

1H

3Ω

12V

1Ω

1/4 F

+ −

12 A

(a)

1Ω 2Ω 1/4 F

1/3 Ω 12 A

1H

5V

+ −

(b)

Chapter 8, Solution 78. The voltage across the igniter is vR = vC since the circuit is a parallel RLC type. vC(0) = 12, and iL(0) = 0. α = 1/(2RC) = 1/(2x3x1/30) = 5 ωo = 1 / LC = 1 / 60 x10 −3 x1 / 30 = 22.36 α < ωo produces an underdamped response. s1, 2 = −α ± α 2 − ω o2 = –5 ± j21.794

vC(t) = e-5t(Acos21.794t + Bsin21.794t)

(1)

vC(0) = 12 = A dvC/dt = –5[(Acos21.794t + Bsin21.794t)e-5t] + 21.794[(–Asin21.794t + Bcos21.794t)e-5t]

(2)

dvC(0)/dt = –5A + 21.794B But,

dvC(0)/dt = –[vC(0) + RiL(0)]/(RC) = –(12 + 0)/(1/10) = –120

Hence,

–120 = –5A + 21.794B, leads to B (5x12 – 120)/21.794 = –2.753

At the peak value, dvC(to)/dt = 0, i.e., 0

= A + Btan21.794to + (A21.794/5)tan21.794to – 21.794B/5 (B + A21.794/5)tan21.794to = (21.794B/5) – A

tan21.794to = [(21.794B/5) – A]/(B + A21.794/5) = –24/49.55 = –0.484 Therefore,

21.7945to = |–0.451|

to = |–0.451|/21.794 = 20.68 ms

Chapter 8, Solution 79. For critical damping of a parallel RLC circuit,

α = ωo

 →

1 = 2 RC

1 LC

Hence,

C=

0.25 L = = 434 µF 2 4 x144 4R

Chapter 8, Solution 80.

t1 = 1/|s1| = 0.1x10-3 leads to s1 = –1000/0.1 = –10,000 t2 = 1/|s2| = 0.5x10-3 leads to s1 = –2,000 s1 = −α − α 2 − ωo2 s 2 = −α + α 2 − ω o2

s1 + s2 = –2α = –12,000, therefore α = 6,000 = R/(2L) L = R/12,000 = 60,000/12,000 = 5H s 2 = −α + α 2 − ωo2 = –2,000 α − α 2 − ωo2 = 2,000 6,000 − α 2 − ωo2 = 2,000 α 2 − ωo2 = 4,000

α2 – ωo2 = 16x106

ωo2 = α2 – 16x106 = 36x106 – 16x106 ωo = 103 20 = 1 / LC C = 1/(20x106x5) = 10 nF

Chapter 8, Solution 81. t = 1/α = 0.25 leads to α = 4 But,

α 1/(2RC) or,

C = 1/(2αR) = 1/(2x4x200) = 625 µF

ωd = ωo2 − α 2

ωo2 = ωd2 + α 2 = (2π4x10 3 ) 2 + 16 ≅ (2π4 x10 3 0 2 = 1/(LC) This results in L = 1/(64π2x106x625x10-6) = 2.533 µH Chapter 8, Solution 82.

For t = 0-, v(0) = 0. For t > 0, the circuit is as shown below. R1

a

+

+ C1

vo

R2

C2

v

−

−

At node a, (vo – v/R1 = (v/R2) + C2dv/dt vo = v(1 + R1/R2) + R1C2 dv/dt 60 = (1 + 5/2.5) + (5x106 x5x10-6)dv/dt 60 = 3v + 25dv/dt v(t) = Vs + [Ae-3t/25] where

3Vs = 60 yields Vs = 20

v(0) = 0 = 20 + A or A = –20 v(t) = 20(1 – e-3t/25)V

Chapter 8, Solution 83.

i = iD + Cdv/dt

(1)

–vs + iR + Ldi/dt + v = 0

(2)

Substituting (1) into (2), vs = RiD + RCdv/dt + Ldi/dt + LCd2v/dt2 + v = 0 LCd2v/dt2 + RCdv/dt + RiD + Ldi/dt = vs d2v/dt2 + (R/L)dv/dt + (R/LC)iD + (1/C)di/dt = vs/LC

Chapter 9, Solution 1. ω = 103 rad/s

(a)

angular frequency

(b)

frequency

f =

ω = 159.2 Hz 2π

(c)

period

T =

1 = 6.283 ms f

(d)

Since sin(A) = cos(A – 90°), vs = 12 sin(103t + 24°) = 12 cos(103t + 24° – 90°) vs in cosine form is vs = 12 cos(103t – 66°) V

(e)

vs(2.5 ms) = 12 sin((10 3 )(2.5 × 10 -3 ) + 24°) = 12 sin(2.5 + 24°) = 12 sin(143.24° + 24°) = 2.65 V

Chapter 9, Solution 2.

(a)

amplitude = 8 A

(b)

ω = 500π = 1570.8 rad/s

(c)

f =

(d)

Is = 8∠-25° A Is(2 ms) = 8 cos((500π )(2 × 10 -3 ) − 25°) = 8 cos(π − 25°) = 8 cos(155°) = -7.25 A

ω = 250 Hz 2π

Chapter 9, Solution 3.

(a)

4 sin(ωt – 30°) = 4 cos(ωt – 30° – 90°) = 4 cos(ωt – 120°)

(b)

-2 sin(6t) = 2 cos(6t + 90°)

(c)

-10 sin(ωt + 20°) = 10 cos(ωt + 20° + 90°) = 10 cos(ωt + 110°)

Chapter 9, Solution 4.

(a)

v = 8 cos(7t + 15°) = 8 sin(7t + 15° + 90°) = 8 sin(7t + 105°)

(b)

i = -10 sin(3t – 85°) = 10 cos(3t – 85° + 90°) = 10 cos(3t + 5°)

Chapter 9, Solution 5. v1 = 20 sin(ωt + 60°) = 20 cos(ωt + 60° − 90°) = 20 cos(ωt − 30°) v2 = 60 cos(ωt − 10°)

This indicates that the phase angle between the two signals is 20° and that v1 lags v2. Chapter 9, Solution 6.

(a)

v(t) = 10 cos(4t – 60°) i(t) = 4 sin(4t + 50°) = 4 cos(4t + 50° – 90°) = 4 cos(4t – 40°) Thus, i(t) leads v(t) by 20°.

(b)

v1(t) = 4 cos(377t + 10°) v2(t) = -20 cos(377t) = 20 cos(377t + 180°) Thus, v2(t) leads v1(t) by 170°.

(c)

x(t) = 13 cos(2t) + 5 sin(2t) = 13 cos(2t) + 5 cos(2t – 90°) X = 13∠0° + 5∠-90° = 13 – j5 = 13.928∠-21.04° x(t) = 13.928 cos(2t – 21.04°) y(t) = 15 cos(2t – 11.8°) phase difference = -11.8° + 21.04° = 9.24° Thus, y(t) leads x(t) by 9.24°.

Chapter 9, Solution 7.

If f(φ) = cosφ + j sinφ, df = -sinφ + j cos φ = j (cos φ + j sin φ) = j f (φ ) dφ

df = j dφ f

Integrating both sides ln f = jφ + ln A f = Aejφ = cosφ + j sinφ f(0) = A = 1 i.e. f(φ) = ejφ = cosφ + j sinφ Chapter 9, Solution 8.

(a)

(b)

(c)

15∠45° 15∠45° + j2 = + j2 5∠ - 53.13° 3 − j4 = 3∠98.13° + j2 = -0.4245 + j2.97 + j2 = -0.4243 + j4.97 (2 + j)(3 – j4) = 6 – j8 + j3 + 4 = 10 – j5 = 11.18∠-26.57° 8∠ - 20° (-5 − j12)(10) 8∠ - 20° 10 + + = 11.18∠ - 26.57° 25 + 144 (2 + j)(3 - j4) - 5 + j12 = 0.7156∠6.57° − 0.2958 − j0.71 = 0.7109 + j0.08188 − 0.2958 − j0.71 = 0.4151 − j0.6281 10 + (8∠50°)(13∠-68.38°) = 10+104∠-17.38° = 109.25 – j31.07

Chapter 9, Solution 9.

(3 + j4)(5 + j8) 3 + j4 = 2+ 25 + 64 5 − j8 15 + j24 + j20 − 32 = 2+ 89 = 1.809 + j0.4944

(a)

2+

(b)

4∠-10° +

1 − j2 2.236 ∠ - 63.43° = 4∠-10° + 3∠6° 3∠6°

= 4∠-10° + 0.7453∠-69.43° = 3.939 – j0.6946 + 0.2619 – j0.6978 = 4.201 – j1.392 (c)

8∠10° + 6 ∠ - 20° 7.879 + j1.3892 + 5.638 − j2.052 = 9∠80° − 4∠50° 1.5628 + j8.863 − 2.571 − j3.064 13.533∠ - 2.81° 13.517 − j0.6629 = = 5.886∠99.86° − 1.0083 + j5.799 = 2.299∠-102.67° = -0.5043 – j2.243

Chapter 9, Solution 10.

(a) z1 = 6 − j8, z 2 = 8.66 − j 5, and z 3 = −4 − j 6.9282 z1 + z 2 + z 3 = 10.66 − j19.93 (b)

z1 z 2 = 9.999 + j 7.499 z3

Chapter 9, Solution 11.

(a)

(b) (c)

z 1 z 2 = (-3 + j4)(12 + j5) = -36 – j15 + j48 – 20 = -56 + j33 z1 - 3 + j4 (-3 + j4)(12 + j5) = -0.3314 + j0.1953 = ∗ = z2 144 + 25 12 − j5 z 1 + z 2 = (-3 + j4) + (12 + j5) = 9 + j9 z 1 − z 2 = (-3 + j4) – (12 + j5) = -15 – j z1 + z 2 9 (1 + j) - 9 (1 + j)(15 - j) - 9 (16 + j14) = = = z1 − z 2 15 2 − 12 226 - (15 + j) = -0.6372 – j0.5575

Chapter 9, Solution 12. (a)

z 1 z 2 = (-3 + j4)(12 + j5) = -36 – j15 + j48 – 20 = -56 + j33 z1 (-3 + j4)(12 + j5) - 3 + j4 = = -0.3314 + j0.1953 ∗ = z2 144 + 25 12 − j5

(b)

z 1 + z 2 = (-3 + j4) + (12 + j5) = 9 + j9 z 1 − z 2 = (-3 + j4) – (12 + j5) = -15 – j z1 + z 2 9 (1 + j) - 9 (1 + j)(15 - j) - 9 (16 + j14) = = = 2 2 z1 − z 2 15 − 1 226 - (15 + j) = -0.6372 – j0.5575

(c)

Chapter 9, Solution 13.

(a) (−0.4324 + j 0.4054)+ (−0.8425 − j 0.2534) = − 1.2749 + j 0.1520

(b)

50∠ − 30 o = − 2.0833 24∠150 o

(c) (2+j3)(8-j5) –(-4) = 35 +j14 Chapter 9, Solution 14.

(a)

3 − j14 = − 0.5751 + j 0.5116 − 15 + j11

(b)

(62.116 + j 231.82 + 138.56 − j80)(60 − j80) 24186 − 6944.9 = = − 1.922 − j11.55 (67 + j84)(16.96 + j10.5983) 246.06 + j 2134.7

(c) (− 2 + j 4 )

2

(260 − j120) = − 256.4 − j 200.89

Chapter 9, Solution 15.

(a)

(b)

(c)

10 + j6 2 − j3 = -10 – j6 + j10 – 6 + 10 – j15 -5 -1 + j = -6 – j11 20∠ − 30° - 4∠ - 10° = 60∠15° + 64∠-10° 16∠0° 3∠45° = 57.96 + j15.529 + 63.03 – j11.114 = 120.99 – j4.415

1− j − j 0 j 1 −j 1 j 1+ j 1− j − j j 1

= 1 + 1 + 0 − 1 − 0 + j2 (1 − j) + j2 (1 + j)

0 −j = 1 − 1 (1 − j + 1 + j) = 1 – 2 = -1

Chapter 9, Solution 16.

(a)

-10 cos(4t + 75°) = 10 cos(4t + 75° − 180°) = 10 cos(4t − 105°) The phasor form is 10∠-105°

(b)

5 sin(20t – 10°) = 5 cos(20t – 10° – 90°) = 5 cos(20t – 100°) The phasor form is 5∠-100°

(c)

4 cos(2t) + 3 sin(2t) = 4 cos(2t) + 3 cos(2t – 90°) The phasor form is 4∠0° + 3∠-90° = 4 – j3 = 5∠-36.87°

Chapter 9, Solution 17.

(a)

Let A = 8∠-30° + 6∠0° = 12.928 – j4 = 13.533∠-17.19° a(t) = 13.533 cos(5t + 342.81°)

(b)

We know that -sinα = cos(α + 90°). Let B = 20∠45° + 30∠(20° + 90°) = 14.142 + j14.142 – 10.261 + j28.19 = 3.881 + j42.33 = 42.51∠84.76° b(t) = 42.51 cos(120πt + 84.76°)

(c)

Let C = 4∠-90° + 3∠(-10° – 90°) = -j4 – 0.5209 – j2.954 = 6.974∠265.72° c(t) = 6.974 cos(8t + 265.72°)

Chapter 9, Solution 18.

(a)

v1 ( t ) = 60 cos(t + 15°)

(b)

V2 = 6 + j8 = 10∠53.13° v 2 ( t ) = 10 cos(40t + 53.13°)

(c)

i1 ( t ) = 2.8 cos(377t – π/3)

(d)

I 2 = -0.5 – j1.2 = 1.3∠247.4° i 2 ( t ) = 1.3 cos(103t + 247.4°)

Chapter 9, Solution 19.

(a)

3∠10° − 5∠-30° = 2.954 + j0.5209 – 4.33 + j2.5 = -1.376 + j3.021 = 3.32∠114.49° Therefore, 3 cos(20t + 10°) – 5 cos(20t – 30°) = 3.32 cos(20t + 114.49°)

(b)

4∠-90° + 3∠-45° = -j40 + 21.21 – j21.21 = 21.21 – j61.21 = 64.78∠-70.89° Therefore, 40 sin(50t) + 30 cos(50t – 45°) = 64.78 cos(50t – 70.89°)

(c)

Using sinα = cos(α − 90°), 20∠-90° + 10∠60° − 5∠-110° = -j20 + 5 + j8.66 + 1.7101 + j4.699 = 6.7101 – j6.641 = 9.44∠-44.7° Therefore, 20 sin(400t) + 10 cos(400t + 60°) – 5 sin(400t – 20°) = 9.44 cos(400t – 44.7°)

Chapter 9, Solution 20. (a) V = 4∠− 60 o − 90 o − 5∠40 o = −3.464 − j 2 − 3.83 − j 3.2139 = 8.966∠ − 4.399 o Hence, v = 8.966 cos(377t − 4.399 o )

(b) I = 10∠0 o + jω 8∠20 o − 90 o ,

ω = 5 , i.e. I = 10 + 40∠20 o = 49.51∠16.04 o

i = 49.51 cos(5t + 16.04 o )

Chapter 9, Solution 21.

(a) F = 5∠15 o − 4∠− 30 o − 90 o = 6.8296 + j 4.758 = 8.3236∠34.86 o f (t ) = 8.324 cos(30t + 34.86 o )

(b) G = 8∠ − 90 o + 4∠50 o = 2.571 − j 4.9358 = 5.565∠ − 62.49 o g (t ) = 5.565 cos(t − 62.49 o )

1 (10∠0 o + 5∠ − 90 o ), ω = 40 jω i.e. H = 0.25∠ − 90 o + 0.125∠ − 180 o = − j 0.25 − 0.125 = 0.2795∠ − 116.6 o (c) H =

h(t ) = 0.2795 cos(40t − 116.6 o )

Chapter 9, Solution 22. t

dv Let f(t) = 10v(t ) + 4 − 2 ∫ v(t )dt dt −∞ 2V F = 10V + jω 4V − , ω = 5, V = 20∠ − 30 o jω F = 10V + j 20V − j 0.4V = (10 − j19.6)(17.32 − j10) = 440.1∠ − 92.97 o f (t ) = 440.1 cos(5t − 92.97 o )

Chapter 9, Solution 23.

(a)

v(t) = 40 cos(ωt – 60°)

(b)

V = -30∠10° + 50∠60° = -4.54 + j38.09 = 38.36∠96.8° v(t) = 38.36 cos(ωt + 96.8°)

(c)

I = j6∠-10° = 6∠(90° − 10°) = 6∠80° i(t) = 6 cos(ωt + 80°)

(d)

2 + 10∠-45° = -j2 + 7.071 – j7.071 j = 11.5∠-52.06° i(t) = 11.5 cos(ωt – 52.06°) I =

Chapter 9, Solution 24.

(a) V = 10∠0°, ω = 1 jω V (1 − j) = 10 10 V= = 5 + j5 = 7.071∠45° 1− j Therefore, v(t) = 7.071 cos(t + 45°) V+

(b) 4V = 20∠(10° − 90°), ω = 4 jω  4 V  j4 + 5 +  = 20 ∠ - 80° j4   20∠ - 80° = 3.43∠ - 110.96° V= 5 + j3 Therefore, v(t) = 3.43 cos(4t – 110.96°) jωV + 5V +

Chapter 9, Solution 25. (a)

2jωI + 3I = 4∠ - 45°, ω = 2 I (3 + j4) = 4∠ - 45° 4∠ - 45° 4∠ - 45° = = 0.8∠ - 98.13° I= 3 + j4 5∠53.13° Therefore, i(t) = 0.8 cos(2t – 98.13°)

(b) I + jωI + 6I = 5∠22°, ω = 5 jω (- j2 + j5 + 6) I = 5∠22° 5∠22° 5∠22° I= = = 0.745∠ - 4.56° 6 + j3 6.708∠26.56° Therefore, i(t) = 0.745 cos(5t – 4.56°) 10

Chapter 9, Solution 26. I = 1∠0°, ω = 2 jω  1 I  j2 + 2 +  = 1 j2   1 = 0.4∠ - 36.87° I= 2 + j1.5 Therefore, i(t) = 0.4 cos(2t – 36.87°) jωI + 2I +

Chapter 9, Solution 27. V = 110∠ - 10°, ω = 377 jω  j100   = 110∠ - 10° V  j377 + 50 −  377  V (380.6∠82.45°) = 110∠ - 10° V = 0.289 ∠ - 92.45° jωV + 50V + 100

Therefore, v(t) = 0.289 cos(377t – 92.45°).

Chapter 9, Solution 28.

i( t ) =

v s ( t ) 110 cos(377 t ) = = 13.75 cos(377t) A. R 8

Chapter 9, Solution 29.

Z=

1 1 = = - j 0.5 6 jωC j (10 )(2 × 10 -6 )

V = IZ = (4∠25°)(0.5∠ - 90°) = 2 ∠ - 65° Therefore

v(t) = 2 sin(106t – 65°) V.

Chapter 9, Solution 30. Z = jωL = j (500)(4 × 10 -3 ) = j2 V 60 ∠ - 65° I= = = 30∠ - 155° Z 2∠90° Therefore, i(t) = 30 cos(500t – 155°) A. Chapter 9, Solution 31.

Thus,

i(t) = 10 sin(ωt + 30°) = 10 cos(ωt + 30° − 90°) = 10 cos(ωt − 60°) I = 10∠-60°

v(t) = -65 cos(ωt + 120°) = 65 cos(ωt + 120° − 180°) = 65 cos(ωt − 60°) Thus, V = 65∠-60°

Z=

V 65∠ - 60° = = 6.5 Ω I 10∠ - 60°

Since V and I are in phase, the element is a resistor with R = 6.5 Ω.

Chapter 9, Solution 32. V = 180∠10°,

Z=

I = 12∠-30°,

ω = 2

V 180∠10° = = 15∠40° = 11.49 + j 9.642 Ω I 12∠ - 30°

One element is a resistor with R = 11.49 Ω. The other element is an inductor with ωL = 9.642 or

L = 4.821 H.

Chapter 9, Solution 33.

110 = v 2R + v 2L v L = 110 2 − v 2R v L = 110 2 − 85 2 = 69.82 V

Chapter 9, Solution 34.

v o = 0 if ωL =

ω=

1 ωC

 → ω =

1 (5 × 10 −3 )(2 × 10 − 3 )

1 LC

= 100 rad/s

Chapter 9, Solution 35. Vs = 5∠0° jωL = j (2)(1) = j2 1 1 = = - j2 jωC j (2)(0.25)

j2 j2 Vs = 5∠0° = (1∠90°)(5∠0°) = 5∠90° 2 − j2 + j2 2 Thus, v o ( t ) = 5 cos(2t + 90°) = -5 sin(2t) V Vo =

Chapter 9, Solution 36.

Let Z be the input impedance at the source.

10 µF

jωL = j 200 x100 x10 −3 = j 20

 →

100 mH

1 1 = = − j 500 jωC j10 x10 −6 x 200

 →

1000//-j500 = 200 –j400 1000//(j20 + 200 –j400) = 242.62 –j239.84 Z = 2242.62 − j 239.84 = 2255∠ − 6.104 o I=

60∠ − 10 o = 26.61∠ − 3.896 o mA o 2255∠ − 6.104

i = 266.1 cos(200t − 3.896 o )

Chapter 9, Solution 37. jωL = j (5)(1) = j5 1 1 = = -j jωC j (5)(0.2) Let Z1 = - j ,

Z 2 = 2 || j5 =

Then,

Ix =

(2)( j5) j10 = 2 + j5 2 + j5

Z2 I , Z1 + Z 2 s

where I s = 2∠0°

j10 j20 2 + j5 Ix = (2) = = 2.12 ∠32° j10 5 + j8 - j+ 2 + j5 Therefore,

i x ( t ) = 2.12 sin(5t + 32°) A

Chapter 9, Solution 38. 1 F  → 6

(a)

1 1 = = - j2 jωC j (3)(1 / 6)

- j2 (10 ∠45°) = 4.472∠ - 18.43° 4 − j2 Hence, i(t) = 4.472 cos(3t – 18.43°) A

I=

V = 4I = (4)(4.472∠ - 18.43°) = 17.89∠ - 18.43° Hence, v(t) = 17.89 cos(3t – 18.43°) V 1 F  → 12

(b)

3H  →

1 1 = = - j3 jωC j (4)(1 / 12)

jωL = j (4)(3) = j12

V 50∠0° = 10∠36.87° = Z 4 − j3 Hence, i(t) = 10 cos(4t + 36.87°) A I=

j12 (50∠0°) = 41.6 ∠33.69° 8 + j12 Hence, v(t) = 41.6 cos(4t + 33.69°) V V=

Chapter 9, Solution 39. Z = 8 + j5 || (- j10) = 8 +

I=

( j5)(- j10) = 8 + j10 j5 − j10

V 40 ∠0° 20 = = = 3.124∠ - 51.34° Z 8 + j10 6.403∠51.34°

I1 =

- j10 I = 2 I = 6.248∠ - 51.34° j5 − j10

I2 =

j5 I = - I = 3.124∠128.66° - j5

Therefore,

i1 ( t ) = 6.248 cos(120πt – 51.34°) A i 2 ( t ) = 3.124 cos(120πt + 128.66°) A

Chapter 9, Solution 40. (a)

For ω = 1 , 1H  →

jωL = j (1)(1) = j 1 1 0.05 F  → = = - j20 jωC j (1)(0.05) - j40 Z = j + 2 || (- j20) = j + = 1.98 + j0.802 2 − j20

V 4 ∠0° 4∠0° = = = 1.872 ∠ - 22.05° Z 1.98 + j0.802 2.136∠22.05° Hence, i o ( t ) = 1.872 cos(t – 22.05°) A Io =

(b)

For ω = 5 , 1H  →

jωL = j (5)(1) = j5 1 1 0.05 F  → = = - j4 jωC j (5)(0.05) - j4 Z = j5 + 2 || (- j4) = j5 + = 1.6 + j4.2 1 − j2

4∠0° 4∠0° V = = = 0.89∠ - 69.14° Z 1.6 + j4 4.494∠69.14° Hence, i o ( t ) = 0.89 cos(5t – 69.14°) A Io =

(c)

For ω = 10 , 1H  → jωL = j (10)(1) = j10 1 1 0.05 F  → = = - j2 jωC j (10)(0.05) - j4 Z = j10 + 2 || (- j2) = j10 + = 1 + j9 2 − j2 V 4∠0° 4 ∠0° = = 0.4417 ∠ - 83.66° = Z 1 + j9 9.055∠83.66° Hence, i o ( t ) = 0.4417 cos(10t – 83.66°) A Io =

Chapter 9, Solution 41.

ω = 1, 1H  →

jωL = j (1)(1) = j 1 1 = = -j jωC j (1)(1)

1F  →

Z = 1 + (1 + j) || (- j) = 1 +

I=

Vs 10 = , Z 2− j

- j+1 = 2− j 1

I c = (1 + j) I

V = (- j)(1 + j) I = (1 − j) I =

Thus,

(1 − j)(10) = 6.325∠ - 18.43° 2− j

v(t) = 6.325 cos(t – 18.43°) V

Chapter 9, Solution 42. ω = 200

50 µF  →

1 1 = = - j100 jωC j (200)(50 × 10 -6 )

0.1 H  →

jωL = j (200)(0.1) = j20

50 || -j100 =

Vo =

(50)(-j100) - j100 = = 40 − j20 50 − j100 1 - j2

j20 j20 (60∠0°) = (60∠0°) = 17.14 ∠90° j20 + 30 + 40 − j20 70

Thus, v o ( t ) = 17.14 sin(200t + 90°) V or

v o ( t ) = 17.14 cos(200t) V

Chapter 9, Solution 43.

ω= 2 1H  → jωL = j (2)(1) = j2 1F  →

Io =

1 1 = = - j0.5 jωC j (2)(1)

j2 − j0.5 j1.5 I= 4∠0° = 3.328∠33.69° j2 − j0.5 + 1 1 + j1.5

Thus, i o ( t ) = 3.328 cos(2t + 33.69°) A Chapter 9, Solution 44. ω = 200 10 mH  → jωL = j (200)(10 × 10 -3 ) = j2

5 mF  →

1 1 = = -j jωC j (200)(5 × 10 -3 )

Y=

1 1 1 3+ j + + = 0.25 − j0.5 + = 0.55 − j0.4 4 j2 3 − j 10

Z=

1 1 = = 1.1892 + j0.865 Y 0.55 − j0.4

I=

6∠0° 6∠0° = = 0.96 ∠ - 7.956° 5 + Z 6.1892 + j0.865

Thus, i(t) = 0.96 cos(200t – 7.956°) A

Chapter 9, Solution 45.

We obtain I o by applying the principle of current division twice. I

I2

Z1

I2

Io

-j2 Ω

Z2

(a) Z 1 = - j2 ,

(b)

Z 2 = j4 + (-j2) || 2 = j4 +

I2 =

Z1 - j2 - j10 I= (5∠0°) = Z1 + Z 2 - j2 + 1 + j3 1+ j

Io =

 - j  - j10  - 10 - j2  = I2 =  = -5 A 2 - j2 1 - j  1 + j  1 + 1

- j4 = 1 + j3 2 - j2

Chapter 9, Solution 46. i s = 5 cos(10 t + 40°)  → I s = 5∠40°

Let

0.1 F  →

1 1 = = -j jωC j (10)(0.1)

0.2 H  →

jωL = j (10)(0.2) = j2

Z1 = 4 || j2 =

2Ω

j8 = 0.8 + j1.6 , 4 + j2

Z2 = 3 − j

Io =

Z1 0.8 + j1.6 (5∠40°) Is = 3.8 + j0.6 Z1 + Z 2

Io =

(1.789∠63.43°)(5∠40°) = 2.325∠94.46° 3.847 ∠8.97°

Thus, i o ( t ) = 2.325 cos(10t + 94.46°) A

Chapter 9, Solution 47. First, we convert the circuit into the frequency domain. Ix

5∠0˚

Ix =

2Ω

j4

+ −

-j10

20 Ω

5 5 5 = = = 0.4607∠52.63° − j10(20 + j4) 2 + 4.588 − j8.626 10.854∠ − 52.63° 2+ − j10 + 20 + j4 is(t) = 0.4607cos(2000t +52.63˚) A

Chapter 9, Solution 48. Converting the circuit to the frequency domain, we get:

10 Ω

V1 30 Ω Ix

20∠-40˚

+ −

We can solve this using nodal analysis.

j20

-j20

V1 − 20∠ − 40° V1 − 0 V −0 =0 + + 1 10 j20 30 − j20 V1(0.1 − j0.05 + 0.02307 + j0.01538) = 2∠ − 40° 2∠40° = 15.643∠ − 24.29° 0.12307 − j0.03462 15.643∠ − 24.29° = = 0.4338∠9.4° 30 − j20 = 0.4338 sin(100 t + 9.4°) A

V1 = Ix ix

Chapter 9, Solution 49. Z T = 2 + j2 || (1 − j) = 2 + I

( j2)(1 − j) =4 1+ j

Ix

1Ω

j2 Ω

-j Ω

j2 j2 I= I, j2 + 1 − j 1+ j 1+ j 1+ j I= Ix = j2 j4

where I x = 0.5∠0° =

Ix =

1 2

1+ j 1+ j (4) = = 1 − j = 1.414∠ - 45° j4 j v s ( t ) = 1.414 sin(200t – 45°) V

Vs = I Z T =

Chapter 9, Solution 50. Since ω = 100, the inductor = j100x0.1 = j10 Ω and the capacitor = 1/(j100x10-3) = -j10Ω.

j10

5∠40˚

Ix +

-j10

20 Ω

vx −

Using the current dividing rule: − j10 5∠40° = − j2.5∠40° = 2.5∠ − 50° − j10 + 20 + j10 Vx = 20I x = 50∠ − 50°

Ix =

v x = 50 cos(100t − 50°) V

Chapter 9, Solution 51.

0.1 F  →

1 1 = = - j5 jωC j (2)(0.1)

0.5 H  →

jωL = j (2)(0.5) = j

The current I through the 2-Ω resistor is Is 1 I= Is = , 1 − j5 + j + 2 3 − j4 I s = (10)(3 − j4) = 50∠ - 53.13°

where I = 10 ∠0°

Therefore, i s ( t ) = 50 cos(2t – 53.13°) A

Chapter 9, Solution 52.

5 || j5 =

j25 j5 = = 2.5 + j2.5 5 + j5 1 + j

Z1 = 10 ,

Z 2 = - j5 + 2.5 + j2.5 = 2.5 − j2.5

I2 IS

Z1

Z2

I2 =

Z1 10 4 Is = Is = I 12.5 − j2.5 5− j s Z1 + Z 2

Vo = I 2 (2.5 + j2.5)  4 8∠30° =  5 −

Is =

 10 (1 + j)  I s (2.5)(1 + j) = I j 5− j s

(8∠30°)(5 − j) = 2.884∠-26.31° A 10 (1 + j)

Chapter 9, Solution 53. Convert the delta to wye subnetwork as shown below. Z1

Io

Z2

2Ω

Z3

+

10 Ω 8Ω

60∠ − 30 V o

-

Z Z1 =

− j 2 x4 = 0.1532 − j 0.7692, 10 − j 2

Z3 =

12 = 1.1538 + j 0.2308 10 − j 2

Z2 =

j6 x4 = −0.4615 + j 2.3077, 10 − j 2

( Z 3 + 8) //( Z 2 + 10) = (9.1538 + j 0.2308) //(9.5385 + j 2.3077) = 4.726 + j 0.6062 Z = 2 + Z 1 + 4.726 + j 0.6062 = 6.878 − j 0.163

Io =

60∠ − 30 o 60∠ − 30 o = = 8.721∠ − 28.64 o A o Z 6.88∠ − 1.3575

Chapter 9, Solution 54. Since the left portion of the circuit is twice as large as the right portion, the equivalent circuit is shown below. + −

Vs −

+ 2Z

Z

V1

V2

+

− V1 = I o (1 − j) = 2 (1 − j) V2 = 2V1 = 4 (1 − j) Vs = V1 + V2 = 6 (1 − j) Vs = 8.485∠-45° V Chapter 9, Solution 55.

12 Ω

-j20 V

I

I1

Z

I2

+ −

-j4 Ω

+ Vo −

I1 =

Vo 4 = = -j0.5 j 8 j8

I2 =

I 1 (Z + j8) (-j0.5)(Z + j8) Z = = +j - j4 - j4 8

I = I 1 + I 2 = -j0.5 +

Z Z + j = + j0.5 8 8

- j20 = 12 I + I 1 (Z + j8)

Z j - j - j20 = 12  +  + (Z + j8)  8 2 2

j8 Ω

3 1 - 4 - j26 = Z  − j  2 2

Z=

- 4 - j26 26.31∠261.25° = = 16.64∠279.68° 3 1 1.5811∠ - 18.43° −j 2 2

Z = 2.798 – j16.403 Ω Chapter 9, Solution 56. 3H

 →

jωL = j 30

3F

 →

1 = − j / 30 jω C

1.5F

1 = − j / 15 jω C

 →

−j 15 = − j 0.06681 j 30 //( − j / 15) = j j 30 − 15 j 30 x

Z=

−j − j 0.033(2 − j 0.06681) //(2 − j 0.06681) = = 6 − j 333 mΩ 30 − j 0.033 + 2 − j 0.06681

Chapter 9, Solution 57. 2H 1F

 →  →

jωL = j 2

1 =−j jω C

Z = 1 + j2 //( 2 − j) = 1 +

j2(2 − j) = 2.6 + j1.2 j2 + 2 − j

Y = 1 = 0.3171 − j0.1463 S Z

Chapter 9, Solution 58. (a)

10 mF  →

1 1 = = - j2 jωC j (50)(10 × 10 -3 )

10 mH  →

jωL = j (50)(10 × 10 -3 ) = j0.5

Z in = j0.5 + 1 || (1 − j2) 1 − j2 Z in = j0.5 + 2 − j2 Z in = j0.5 + 0.25 (3 − j) Z in = 0.75 + j0.25 Ω

(b)

0.4 H  →

jωL = j (50)(0.4) = j20

0.2 H  →

jωL = j (50)(0.2) = j10 1 1 = = - j20 jωC j (50)(1 × 10 -3 )

1 mF  →

For the parallel elements, 1 1 1 1 = + + Z p 20 j10 - j20

Z p = 10 + j10 Then, Z in = 10 + j20 + Z p = 20 + j30 Ω

Chapter 9, Solution 59. Z eq = 6 + (1 − j2) || (2 + j4) Z eq = 6 +

(1 − j2)(2 + j4) (1 − j2) + (2 + j4)

Z eq = 6 + 2.308 − j1.5385 Z eq = 8.308 – j1.5385 Ω

Chapter 9, Solution 60. Z = (25 + j15) + (20 − j 50) //(30 + j10) = 25 + j15 + 26.097 − j 5.122 = 51.1 + j 9.878Ω

Chapter 9, Solution 61. All of the impedances are in parallel. 1 1 1 1 1 = + + + Z eq 1 − j 1 + j2 j5 1 + j3 1 = (0.5 + j0.5) + (0.2 − j0.4) + (- j0.2) + (0.1 − j0.3) = 0.8 − j0.4 Z eq

Z eq =

1 = 1 + j0.5 Ω 0.8 − j0.4

Chapter 9, Solution 62.

2 mH  →

jωL = j (10 × 10 3 )(2 × 10 -3 ) = j20 1 1 1 µF  → = = - j100 3 jωC j (10 × 10 )(1 × 10 -6 ) 50 Ω +

1∠0° A

+

V

j20 Ω

−

Vin

+

− -j100 Ω

V = (1∠0°)(50) = 50 Vin = (1∠0°)(50 + j20 − j100) + (2)(50) Vin = 50 − j80 + 100 = 150 − j80

Z in =

Vin = 150 – j80 Ω 1∠0°

2V

Chapter 9, Solution 63.

First, replace the wye composed of the 20-ohm, 10-ohm, and j15-ohm impedances with the corresponding delta. 200 + j150 + j300 = 20 + j45 10 200 + j450 200 + j450 z2 = = 30 − j13.333, z3 = = 10 + j22.5 j15 20 z1 =

8Ω

–j12 Ω

–j16 Ω

z2

10 Ω

z1

ZT

z3 –j16 Ω

10 Ω

Now all we need to do is to combine impedances. z 2 (10 − j16) =

(30 − j13.333)(10 − j16) = 8.721 − j8.938 40 − j29.33

z3 (10 − j16) = 21.70 − j3.821 ZT = 8 − j12 + z1 (8.721 − j8.938 + 21.7 − j3.821) = 34.69 − j6.93Ω

Chapter 9, Solution 64. − j10(6 + j8) = 19 − j5Ω 6 − j2 30∠90° I= = −0.3866 + j1.4767 = 1.527∠104.7° A ZT ZT = 4 +

Chapter 9, Solution 65. Z T = 2 + (4 − j6) || (3 + j4)

ZT = 2 +

(4 − j6)(3 + j4) 7 − j2

Z T = 6.83 + j1.094 Ω = 6.917∠9.1° Ω

I=

V 120 ∠10° = = 17.35∠0.9° A Z T 6.917 ∠9.1°

Chapter 9, Solution 66.

Z T = (20 − j5) || (40 + j10) =

(20 − j5)(40 + j10) 170 = (12 − j) 60 + j5 145

Z T = 14.069 – j1.172 Ω = 14.118∠-4.76°

I=

V 60∠90° = = 4.25∠94.76° Z T 14.118∠ - 4.76° I I1

I2

20 Ω

j10 Ω

+

I1 =

40 + j10 8 + j2 I= I 60 + j5 12 + j

I2 =

20 − j5 4− j I= I 60 + j5 12 + j

Vab = -20 I 1 + j10 I 2

Vab

−

Vab =

- (160 + j40) 10 + j40 I+ I 12 + j 12 + j

Vab =

- 150 (-12 + j)(150) I= I 12 + j 145

Vab = (12.457 ∠175.24°)(4.25∠97.76°) Vab = 52.94∠273° V Chapter 9, Solution 67.

(a)

20 mH  →

jωL = j (10 3 )(20 × 10 -3 ) = j20 1 1 12.5 µF  → = = - j80 3 jωC j (10 )(12.5 × 10 -6 ) Z in = 60 + j20 || (60 − j80) ( j20)(60 − j80) Z in = 60 + 60 − j60 Z in = 63.33 + j23.33 = 67.494 ∠20.22°

Yin =

(b)

1 = 0.0148∠-20.22° S Z in

10 mH  → 20 µF  →

jωL = j (10 3 )(10 × 10 -3 ) = j10 1 1 = = - j50 jωC j (10 3 )(20 × 10 -6 )

30 || 60 = 20 Z in = - j50 + 20 || (40 + j10) (20)(40 + j10) Z in = - j50 + 60 + j10 Z in = 13.5 − j48.92 = 50.75∠ - 74.56°

Yin =

1 = 0.0197∠74.56° S = 5.24 + j18.99 mS Z in

Chapter 9, Solution 68.

Yeq =

1 1 1 + + 5 − j2 3 + j - j4

Yeq = (0.1724 + j0.069) + (0.3 − j0.1) + ( j0.25)

Yeq = 0.4724 + j0.219 S

Chapter 9, Solution 69. 1 1 1 1 = + = (1 + j2) Yo 4 - j2 4

Yo =

4 (4)(1 − j2) = = 0.8 − j1.6 1 + j2 5

Yo + j = 0.8 − j0.6 1

1 1 1 = + + = (1) + ( j0.333) + (0.8 + j0.6) Yo ′ 1 - j3 0.8 − j0.6

1

Yo ′

= 1.8 + j0.933 = 2.028∠27.41°

Yo ′ = 0.4932∠ - 27.41° = 0.4378 − j0.2271 Yo ′ + j5 = 0.4378 + j4.773 1 1 1 0.4378 − j4.773 = + = 0.5 + Yeq 2 0.4378 + j4.773 22.97 1 = 0.5191 − j0.2078 Yeq

Yeq =

0.5191 − j0.2078 = 1.661 + j0.6647 S 0.3126

Chapter 9, Solution 70. Make a delta-to-wye transformation as shown in the figure below. a Zan

Zbn

Zeq

n

Zcn

b

c 8Ω

2Ω -j5 Ω

Z an =

(- j10)(10 + j15) (10)(15 − j10) = = 7 − j9 5 − j10 + 10 + j15 15 + j5

Z bn =

(5)(10 + j15) = 4.5 + j3.5 15 + j5

Z cn =

(5)(- j10) = -1 − j3 15 + j5

Z eq = Z an + (Z bn + 2) || (Z cn + 8 − j5) Z eq = 7 − j9 + (6.5 + j3.5) || (7 − j8) Z eq = 7 − j9 +

(6.5 + j3.5)(7 − j8) 13.5 − j4.5

Z eq = 7 − j9 + 5.511 − j0.2 Z eq = 12.51 − j9.2 = 15.53∠-36.33° Ω

Chapter 9, Solution 71. We apply a wye-to-delta transformation.

j4 Ω Zab

b

a Zac

Zbc

Zeq -j2 Ω

1Ω c Z ab =

2 − j2 + j4 2 + j2 = = 1− j j2 j2

Z ac =

2 + j2 = 1+ j 2

Z bc =

2 + j2 = -2 + j2 -j

j4 || Z ab = j4 || (1 − j) =

1 || Z ac = 1 || (1 + j) =

( j4)(1 − j) = 1.6 − j0.8 1 + j3

(1)(1 + j) = 0.6 + j0.2 2+ j

j4 || Z ab + 1 || Z ac = 2.2 − j0.6 1 1 1 1 = + + Z eq - j2 - 2 + j2 2.2 − j0.6

= j0.5 − 0.25 − j0.25 + 0.4231 + j0.1154

= 0.173 + j0.3654 = 0.4043∠64.66° Z eq = 2.473∠-64.66° Ω = 1.058 – j2.235 Ω

Chapter 9, Solution 72.

Transform the delta connections to wye connections as shown below. a j2 Ω

j2 Ω -j18 Ω

-j9 Ω

j2 Ω

R1

R2

R3 b

- j9 || - j18 = - j6 , R1 =

(20)(20) = 8 Ω, 20 + 20 + 10

R2 =

Z ab = j2 + ( j2 + 8) || (j2 − j6 + 4) + 4 Z ab = 4 + j2 + (8 + j2) || (4 − j4) Z ab = 4 + j2 +

(8 + j2)(4 − j4) 12 - j2

Z ab = 4 + j2 + 3.567 − j1.4054 Z ab = 7.567 + j0.5946 Ω

(20)(10) = 4Ω, 50

R3 =

(20)(10) = 4Ω 50

Chapter 9, Solution 73.

Transform the delta connection to a wye connection as in Fig. (a) and then transform the wye connection to a delta connection as in Fig. (b). a j2 Ω

j2 Ω -j18 Ω

-j9 Ω

j2 Ω

R1

R2

R3 b

( j8)(- j6) 48 = = - j4.8 j8 + j8 − j6 j10 Z 2 = Z1 = -j4.8 ( j8)( j8) - 64 Z3 = = = j6.4 j10 j10 Z1 =

(2 + Z1 )(4 + Z 2 ) + (4 + Z 2 )(Z 3 ) + (2 + Z1 )(Z 3 ) = (2 − j4.8)(4 − j4.8) + (4 − j4.8)( j6.4) + (2 − j4.8)( j6.4) = 46.4 + j9.6 46.4 + j9.6 = 1.5 − j7.25 j6.4 46.4 + j9.6 = 3.574 + j6.688 Zb = 4 − j4.8 46.4 + j9.6 = 1.727 + j8.945 Zc = 2 − j4.8 Za =

(6∠90°)(7.583∠61.88°) = 07407 + j3.3716 3.574 + j12.688 (-j4)(1.5 − j7.25) - j4 || Z a = = 0.186 − j2.602 1.5 − j11.25 j6 || Z b =

j12 || Z c =

(12∠90°)(9.11∠79.07°) = 0.5634 + j5.1693 1.727 + j20.945

Z eq = ( j6 || Z b ) || (- j4 || Z a + j12 || Z c )

Z eq = (0.7407 + j3.3716) || (0.7494 + j2.5673) Z eq = 1.508∠75.42° Ω = 0.3796 + j1.46 Ω

Chapter 9, Solution 74. One such RL circuit is shown below. 20 Ω

V

20 Ω +

+ j20 Ω

Vi = 1∠0°

j20 Ω

Vo −

Z

We now want to show that this circuit will produce a 90° phase shift. Z = j20 || (20 + j20) =

V=

( j20)(20 + j20) - 20 + j20 = = 4 (1 + j3) 20 + j40 1 + j2

Z 4 + j12 1 + j3 1 Vi = (1∠0°) = = (1 + j) Z + 20 24 + j12 6 + j3 3

Vo =

 j  1 j20  (1 + V = 20 + j20 1 + j  3

 j j)  = = 0.3333∠90°  3

This shows that the output leads the input by 90°. Chapter 9, Solution 75. Since cos(ωt ) = sin(ωt + 90°) , we need a phase shift circuit that will cause the output to lead the input by 90°. This is achieved by the RL circuit shown below, as explained in the previous problem.

10 Ω

10 Ω

+

+ j10 Ω

Vi

j10 Ω

−

Vo

−

This can also be obtained by an RC circuit. Chapter 9, Solution 76.

Let Z = R – jX, where X = | Z |= R 2 + X 2

X = | Z |2 − R 2 = 1162 = 662 = 95.394

 →

C=

1 1 = ωC 2πfC

1 1 = 27.81µF = 2πfX 2πx 60x95.394

Chapter 9, Solution 77.

(a)

- jX c V R − jX c i 1 1 where X c = = = 3.979 ωC (2π)(2 × 10 6 )(20 × 10 -9 ) Vo =

Vo - j3.979 = = Vi 5 - j3.979

Vo = Vi

3.979 25 + 15.83

3.979 5 + 3.979 2

2

∠(-90° + tan -1 (3.979 5))

∠(-90° − 38.51°)

Vo = 0.6227 ∠ - 51.49° Vi

Therefore, the phase shift is 51.49° lagging

(b)

θ = -45° = -90° + tan -1 (X c R ) 45° = tan -1 (X c R )  → R = X c = ω = 2πf = f=

1 ωC

1 RC

1 1 = = 1.5915 MHz 2πRC (2π )(5)(20 × 10 -9 )

Chapter 9, Solution 78.

8+j6 R Z -jX

Z = R //[8 + j (6 − X )] =

R[8 + j (6 − X )] =5 R + 8 + j (6 − X )

i.e 8R + j6R – jXR = 5R + 40 + j30 –j5X Equating real and imaginary parts: 8R = 5R + 40 which leads to R=13.33Ω 6R-XR =30-5 which leads to X=4.125Ω. Chapter 9, Solution 79.

(a)

Consider the circuit as shown. 20 Ω

V2

40 Ω

V1

30 Ω

+ Vi

+ j10 Ω

j30 Ω

−

j60 Ω

Vo

− Z2

Z1

( j30)(30 + j60) = 3 + j21 30 + j90 ( j10)(43 + j21) Z 2 = j10 || (40 + Z1 ) = = 1.535 + j8.896 = 9.028∠80.21° 43 + j31

Z1 = j30 || (30 + j60) =

Let Vi = 1∠0° . Z2 (9.028∠80.21°)(1∠0°) Vi = 21.535 + j8.896 Z 2 + 20 V2 = 0.3875∠57.77°

V2 =

Z1 3 + j21 (21.213∠81.87°)(0.3875∠57.77°) V2 = V2 = Z1 + 40 43 + j21 47.85∠26.03° V1 = 0.1718∠113.61°

V1 =

j60 j2 2 V1 = V1 = (2 + j)V1 30 + j60 1 + j2 5 Vo = (0.8944∠26.56°)(0.1718∠113.6°) Vo = 0.1536∠140.2° Vo =

Therefore, the phase shift is 140.2° (b)

The phase shift is leading.

(c)

If Vi = 120 V , then Vo = (120)(0.1536∠140.2°) = 18.43∠140.2° V and the magnitude is 18.43 V.

Chapter 9, Solution 80.

200 mH  → Vo =

(a)

jωL = j (2π )(60)(200 × 10 -3 ) = j75.4 Ω

j75.4 j75.4 Vi = (120∠0°) R + 50 + j75.4 R + 50 + j75.4 When R = 100 Ω , j75.4 (75.4∠90°)(120∠0°) (120 ∠0°) = Vo = 150 + j75.4 167.88∠26.69° Vo = 53.89∠63.31° V

(b)

When R = 0 Ω , j75.4 (75.4∠90°)(120 ∠0°) (120∠0°) = Vo = 50 + j75.4 90.47 ∠56.45° Vo = 100∠33.55° V

(c)

To produce a phase shift of 45°, the phase of Vo = 90° + 0° − α = 45°. Hence, α = phase of (R + 50 + j75.4) = 45°. For α to be 45°, R + 50 = 75.4 Therefore, R = 25.4 Ω

Chapter 9, Solution 81.

Let

Z1 = R 1 ,

Z2 = R 2 +

1 , jωC 2

Zx =

Z3 Z Z1 2

Rx +

R3  1 1  R 2 +  = jωC x R 1  jωC 2 

Rx =

R3 1200 R2 = (600) = 1.8 kΩ R1 400

Z 3 = R 3 , and Z x = R x +

R1  400  1 R3   1  (0.3 × 10 -6 ) = 0.1 µF → C x = C2 =  =     1200  Cx  R1   C2  R3

Chapter 9, Solution 82.

Cx =

R1  100  (40 × 10 -6 ) = 2 µF Cs =   2000  R2

Chapter 9, Solution 83.

Lx =

R2  500  (250 × 10 -3 ) = 104.17 mH Ls =  1200  R1

1 . jωC x

Chapter 9, Solution 84.

Let

1 Z2 = R 2 , , jωC s R1 jωC s R1 Z1 = = 1 jωR 1C s + 1 R1 + jωC s Z1 = R 1 ||

Since Z x =

Z 3 = R 3 , and Z x = R x + jωL x .

Z3 Z , Z1 2

R x + jωL x = R 2 R 3

jωR 1C s + 1 R 2 R 3 = (1 + jωR 1C s ) R1 R1

Equating the real and imaginary components, R 2R 3 Rx = R1 ωL x =

R 2R 3 (ωR 1C s ) implies that R1 L x = R 2 R 3Cs

Given that R 1 = 40 kΩ , R 2 = 1.6 kΩ , R 3 = 4 kΩ , and C s = 0.45 µF R 2 R 3 (1.6)(4) = kΩ = 0.16 kΩ = 160 Ω R1 40 L x = R 2 R 3 C s = (1.6)(4)(0.45) = 2.88 H

Rx =

Chapter 9, Solution 85.

Let

1 , jωC 2 R4 - jR 4 Z4 = = jωR 4 C 4 + 1 ωR 4 C 4 − j

Z1 = R 1 ,

Since Z 4 =

Z3 Z Z1 2

Z2 = R 2 +

 → Z1 Z 4 = Z 2 Z 3 ,

Z 3 = R 3 , and Z 4 = R 4 ||

1 . jωC 4

 - jR 4 R 1 j   = R 3 R 2 − ωR 4 C 4 − j ωC 2   jR 3 - jR 4 R 1 (ωR 4 C 4 + j) = R 3R 2 − 2 2 2 ω R 4C4 + 1 ωC 2

Equating the real and imaginary components, R 1R 4 = R 2R 3 2 ω R 24 C 24 + 1 (1) 2 R3 ωR 1 R 4 C 4 = 2 2 2 ω R 4 C 4 + 1 ωC 2 (2) Dividing (1) by (2), 1 = ωR 2 C 2 ωR 4 C 4 1 ω2 = R 2C2R 4C4 1 ω = 2πf = R 2C2 R 4C4 f=

1 2π R 2 R 4 C 2 C 4

Chapter 9, Solution 86.

Y=

1 1 1 + + 240 j95 - j84

Y = 4.1667 × 10 -3 − j0.01053 + j0.0119 Z=

1 1000 1000 = = Y 4.1667 + j1.37 4.3861∠18.2°

Z = 228∠-18.2° Ω

Chapter 9, Solution 87.

Z1 = 50 +

-j 1 = 50 + (2π)(2 × 10 3 )(2 × 10 -6 ) jωC

Z1 = 50 − j39.79

Z 2 = 80 + jωL = 80 + j (2π)(2 × 10 3 )(10 × 10 -3 ) Z 2 = 80 + j125.66

Z 3 = 100

1 1 1 1 = + + Z Z1 Z 2 Z 3 1 1 1 1 = + + Z 100 50 − j39.79 80 + j125.66 1 = 10 -3 (10 + 12.24 + j9.745 + 3.605 − j5.663) Z = (25.85 + j4.082) × 10 -3 = 26.17 × 10 -3 ∠8.97° Z = 38.21∠-8.97° Ω Chapter 9, Solution 88. (a)

(b)

Z = - j20 + j30 + 120 − j20 Z = 120 – j10 Ω 1 1 = would cause the capacitive ωC 2πf C impedance to double, while ωL = 2πf L would cause the inductive impedance to halve. Thus, Z = - j40 + j15 + 120 − j40 Z = 120 – j65 Ω If the frequency were halved,

Chapter 9, Solution 89.  1   Z in = jωL ||  R + jωC  

 1  L  jωL  R + + jωL R jωC   C = Z in = 1  1  R + jωL +  R + jωL − jωC  ωC 

Z in =

  L 1    + jωL R  R − jωL −   C ωC   1   R + ωL −  ωC 

2

2

To have a resistive impedance, Im(Z in ) = 0 . Hence,  L  1  =0 ωL R 2 −  ωL −  C  ωC 

ωR 2 C = ωL −

1 ωC

ω2 R 2 C 2 = ω2 LC − 1 ω2 R 2 C 2 + 1 L= ω2 C

(1) Ignoring the +1 in the numerator in (1), L = R 2 C = (200) 2 (50 × 10 -9 ) = 2 mH

Chapter 9, Solution 90.

Let

Vs = 145∠0° , I=

X = jωL = j (2π)(60) L = j377 L

Vs 145∠0° = 80 + R + jX 80 + R + jX

V1 = 80 I =

50 =

(80)(145) 80 + R + jX

(80)(145) 80 + R + jX

Vo = (R + jX) I =

110 =

(1) (R + jX)(145∠0°) 80 + R + jX

(R + jX)(145) 80 + R + jX

(2)

From (1) and (2), 50 80 = 110 R + jX

11  R + jX = (80)   5 R 2 + X 2 = 30976 From (1), (80)(145) 80 + R + jX = = 232 50

(3)

6400 + 160R + R 2 + X 2 = 53824 160R + R 2 + X 2 = 47424

(4)

Subtracting (3) from (4), 160R = 16448  → R = 102.8 Ω From (3), X 2 = 30976 − 10568 = 20408 X = 142.86 = 377 L  → L = 0.3789 H

Chapter 9, Solution 91.

Z in =

1 + R || jωL jωC

Z in =

-j jωLR + ωC R + jωL

- j ω 2 L2 R + jωLR 2 = + ωC R 2 + ω 2 L2 To have a resistive impedance, Im(Z in ) = 0 . Hence, -1 ωLR 2 =0 + 2 ωC R + ω2 L2 1 ωLR 2 = 2 ωC R + ω2 L2 R 2 + ω2 L2 C= ω2 LR 2

where ω = 2π f = 2π × 10 7 C=

9 × 10 4 + (4π 2 × 1014 )(400 × 10 −12 ) (4π 2 × 1014 )(20 × 10 − 6 )(9 × 10 4 )

C=

9 + 16π 2 nF 72π 2

C = 235 pF

Chapter 9, Solution 92.

(a) Z o =

Z = Y

100∠75 o = 471.4∠13.5 o Ω o −6 450∠48 x10

(b) γ = ZY = 100∠75 o x 450∠48 o x10 −6 = 0.2121∠61.5 o

Chapter 9, Solution 93. Z = Zs + 2 ZA + ZL Z = (1 + 0.8 + 23.2) + j(0.5 + 0.6 + 18.9) Z = 25 + j20 IL =

VS 115∠0° = Z 32.02 ∠38.66°

I L = 3.592∠-38.66° A

Chapter 10, Solution 1.

ω=1 10 cos( t − 45°)  → 10∠ - 45° 5 sin( t + 30°)  → 5∠ - 60°

1H  → 1F  →

jωL = j 1 = -j jωC

The circuit becomes as shown below. 3Ω

10∠-45° V

+ −

Vo

jΩ

2 Io

+ −

5∠-60° V

Applying nodal analysis, (10∠ - 45°) − Vo (5∠ - 60°) − Vo Vo + = 3 j -j j10∠ - 45° + 15∠ - 60° = j Vo Vo = 10 ∠ - 45° + 15∠ - 150° = 15.73∠247.9° Therefore, v o ( t ) = 15.73 cos(t + 247.9°) V Chapter 10, Solution 2. ω = 10 4 cos(10t − π 4)  → 4∠ - 45°

20 sin(10 t + π 3)  → 20 ∠ - 150° 1H  →

jωL = j10 1 1 0.02 F  → = = - j5 jωC j 0.2

The circuit becomes that shown below.

10 Ω

Vo Io

20∠-150° V

+ −

j10 Ω

4∠-45° A

-j5 Ω

Applying nodal analysis, Vo Vo (20∠ - 150°) − Vo + 4∠ - 45° = + 10 j10 - j5 20 ∠ - 150° + 4∠ - 45° = 0.1(1 + j) Vo Io =

Therefore,

Vo 2 ∠ - 150° + 4 ∠ - 45° = 2.816 ∠150.98° = j10 j (1 + j) i o ( t ) = 2.816 cos(10t + 150.98°) A

Chapter 10, Solution 3. ω= 4 2 cos(4t )  → 2∠0° 16 sin(4 t )  → 16∠ - 90° = -j16 2H  →

jωL = j8 1 1 1 12 F  → = = - j3 jωC j (4)(1 12) The circuit is shown below.

4Ω

-j16 V

+ −

-j3 Ω

Vo

1Ω

j8 Ω

2∠0° A

6Ω

Applying nodal analysis,

Vo Vo - j16 − Vo +2= + 4 − j3 1 6 + j8  - j16 1 1  V + 2 = 1 + + 4 − j3  4 − j3 6 + j8  o Vo =

3.92 − j2.56 4.682∠ - 33.15° = = 3.835∠ - 35.02° 1.22 + j0.04 1.2207 ∠1.88° v o ( t ) = 3.835 cos(4t – 35.02°) V

Therefore,

Chapter 10, Solution 4. 16 sin(4 t − 10°)  → 16∠ - 10°, ω = 4 1H  →

jωL = j4

0.25 F  → Ix

16∠-10° V

+ −

1 1 = = -j jωC j (4)(1 4) j4 Ω

V1

-j Ω +

0.5 Ix

1Ω

Vo −

(16∠ - 10°) − V1 1 V + Ix = 1 j4 2 1− j But Ix =

So,

(16∠ - 10°) − V1 j4

3 ((16∠ - 10°) − V1 ) V = 1 j8 1− j

V1 =

48∠ - 10° - 1 + j4

Using voltage division, 1 48∠ - 10° Vo = V1 = = 8.232∠ - 69.04° 1− j (1 - j)(-1 + j4) v o ( t ) = 8.232 sin(4t – 69.04°) V

Therefore, Chapter 10, Solution 5.

Let the voltage across the capacitor and the inductor be Vx and we get: Vx − 0.5I x − 10∠30° Vx Vx + + =0 4 − j2 j3 (3 + j6 − j4)Vx − 1.5I x = 30∠30° but I x =

Vx = j0.5Vx − j2

Combining these equations we get: (3 + j2 − j0.75)Vx = 30∠30° or Vx = I x = j0.5

30∠30° 3 + j1.25

30∠30° = 4.615∠97.38° A 3 + j1.25

Chapter 10, Solution 6. Let Vo be the voltage across the current source. Using nodal analysis we get: Vo − 4Vx Vo 20 −3+ = 0 where Vx = Vo 20 + j10 20 20 + j10 Combining these we get: Vo 4Vo Vo − −3+ = 0 → (1 + j0.5 − 3)Vo = 60 + j30 20 20 + j10 20 + j10 Vo =

60 + j30 20(3) or Vx = = 29.11∠–166˚ V. − 2 + j0.5 − 2 + j0.5

Chapter 10, Solution 7. At the main node, 120∠ − 15 o − V V V = 6∠30 o + + − j30 50 40 + j20

 →

115.91 − j31.058 − 5.196 − j3 = 40 + j20

 1 j 1 + +  V  40 + j20 30 50  V=

− 3.1885 − j4.7805 = 124.08∠ − 154 o V 0.04 + j0.0233

Chapter 10, Solution 8.

ω = 200, 100mH 50µF

 →  →

jωL = j200x 0.1 = j20 1 1 = = − j100 jωC j200x 50x10 − 6

The frequency-domain version of the circuit is shown below. 0.1 Vo 40 Ω

V1

6∠15

o

20 Ω

+ Vo -

Io

V2

-j100 Ω

j20 Ω

At node 1,

or

V V1 V − V2 6∠15 o + 0.1V1 = 1 + + 1 20 − j100 40 5.7955 + j1.5529 = (−0.025 + j 0.01)V1 − 0.025V2

(1)

At node 2, V V1 − V2 = 0.1V1 + 2 j20 40 From (1) and (2),

→

0 = 3V1 + (1 − j2)V2

(−0.025 + j0.01) − 0.025 V1   (5.7955 + j1.5529)   =   3 (1 − j2)  V2   0  

or

(2)

AV = B

Using MATLAB, V = inv(A)*B V2 = −110.3 + j161.09 leads to V1 = −70.63 − j127.23, V − V2 Io = 1 = 7.276∠ − 82.17 o 40

Thus, i o ( t ) = 7.276 cos(200 t − 82.17 o ) A

Chapter 10, Solution 9. 10 cos(10 3 t )  → 10 ∠0°, ω = 10 3 10 mH  →

jωL = j10

50 µF  →

1 1 = = - j20 3 jωC j (10 )(50 × 10 -6 )

Consider the circuit shown below.

20 Ω

V1

-j20 Ω

V2

j10 Ω

Io

10∠0° V

+ −

20 Ω

+ 4 Io

30 Ω

Vo −

At node 1, 10 − V1 V1 V1 − V2 = + 20 20 - j20 10 = (2 + j) V1 − jV2

(1)

At node 2,

V1 − V2 V V2 V , where I o = 1 has been substituted. = (4) 1 + 20 - j20 20 30 + j10 (-4 + j) V1 = (0.6 + j0.8) V2

V1 =

0.6 + j0.8 V2 -4+ j

(2)

Substituting (2) into (1) (2 + j)(0.6 + j0.8) 10 = V2 − jV2 -4+ j or

V2 =

170 0.6 − j26.2

Vo =

30 3 170 V2 = ⋅ = 6.154 ∠70.26° 30 + j10 3 + j 0.6 − j26.2 v o ( t ) = 6.154 cos(103 t + 70.26°) V

Therefore,

Chapter 10, Solution 10.

50 mH 2µF

 →  →

jωL = j2000x50 x10 − 3 = j100,

ω = 2000

1 1 = = − j250 jωC j2000 x 2x10 − 6

Consider the frequency-domain equivalent circuit below. V1

36<0o

2k Ω

-j250

j100

V2

0.1V1

4k Ω

At node 1, 36 =

V1 V V − V2 + 1 + 1 2000 j100 − j250

 →

36 = (0.0005 − j0.006)V1 − j0.004V2

(1)

At node 2, V V1 − V2 = 0.1V1 + 2 4000 − j250

→

0 = (0.1 − j0.004)V1 + (0.00025 + j0.004)V2 (2)

Solving (1) and (2) gives Vo = V2 = −535.6 + j893.5 = 8951.1∠93.43o vo (t) = 8.951 sin(2000t +93.43o) kV

Chapter 10, Solution 11. cos(2t )  → 1∠0°, ω = 2 8 sin( 2t + 30°)  → 8∠ - 60° 1H  →

jωL = j2

12F  →

1 1 = = -j jωC j (2)(1 2)

2H  →

jωL = j4

14F  →

1 1 = = - j2 jωC j (2)(1 4)

Consider the circuit below.

2 Io

2 Io

2 Io

2 -j Ω 2

2 Io

2 I

-j Ω

2 2 Io

At node 1, (8∠ - 60°) − V1 V1 V1 − V2 = + 2 -j j2 8∠ - 60° = (1 + j) V1 + j V2

(1)

At node 2, 1+

V1 − V2 (8∠ - 60°) − V2 =0 + j2 j4 − j2

V2 = 4 ∠ - 60° + j + 0.5 V1 Substituting (2) into (1), 1 + 8∠ - 60° − 4 ∠30° = (1 + j1.5) V1

Therefore,

(2)

V1 =

1 + 8∠ - 60° − 4∠30° 1 + j1.5

Io =

V1 1 + 8∠ - 60° − 4 ∠30° = 5.024∠ - 46.55° = -j 1.5 − j

i o ( t ) = 5.024 cos(2t – 46.55°)

Chapter 10, Solution 12. 20 sin(1000t )  → 20 ∠0°, ω = 1000 10 mH  →

jωL = j10

50 µF  →

1 1 = = - j20 3 jωC j (10 )(50 × 10 -6 )

The frequency-domain equivalent circuit is shown below.

2 Io

V1

10 Ω

V2 Io

20∠0° A

20 Ω

-j20 Ω

j10 Ω

At node 1, 20 = 2 I o +

V1 V1 − V2 + , 20 10

where

Io =

V2 j10

20 =

2V2 V1 V1 − V2 + + j10 20 10

400 = 3V1 − (2 + j4) V2

(1)

At node 2, 2V2 V1 − V2 V V + = 2 + 2 j10 10 - j20 j10 j2 V1 = (-3 + j2) V2 V1 = (1 + j1.5) V2 or Substituting (2) into (1), 400 = (3 + j4.5) V2 − (2 + j4) V2 = (1 + j0.5) V2

Therefore,

V2 =

400 1 + j0.5

Io =

V2 40 = = 35.74 ∠ - 116.6° j10 j (1 + j0.5)

(2)

i o ( t ) = 35.74 sin(1000t – 116.6°) A

Chapter 10, Solution 13. Nodal analysis is the best approach to use on this problem. We can make our work easier by doing a source transformation on the right hand side of the circuit.

–j2 Ω

40∠30º V

+ −

18 Ω

j6 Ω

+ Vx −

3Ω

50∠0º V

+ −

Vx − 40∠30° Vx Vx − 50 + + =0 − j2 3 18 + j6 which leads to Vx = 29.36∠62.88˚ A.

Chapter 10, Solution 14. At node 1, 0 − V1 0 − V1 V2 − V1 + + = 20∠30° - j2 10 j4 - (1 + j2.5) V1 − j2.5 V2 = 173.2 + j100

(1)

At node 2,

V2 V2 V2 − V1 + + = 20∠30° j2 - j5 j4 - j5.5 V2 + j2.5 V1 = 173.2 + j100

Equations (1) and (2) can be cast into matrix form as 1 + j2.5 j2.5  V1   - 200 ∠30° =  j2.5 - j5.5 V2   200 ∠30°   ∆=

1 + j2.5 j2.5 = 20 − j5.5 = 20.74∠ - 15.38° j2.5 - j5.5

∆1 =

∆2 =

- 200 ∠30° j2.5 = j3 (200∠30°) = 600∠120° 200 ∠30° - j5.5 1 + j2.5 - 200∠30° j2.5

200∠30°

V1 =

∆1 = 28.93∠135.38° ∆

V2 =

∆2 = 49.18∠124.08° ∆

= (200 ∠30°)(1 + j5) = 1020∠108.7°

(2)

Chapter 10, Solution 15. We apply nodal analysis to the circuit shown below.

5A

2Ω

V2

I

+ −

-j20 V

jΩ

V1

-j2 Ω

2I

4Ω

At node 1,

V V − V2 - j20 − V1 = 5+ 1 + 1 2 - j2 j - 5 − j10 = (0.5 − j0.5) V1 + j V2

At node 2, 5 + 2I +

V1 − V2 V2 , = j 4

where I =

V2 =

V1 - j2

5 V1 0.25 − j (2)

Substituting (2) into (1), - 5 − j10 −

j5 = 0.5 (1 − j) V1 0.25 − j

(1 − j) V1 = -10 − j20 −

j40 1 − j4

( 2 ∠ - 45°) V1 = -10 − j20 + V1 = 15.81∠313.5°

160 j40 − 17 17

(1)

I=

V1 = (0.5∠90°)(15.81∠313.5°) - j2

I = 7.906∠43.49° A Chapter 10, Solution 16.

At node 1, V1 V1 − V2 V1 − V2 + + 20 10 - j5 j40 = (3 + j4) V1 − (2 + j4) V2

j2 =

At node 2, V1 − V2 V1 − V2 V + +1+ j = 2 10 - j5 j10 10 (1 + j) = - (1 + j2) V1 + (1 + j) V2

Thus,  j40 10 (1 + 

  3 + j4 - 2 (1 + j2)  V1  = j)   - (1 + j2) 1 + j  V2 

∆=

3 + j4 - 2 (1 + j2) = 5 − j = 5.099 ∠ - 11.31° - (1 + j2) 1+ j

∆1 =

j40 - 2 (1 + j2) = −60 + j100 = 116.62 ∠120.96° 10 (1 + j) 1+ j

∆2 =

3 + j4

j40

- (1 + j2) 10 (1 + j)

= -90 + j110 = 142.13∠129.29°

∆1 = 22.87∠132.27° V ∆ ∆2 V2 = = 27.87∠140.6° V ∆ V1 =

Chapter 10, Solution 17. Consider the circuit below.

j4 Ω 100∠20° V

1Ω Io

+ −

2Ω

V1

V2

3Ω

-j2 Ω

At node 1, 100∠20° − V1 V1 V1 − V2 = + j4 3 2 100 ∠20° =

V1 (3 + j10) − j2 V2 3

(1)

At node 2, 100∠20° − V2 V1 − V2 V2 + = 1 2 - j2 100 ∠20° = -0.5 V1 + (1.5 + j0.5) V2 From (1) and (2), 100∠20°  - 0.5 0.5 (3 + j)  V1  = 100∠20° 1 + j10 3 - j2  V2    

∆=

- 0.5 1.5 + j0.5 = 0.1667 − j4.5 1 + j10 3 - j2

∆1 =

∆2 =

100∠20° 1.5 + j0.5 = -55.45 − j286.2 100∠20° - j2 - 0.5

100∠20°

1 + j10 3 100∠20°

= -26.95 − j364.5

(2)

V1 =

∆1 = 64.74 ∠ - 13.08° ∆

V2 =

∆2 = 81.17 ∠ - 6.35° ∆

Io =

V1 − V2 ∆ 1 − ∆ 2 - 28.5 + j78.31 = = 2 2∆ 0.3333 − j 9

I o = 9.25∠-162.12° Chapter 10, Solution 18. Consider the circuit shown below.

8Ω

V1

j6 Ω V 2

4Ω

j5 Ω

+ 4∠45° A

2Ω

+ 2 Vx

Vx

-j Ω

-j2 Ω

−

−

At node 1, 4∠45° =

V1 V1 − V2 + 2 8 + j6

200 ∠45° = (29 − j3) V1 − (4 − j3) V2

(1)

At node 2,

V1 − V2 V V2 , + 2Vx = 2 + 8 + j6 - j 4 + j5 − j2

where Vx = V1

(104 − j3) V1 = (12 + j41) V2

12 + j41 V 104 − j3 2 Substituting (2) into (1),

V1 =

200∠45° = (29 − j3)

Vo

(2) (12 + j41) V − (4 − j3) V2 104 − j3 2

200 ∠45° = (14.21∠89.17°) V2

V2 =

200∠45° 14.21∠89.17°

Vo =

- j2 - j2 - 6 − j8 V2 = V2 = V2 4 + j5 − j2 4 + j3 25

Vo =

10∠233.13° 200∠45° ⋅ 25 14.21∠89.17°

Vo = 5.63∠189° V Chapter 10, Solution 19. We have a supernode as shown in the circuit below. j2 Ω

V1

V2

4Ω V3

+ 2Ω

Vo

-j4 Ω

0.2 Vo

−

Vo = V1 . Notice that At the supernode, V3 − V2 V2 V1 V1 − V3 = + + 4 - j4 2 j2 0 = (2 − j2) V1 + (1 + j) V2 + (-1 + j2) V3

(1)

At node 3, 0.2V1 +

V1 − V3 V3 − V2 = j2 4

(0.8 − j2) V1 + V2 + (-1 + j2) V3 = 0 Subtracting (2) from (1),

(2)

0 = 1.2V1 + j V2 But at the supernode, V1 = 12 ∠0° + V2 V2 = V1 − 12 or Substituting (4) into (3), 0 = 1.2V1 + j (V1 − 12)

V1 =

j12 = Vo 1.2 + j

Vo =

12∠90° 1.562∠39.81°

(3)

(4)

Vo = 7.682∠50.19° V Chapter 10, Solution 20. The circuit is converted to its frequency-domain equivalent circuit as shown below. R + Vm∠0°

+ −

jωL

Vo

1 jωC

−

Let

Z = jωL ||

1 = jωC

L C

1 jωL + jωC

=

jωL 1 − ω2 LC

jωL Z jωL 1 − ω2 LC Vo = Vm = Vm = V jωL R+Z R (1 − ω2 LC) + jωL m R+ 1 − ω2 LC

Vo =

  ωL  90° − tan -1 ∠ 2 R (1 − ω LC)  R 2 (1 − ω2 LC) 2 + ω2 L2  ωL Vm

If

Vo = A∠φ , then

A=

and

ωL Vm R 2 (1 − ω 2 LC) 2 + ω 2 L2

φ = 90° − tan -1

ωL R (1 − ω 2 LC)

Chapter 10, Solution 21.

(a)

Vo = Vi

1 jωC R + jωL +

1 jωC

As ω → ∞ ,

(b)

Vo = Vi

1 LC

Vo = Vi

,

jωL R + jωL +

As ω → ∞ , 1 LC

1 jωC

1 jRC ⋅

1

=

-j L R C

LC

− ω2 LC = 1 − ω2 LC + jωRC Vo = 0 Vi Vo 1 = = 1 Vi 1

At ω = 0 ,

At ω =

1 1 − ω LC + jωRC 2

Vo 1 = = 1 Vi 1 Vo = 0 Vi

At ω = 0 ,

At ω =

=

,

Vo = Vi

−1 jRC ⋅

1 LC

=

j L R C

Chapter 10, Solution 22. Consider the circuit in the frequency domain as shown below. R1

R2 Vs

+ −

1 jωC

jωL

Let

Z = (R 2 + jωL) ||

+ Vo

−

1 jωC

1 (R + jωL) R 2 + jωL jωC 2 Z= = 1 1 + jωR 2 − ω2 LC R 2 + jωL + jωC R 2 + jωL Vo 1 − ω2 LC + jωR 2 C Z = = R 2 + jωL Vs Z + R 1 R1 + 1 − ω2 LC + jωR 2 C Vo R 2 + jωL = 2 Vs R 1 + R 2 − ω LCR 1 + jω (L + R 1 R 2 C)

Chapter 10, Solution 23.

V − Vs V + + jωCV = 0 1 R jωL + jω C V+

jωRCV − ω2LC + 1

+ jωRCV = Vs

 1 − ω2LC + jωRC + jωRC − jω3RLC2    V = Vs 2   − ω 1 LC  

V=

(1 − ω2 LC)Vs

1 − ω2LC + jωRC(2 − ω2LC)

Chapter 10, Solution 24.

For mesh 1,  1 1  1  I1 − + Vs =  I jωC 2 2  jωC1 jωC 2 

(1)

For mesh 2,  1  −1 I I 1 +  R + jωL + jωC 2 jωC 2  2  Putting (1) and (2) into matrix form, −1 1  1  +    I1  Vs  jωC1 jωC 2 jωC 2   = 0 −1 1  I 2     R + jωL +  jω C 2 jωC 2  0=

(2)

 1 1  1  1 + 2  R + jωL + ∆ = + jωC 2  ω C1C 2  jωC1 jωC 2   1   ∆ 1 = Vs  R + jωL + jωC 2  

and

 1   Vs  R + jωL + jωC 2   ∆1 I1 = = ∆  1 1  1 1    R + jωL + + 2 + jωC 2  ω C1 C 2  jωC 1 jωC 2 

I2 =

Vs jωC 2

∆2 = ∆  1 1  1  1 + 2   R + jωL + + jωC 2  ω C1 C 2  jωC 1 jωC 2 

Chapter 10, Solution 25.

ω= 2 10 cos(2t )  → 10∠0°

∆2 =

Vs jωC 2

6 sin(2t )  → 6 ∠ - 90° = -j6

2H  →

jωL = j4 1 1 0.25 F  → = = - j2 jωC j (2)(1 4) The circuit is shown below.

4Ω

j4 Ω Io

10∠0° V

+ −

I1

-j2 Ω

I2

+ −

6∠-90° V

For loop 1, - 10 + (4 − j2) I 1 + j2 I 2 = 0 5 = (2 − j) I 1 + j I 2

(1)

For loop 2, j2 I 1 + ( j4 − j2) I 2 + (- j6) = 0 I1 + I 2 = 3 In matrix form (1) and (2) become  2 − j j   I 1   5  1 1   I  =  3  2    

∆ = 2 (1 − j) ,

∆ 1 = 5 − j3 ,

(2)

∆ 2 = 1 − j3

∆1 − ∆ 2 4 = = 1 + j = 1.414 ∠45° 2 (1 − j) ∆ i o ( t ) = 1.414 cos(2t + 45°) A I o = I1 − I 2 =

Therefore,

Chapter 10, Solution 26. We apply mesh analysis to the circuit shown below. For mesh 1, - 10 + 40 I 1 − 20 I 2 = 0 1 = 4 I1 − 2 I 2 For the supermesh, (20 − j20) I 2 − 20 I 1 + (30 + j10) I 3 = 0

- 2 I 1 + (2 − j2) I 2 + (3 + j) I 3 = 0

(1)

(2)

At node A,

I o = I1 − I 2

(3)

At node B,

I2 = I3 + 4Io Substituting (3) into (4) I 2 = I 3 + 4 I1 − 4 I 2 I 3 = 5 I 2 − 4 I1 Substituting (5) into (2) gives 0 = -(14 + j4) I 1 + (17 + j3) I 2 From (1) and (6), 1   4 - 2  I 1   0 =  - (14 + j4) 17 + j3 I      2 

(4)

(5) (6)

∆ = 40 + j4 ∆1 =

1

-2

0 17 + j3

I 3 = 5 I 2 − 4 I1 =

Vo = 30 I 3 = Therefore,

= 17 + j3 ,

∆2 =

4

1

- (14 + j4) 0

= 14 + j4

5 ∆ 2 − 4 ∆1 2 + j8 = ∆ 40 + j4

15 (1 + j4) = 6.154∠70.25° 10 + j

v o ( t ) = 6.154 cos(103 t + 70.25°) V

Chapter 10, Solution 27. For mesh 1, - 40 ∠30° + ( j10 − j20) I 1 + j20 I 2 = 0 4 ∠30° = - j I 1 + j2 I 2

(1)

For mesh 2, 50 ∠0° + (40 − j20) I 2 + j20 I 1 = 0 5 = - j2 I 1 − (4 − j2) I 2 From (1) and (2),  4∠30°  - j j2  I 1  =  5   - j2 - (4 − j2)  I      2 

∆ = -2 + 4 j = 4.472∠116.56°

(2)

∆ 1 = -(4 ∠30°)(4 − j2) − j10 = 21.01∠211.8° ∆ 2 = - j5 + 8∠120° = 4.44 ∠154.27° I1 =

∆1 = 4.698∠95.24° A ∆

I2 =

∆2 = 0.9928∠37.71° A ∆

Chapter 10, Solution 28. 1H

 →

jωL = j4,

 →

1F

1 1 = = − j0.25 jωC j1x 4

The frequency-domain version of the circuit is shown below, where V1 = 10∠0 o ,

V2 = 20∠ − 30 o .

1

j4

j4

1

-j0.25 +

+

V1 -

I1

V1 = 10∠0 o ,

1

I2

V2 -

V2 = 20∠ − 30 o

Applying mesh analysis, 10 = (2 + j3.75)I1 − (1 − j0.25)I 2

(1)

− 20∠ − 30 o = −(1 − j0.025)I1 + (2 + j3.75)I 2

(2)

From (1) and (2), we obtain

10   2 + j3.75 − 1 + j0.25  I1      =    − 17.32 + j10   − 1 + j0.25 2 + j3.75  I 2 

Solving this leads to I1 = 1.3602 − j0.9769 = 1.6747∠ − 35.69 o ,

I 2 = −4.1438 + j2.111 = 4.6505∠153o

Hence, i1 = 1.675 cos(4t − 35.69 o ) A,

i 2 = 4.651cos(46 + 153o ) A

Chapter 10, Solution 29. For mesh 1, (5 + j5) I 1 − (2 + j) I 2 − 30 ∠20° = 0 30 ∠20° = (5 + j5) I 1 − (2 + j) I 2

(1)

For mesh 2, (5 + j3 − j6) I 2 − (2 + j) I 1 = 0 0 = - (2 + j) I 1 + (5 − j3) I 2 From (1) and (2), 30∠20°  5 + j5 - (2 + j)  I 1   0  =  - (2 + j) 5 - j3  I      2 

∆ = 37 + j6 = 37.48∠9.21° ∆ 1 = (30 ∠20°)(5.831∠ - 30.96°) = 175∠ - 10.96° ∆ 2 = (30 ∠20°)(2.356 ∠26.56°) = 67.08∠46.56° I1 =

∆1 = 4.67∠-20.17° A ∆

I2 =

∆2 = 1.79∠37.35° A ∆

(2)

Chapter 10, Solution 30. Consider the circuit shown below.

I2 j4 Ω 10∠0° V

+ −

1Ω Io

I1 3Ω

2Ω

I3

-j2 Ω

For mesh 1, 100 ∠20° = (3 + j4) I 1 − j4 I 2 − 3 I 3

(1)

0 = - j4 I 1 + (3 + j4) I 2 − j2 I 3

(2)

For mesh 2, For mesh 3, 0 = -3 I 1 − 2 I 2 + (5 − j2) I 3 Put (1), (2), and (3) into matrix form. 3 + j4 - j4 - 3  I 1  100∠20°      j4 3 + j4 j2 0 = I 2        - 3  -2 5 - j2  I 3   0

3 + j4 - j4 -3 ∆ = - j4 3 + j4 - j2 = 106 + j30 -3 -2 5 - j2 3 + j4 100∠20° - 3 ∆ 2 = - j4 0 - j2 = (100∠20°)(8 + j26) -3 0 5 - j2 3 + j4 - j4 100∠20° ∆ 3 = - j4 3 + j4 0 = (100∠20°)(9 + j20) -3 -2 0 Io = I3 − I2 =

∆ 3 − ∆ 2 (100∠20°)(1 − j6) = ∆ 106 + j30

I o = 5.521∠-76.34° A

(3)

Chapter 10, Solution 31. Consider the network shown below.

80 Ω

100∠120° V

+ −

I1

Io

-j40 Ω

j60 Ω

I2

-j40 Ω

20 Ω

I3

+ −

60∠-30° V

For loop 1, - 100 ∠20° + (80 − j40) I 1 + j40 I 2 = 0 10 ∠20° = 4 (2 − j) I 1 + j4 I 2

(1)

j40 I 1 + ( j60 − j80) I 2 + j40 I 3 = 0 0 = 2 I1 − I 2 + 2 I 3

(2)

60 ∠ - 30° + (20 − j40) I 3 + j40 I 2 = 0 - 6 ∠ - 30° = j4 I 2 + 2 (1 − j2) I 3

(3)

For loop 2,

For loop 3,

From (2), 2 I 3 = I 2 − 2 I1 Substituting this equation into (3), - 6 ∠ - 30° = -2 (1 − j2) I 1 + (1 + j2) I 2

(4)

From (1) and (4),  10∠120°   4 (2 − j) j4  I 1   - 6∠ - 30° =  - 2 (1 − j2) 1 + j2 I   2    

∆=

∆2 =

8 − j4 - j4 = 32 + j20 = 37.74∠32° - 2 + j4 1 + j2 8 − j4 10∠120° = -4.928 + j82.11 = 82.25∠93.44° - 2 + j4 - 6∠ - 30°

Io = I2 =

∆2 = 2.179∠61.44° A ∆

Chapter 10, Solution 32. Consider the circuit below.

j4 Ω Io

+ 2Ω

4∠-30° V

Vo

I1

+ −

3 Vo

− For mesh 1, where

(2 + j4) I 1 − 2 (4∠ - 30°) + 3 Vo = 0 Vo = 2 (4∠ - 30° − I 1 )

Hence, (2 + j4) I 1 − 8∠ - 30° + 6 (4 ∠ - 30° − I 1 ) = 0 4 ∠ - 30° = (1 − j) I 1

I 1 = 2 2 ∠15°

or

Io =

3 Vo 3 = (2)(4∠ - 30° − I 1 ) - j2 - j2

I o = j3 (4 ∠ - 30° − 2 2 ∠15°)

I o = 8.485∠15° A Vo =

- j2 I o = 5.657∠-75° V 3

Chapter 10, Solution 33. Consider the circuit shown below.

I2

-j2 Ω

5A

I4 2Ω

jΩ I

-j20 V

+ −

I1

I2

-j2 Ω

2I

I3

4Ω

For mesh 1, j20 + (2 − j2) I 1 + j2 I 2 = 0 (1 − j) I 1 + j I 2 = - j10

(1)

For the supermesh, ( j − j2) I 2 + j2 I 1 + 4 I 3 − j I 4 = 0

(2)

Also, I 3 − I 2 = 2 I = 2 (I 1 − I 2 ) I 3 = 2 I1 − I 2

(3)

I4 = 5

(4)

For mesh 4,

Substituting (3) and (4) into (2), (8 + j2) I 1 − (- 4 + j) I 2 = j5

(5)

Putting (1) and (5) in matrix form,  1− j j  I 1   - j10  8 + j2 4 − j I  =  j5     2  

∆ = -3 − j5 , I = I1 − I 2 =

∆ 1 = -5 + j40 ,

∆ 2 = -15 + j85

∆ 1 − ∆ 2 10 − j45 = = 7.906∠43.49° A ∆ - 3 − j5

Chapter 10, Solution 34.

The circuit is shown below.

Io I2

5Ω

3A 20 Ω

8Ω 40∠90° V

+ −

-j2 Ω

I3

10 Ω I1

j15 Ω

j4 Ω

For mesh 1, - j40 + (18 + j2) I 1 − (8 − j2) I 2 − (10 + j4) I 3 = 0 For the supermesh, (13 − j2) I 2 + (30 + j19) I 3 − (18 + j2) I 1 = 0

(1)

(2)

Also, I2 = I3 − 3

Adding (1) and (2) and incorporating (3), - j40 + 5 (I 3 − 3) + (20 + j15) I 3 = 0 3 + j8 I3 = = 1.465∠38.48° 5 + j3 I o = I 3 = 1.465∠38.48° A

(3)

Chapter 10, Solution 35.

Consider the circuit shown below.

4Ω

j2 Ω I3

8Ω

1Ω

-j3 Ω

10 Ω 20 V

+ −

I1

-j4 A

I2 -j5 Ω

For the supermesh, - 20 + 8 I 1 + (11 − j8) I 2 − (9 − j3) I 3 = 0

(1)

Also, I 1 = I 2 + j4

(2)

(13 − j) I 3 − 8 I 1 − (1 − j3) I 2 = 0

(3)

For mesh 3,

Substituting (2) into (1), (19 − j8) I 2 − (9 − j3) I 3 = 20 − j32

(4)

Substituting (2) into (3), - (9 − j3) I 2 + (13 − j) I 3 = j32

(5)

From (4) and (5),  19 − j8 - (9 − j3)  I 2   20 − j32   - (9 − j3) 13 − j  I  =  j32    3   

∆ = 167 − j69 ,

∆ 2 = 324 − j148

∆ 2 324 − j148 356.2∠ - 24.55° = = ∆ 167 − j69 180.69∠ - 22.45° I 2 = 1.971∠-2.1° A I2 =

Chapter 10, Solution 36. Consider the circuit below. j4 Ω

-j3 Ω +

I1

4∠90° A

2Ω

Vo

I2

+ −

− 2Ω

2Ω I3

2∠0° A Clearly, I 1 = 4 ∠90° = j4

and

I 3 = -2

For mesh 2, (4 − j3) I 2 − 2 I 1 − 2 I 3 + 12 = 0 (4 − j3) I 2 − j8 + 4 + 12 = 0 - 16 + j8 = -3.52 − j0.64 I2 = 4 − j3 Thus, Vo = 2 (I 1 − I 2 ) = (2)(3.52 + j4.64) = 7.04 + j9.28 Vo = 11.648∠52.82° V Chapter 10, Solution 37. I1 + 120∠ − 90 o V -

Ix Z

Z=80-j35 Ω

I2 Iy

Iz

12∠0° V

120∠ − 30 o V +

Z I3

For mesh x, ZI x − ZI z = − j120

(1)

ZI y − ZI z = −120∠30 o = −103.92 + j60

(2)

− ZI x − ZI y + 3ZI z = 0

(3)

For mesh y,

For mesh z,

Putting (1) to (3) together leads to the following matrix equation: 0 (−80 + j35)  I x   − j120  (80 − j35)       0 (80 − j35) (−80 + j35)  I y  =  − 103.92 + j60    (−80 + j35) (−80 + j35) (240 − j105)  I    0   z   

 →

Using MATLAB, we obtain  - 1.9165 + j1.4115    I = inv(A) * B =  - 2.1806 - j0.954   - 1.3657 + j0.1525   

I1 = I x = −1.9165 + j1.4115 = 2.3802∠143.6 o A I 2 = I y − I x = −0.2641 − j2.3655 = 2.3802∠ − 96.37 o A I 3 = − I y = 2.1806 + j0.954 = 2.3802∠23.63o A

Chapter 10, Solution 38. Consider the circuit below.

AI = B

Io I1

2∠0° A

2Ω

j2 Ω

1Ω

I2

+ −

-j4 Ω

4∠0° A

I3

I4

10∠90° V

1Ω

A Clearly, I1 = 2

(1)

For mesh 2, (2 − j4) I 2 − 2 I 1 + j4 I 4 + 10 ∠90° = 0 Substitute (1) into (2) to get (1 − j2) I 2 + j2 I 4 = 2 − j5 For the supermesh, (1 + j2) I 3 − j2 I 1 + (1 − j4) I 4 + j4 I 2 = 0

j4 I 2 + (1 + j2) I 3 + (1 − j4) I 4 = j4

(2)

(3)

At node A, I3 = I4 − 4 Substituting (4) into (3) gives j2 I 2 + (1 − j) I 4 = 2 (1 + j3) From (2) and (5), 1 − j2 j2  I 2   2 − j5  j2 1 − j I  =  2 + j6   4    ∆ = 3 − j3 ,

∆ 1 = 9 − j11

- ∆ 1 - (9 − j11) 1 = = (-10 + j) ∆ 3 − j3 3 I o = 3.35∠174.3° A

Io = -I2 =

(4) (5)

Chapter 10, Solution 39. For mesh 1, (28 − j15)I1 − 8I 2 + j15I 3 = 12∠64 o

(1)

− 8I1 + (8 − j9)I 2 − j16I 3 = 0

(2)

j15I1 − j16I 2 + (10 + j)I 3 = 0

(3)

For mesh 2, For mesh 3,

In matrix form, (1) to (3) can be cast as j15  I1  12∠64 o  −8  (28 − j15)    (8 − j9) − j16  I 2  =  0   −8    j15 − j16 (10 + j)  I 3   0     Using MATLAB, I = inv(A)*B I1 = −0.128 + j0.3593 = 0.3814∠109.6 o A I 2 = −0.1946 + j0.2841 = 0.3443∠124.4 o A

I 3 = 0.0718 − j0.1265 = 0.1455∠ − 60.42 o A

or

AI = B

I x = I1 − I 2 = 0.0666 + j0.0752 = 0.1005∠48.5 o A Chapter 10, Solution 40.

Let i O = i O1 + i O 2 , where i O1 is due to the dc source and i O 2 is due to the ac source. For i O1 , consider the circuit in Fig. (a). 4Ω

2Ω iO1

+ −

8V

(a)

Clearly, i O1 = 8 2 = 4 A For i O 2 , consider the circuit in Fig. (b). 4Ω

2Ω IO2

10∠0° V

+ −

j4 Ω

(b)

If we transform the voltage source, we have the circuit in Fig. (c), where 4 || 2 = 4 3 Ω . IO2 2.5∠0° A

4Ω

(c)

By the current division principle, 43 I O2 = (2.5∠0°) 4 3 + j4 I O 2 = 0.25 − j0.75 = 0.79∠ - 71.56°

2Ω

j4 Ω

i O 2 = 0.79 cos(4t − 71.56°) A

Thus, Therefore,

i O = i O1 + i O 2 = 4 + 0.79 cos(4t – 71.56°) A Chapter 10, Solution 41.

Let vx = v1 + v2. For v1 we let the DC source equal zero. 5Ω +

+ –

20∠0˚

1Ω

–j

V1

−

V1 − 20 V1 V1 + + = 0 which simplifies to (1j − 5 + 5 j)V1 = 100 j 5 −j 1 V1 = 2.56∠–39.8˚ or v1 = 2.56sin(500t – 39.8˚) V Setting the AC signal to zero produces: 5Ω

1Ω + V2

+ –

6V

− The 1-ohm resistor in series with the 5-ohm resistor creating a simple voltage divider yielding: v2 = (5/6)6 = 5 V. vx = {2.56sin(500t – 39.8˚) + 5} V.

Chapter 10, Solution 42.

Let ix = i1 + i2, where i1 and i2 which are generated by is and vs respectively. For i1 we let is = 6sin2t A becomes Is = 6∠0˚, where ω =2. 2 − j4 1 − j2 6 = 12 = 3.724 − j3.31 = 4.983∠ − 41.63° 3 + j2 + 2 − j4 5 − j2 i1= 4.983sin(2t – 41.63˚) A I1 =

–j4

2Ω

i1 3Ω is j2

For i2, we transform vs = 12cos(4t – 30˚) into the frequency domain and get Vs = 12∠–30˚. Thus, I 2 =

12∠ − 30° = 5.385∠8.2° or i2 = 5.385cos(4t + 8.2˚) A 2 − j2 + 3 + j4 –j2

2Ω

i2 3Ω Vs

+ −

j4

ix = [5.385cos(4t + 8.2˚) + 4.983sin(2t – 41.63˚)] A.

Chapter 10, Solution 43. Let i O = i O1 + i O 2 , where i O1 is due to the dc source and i O 2 is due to the ac source. For i O1 , consider the circuit in Fig. (a).

4Ω

2Ω iO1

+ −

8V

(a) Clearly, i O1 = 8 2 = 4 A For i O 2 , consider the circuit in Fig. (b).

4Ω

2Ω IO2

10∠0° V

+ −

j4 Ω

(b) If we transform the voltage source, we have the circuit in Fig. (c), where 4 || 2 = 4 3 Ω .

IO2 2.5∠0° A

4Ω

2Ω

(c) By the current division principle, 43 I O2 = (2.5∠0°) 4 3 + j4 I O 2 = 0.25 − j0.75 = 0.79∠ - 71.56° Thus, i O 2 = 0.79 cos(4t − 71.56°) A Therefore, i O = i O1 + i O 2 = 4 + 0.79 cos (89)(4t – 71.56°) A

j4 Ω

Chapter 10, Solution 44. Let v x = v1 + v 2 , where v1 and v2 are due to the current source and voltage source respectively. For v1 , ω = 6 , 5 H

 →

jωL = j30

The frequency-domain circuit is shown below.

20 Ω

16 Ω

Is

Let Z = 16 //( 20 + j30) =

+ V1 -

16(20 + j30) = 11.8 + j3.497 = 12.31∠16.5 o 36 + j30

V1 = I s Z = (12∠10 o )(12.31∠16.5 o ) = 147.7∠26.5 o

For v2 , ω = 2 , 5 H

j30

 →

 →

v1 = 147.7 cos(6 t + 26.5 o ) V

jωL = j10

The frequency-domain circuit is shown below. 20 Ω

16 Ω

-

j10

+ V2 -

+

Vs -

Using voltage division, 16 16(50∠0 o ) V2 = Vs = = 21.41∠ − 15.52 o 16 + 20 + j10 36 + j10

 →

v 2 = 21.41sin(2t − 15.52 o ) V

Thus, v x = 147.7 cos(6 t + 26.5 o ) + 21.41sin( 2 t − 15.52 o ) V

Chapter 10, Solution 45. Let I o = I 1 + I 2 , where I 1 is due to the voltage source and I 2 is due to the current source. For I 1 , consider the circuit in Fig. (a).

10 Ω

IT I1

20∠-150° V

+ −

j10 Ω

-j5 Ω

(a)

j10 || - j5 = - j10 20 ∠ - 150° 2∠ - 150° IT = = 10 − j10 1− j Using current division, - j5 - j5 2 ∠ - 150° I1 = IT = ⋅ = - (1 + j) ∠ - 150° j10 − j5 j5 1− j For I 2 , consider the circuit in Fig. (b). I2 10 Ω

j10 Ω (b)

10 || - j5 = Using current division,

- j10 2− j

-j5 Ω

4∠-45° A

I2 =

- j10 (2 − j) (4∠ - 45°) = -2 (1 + j) ∠ - 45° - j10 (2 − j) + j10

I o = I 1 + I 2 = - 2 ∠ - 105° − 2 2 ∠0°

I o = -2.462 + j1.366 = 2.816∠150.98°

i o = 2.816 cos(10t + 150.98°) A

Therefore,

Chapter 10, Solution 46.

Let v o = v1 + v 2 + v 3 , where v1 , v 2 , and v 3 are respectively due to the 10-V dc source, the ac current source, and the ac voltage source. For v1 consider the circuit in Fig. (a). 6Ω

2H +

1/12 F

+ −

v1

10 V

− (a)

The capacitor is open to dc, while the inductor is a short circuit. Hence, v1 = 10 V For v 2 , consider the circuit in Fig. (b). ω= 2 2H  → jωL = j4 1 1 1 F  → = = - j6 12 jωC j (2)(1 / 12) + 6Ω

-j6 Ω

4∠0° A

V2

− (b)

Applying nodal analysis, V V V 1 j j  4 = 2 + 2 + 2 =  + −  V2 6 - j6 j4  6 6 4  V2 =

24 = 21.45∠26.56° 1 − j0.5

j4 Ω

Hence,

v 2 = 21.45 sin( 2 t + 26.56°) V

For v 3 , consider the circuit in Fig. (c). ω=3 2H  → jωL = j6 1 1 1 F  → = = - j4 12 jωC j (3)(1 / 12) 6Ω

12∠0° V

j6 Ω +

+ −

-j4 Ω

V3

− (c)

At the non-reference node, 12 − V3 V3 V3 = + 6 - j4 j6 12 V3 = = 10.73∠ - 26.56° 1 + j0.5 Hence, v 3 = 10.73 cos(3t − 26.56°) V Therefore,

v o = 10 + 21.45 sin(2t + 26.56°) + 10.73 cos(3t – 26.56°) V

Chapter 10, Solution 47.

Let i o = i1 + i 2 + i 3 , where i1 , i 2 , and i 3 are respectively due to the 24-V dc source, the ac voltage source, and the ac current source. For i1 , consider the circuit in Fig. (a). 1Ω

24 V

1/6 F

− +

2Ω

(a)

Since the capacitor is an open circuit to dc,

2H i1 4Ω

i1 =

24 =4A 4+2

For i 2 , consider the circuit in Fig. (b). ω=1 2H  → jωL = j2 1 1 F  → = - j6 6 jωC 1Ω

j2 Ω

-j6 Ω

I2 10∠-30° V

+ −

I1

2Ω

I2

4Ω

(b)

For mesh 1, - 10 ∠ - 30° + (3 − j6) I 1 − 2 I 2 = 0 10 ∠ - 30° = 3 (1 − 2 j) I 1 − 2 I 2 (1)

For mesh 2, 0 = -2 I 1 + (6 + j2) I 2 I 1 = (3 + j) I 2

(2) Substituting (2) into (1) 10 ∠ - 30° = 13 − j15 I 2 I 2 = 0.504 ∠19.1° Hence, i 2 = 0.504 sin( t + 19.1°) A For i 3 , consider the circuit in Fig. (c). ω=3 2H  → jωL = j6 1 1 1 F  → = = - j2 jωC j (3)(1 / 6) 6 1Ω

j6 Ω

-j2 Ω

I3 2Ω

(c)

2∠0° A

4Ω

2 || (1 − j2) =

2 (1 − j2) 3 − j2

Using current division, 2 (1 − j2) ⋅ (2∠0°) 2 (1 − j2) 3 − j2 = I3 = 2 (1 − j2) 13 + j3 4 + j6 + 3 − j2 I 3 = 0.3352 ∠ - 76.43° Hence i 3 = 0.3352 cos(3t − 76.43°) A Therefore,

i o = 4 + 0.504 sin(t + 19.1°) + 0.3352 cos(3t – 76.43°) A

Chapter 10, Solution 48.

Let i O = i O1 + i O 2 + i O 3 , where i O1 is due to the ac voltage source, i O 2 is due to the dc voltage source, and i O3 is due to the ac current source. For i O1 , consider the circuit in Fig. (a). ω = 2000 50 cos(2000t )  → 50∠0° → 40 mH 

jωL = j (2000)(40 × 10 -3 ) = j80 1 1 = = - j25 jωC j (2000)(20 × 10 -6 )

→ 20 µF  I

50∠0° V

-j25 Ω

IO1

+ −

80 Ω

j80 Ω

(a)

80 || (60 + 100) = 160 3 50 30 = I= 160 3 + j80 − j25 32 + j33

Using current division,

60 Ω

100 Ω

- 80 I -1 10∠180° = I= 80 + 160 3 46∠45.9° = 0.217 ∠134.1° i O1 = 0.217 cos(2000 t + 134.1°) A

I O1 = I O1 Hence,

For i O 2 , consider the circuit in Fig. (b). iO2 80 Ω

100 Ω

60 Ω

+ −

24 V

(b)

i O2 =

24 = 0.1 A 80 + 60 + 100

For i O3 , consider the circuit in Fig. (c). ω = 4000 2 cos(4000t )  → 2∠0° → 40 mH  20 µF  →

jωL = j (4000)(40 × 10 -3 ) = j160 1 1 = = - j12.5 jωC j (4000)(20 × 10 -6 ) -j12.5 Ω

I2

IO3

80 Ω j160 Ω I3 2∠0° A

I1

100 Ω

60 Ω

(c)

For mesh 1, I1 = 2

For mesh 2,

(1)

(80 + j160 − j12.5) I 2 − j160 I 1 − 80 I 3 = 0 Simplifying and substituting (1) into this equation yields (8 + j14.75) I 2 − 8 I 3 = j32 For mesh 3, 240 I 3 − 60 I 1 − 80 I 2 = 0 Simplifying and substituting (1) into this equation yields I 2 = 3 I 3 − 1.5 Substituting (3) into (2) yields (16 + j44.25) I 3 = 12 + j54.125 12 + j54.125 I3 = = 1.1782∠7.38° 16 + j44.25

(2)

(3)

Hence,

I O 3 = - I 3 = -1.1782∠7.38° i O 3 = -1.1782 sin( 4000t + 7.38°) A

Therefore,

i O = 0.1 + 0.217 cos(2000t + 134.1°) – 1.1782 sin(4000t + 7.38°) A

Chapter 10, Solution 49.

8 sin( 200t + 30°)  → 8∠30°, ω = 200 5 mH  → 1 mF  →

jωL = j (200)(5 × 10 -3 ) = j 1 1 = = - j5 jωC j (200)(1 × 10 -3 )

After transforming the current source, the circuit becomes that shown in the figure below. 5Ω

40∠30° V

3Ω

I

+ −

40 ∠30° 40 ∠30° = = 4.472∠56.56° 5 + 3 + j − j5 8 − j4 i = 4.472 sin(200t + 56.56°) A I=

jΩ -j5 Ω

Chapter 10, Solution 50.

50 cos(10 5 t )  → 50 ∠0°, ω = 10 5 0.4 mH  → 0.2 µF  →

jωL = j (10 5 )(0.4 × 10 -3 ) = j40 1 1 = = - j50 5 jωC j (10 )(0.2 × 10 -6 )

After transforming the voltage source, we get the circuit in Fig. (a). j40 Ω + 20 Ω

2.5∠0° A

-j50 Ω

80 Ω

Vo −

(a)

Let

Z = 20 || - j 50 =

and

- j100 2 − j5

Vs = (2.5∠0°) Z =

- j250 2 − j5

With these, the current source is transformed to obtain the circuit in Fig.(b). j40 Ω

Z

Vs

+

+ −

80 Ω

Vo −

(b)

By voltage division, 80 80 - j250 Vs = ⋅ - j100 Z + 80 + j40 2 − j5 + 80 + j40 2 − j5 8 (- j250) Vo = = 36.15∠ - 40.6° 36 − j42 v o = 36.15 cos(105 t – 40.6°) V

Vo =

Therefore,

Chapter 10, Solution 51. The original circuit with mesh currents and a node voltage labeled is shown below.

Io j10 Ω

4∠-60° V

-j20 Ω

40 Ω

1.25∠0° A

The following circuit is obtained by transforming the voltage sources.

Io 4∠-60° V

j10 Ω

-j20 Ω

40 Ω

Use nodal analysis to find Vx .

 1 1 1 4 ∠ - 60° + 1.25∠0° =  + +  Vx  j10 - j20 40  3.25 − j3.464 = (0.025 − j0.05) Vx 3.25 − j3.464 Vx = = 81.42 + j24.29 = 84.97 ∠16.61° 0.025 − j0.05 Thus, from the original circuit, 40 ∠30° − Vx (34.64 + j20) − (81.42 + j24.29) I1 = = j10 j10 - 46.78 − j4.29 I1 = = -0.429 + j4.678 = 4.698∠95.24° A j10 Vx − 50 ∠0° 31.42 + j24.29 = 40 40 I 2 = 0.7855 + j0.6072 = 0.9928∠37.7° = 0.9928∠37.7° A

I2 =

Chapter 10, Solution 52. We transform the voltage source to a current source. 60∠0° = 6 − j12 Is = 2 + j4

1.25∠0° A

The new circuit is shown in Fig. (a). -j2 Ω Ix 2Ω Is = 6 – j12 A

4Ω

6Ω j4 Ω

5∠90° A

-j3 Ω (a)

Let

6 (2 + j4) = 2.4 + j1.8 8 + j4 Vs = I s Z s = (6 − j12)(2.4 + j1.8) = 36 − j18 = 18 (2 − j)

Z s = 6 || (2 + j4) =

With these, we transform the current source on the left hand side of the circuit to a voltage source. We obtain the circuit in Fig. (b). Zs

-j2 Ω Ix

Vs

4Ω

+ −

j5 A

-j3 Ω (b) Let

Z o = Z s − j2 = 2.4 − j0.2 = 0.2 (12 − j) Vs 18 (2 − j) Io = = = 15.517 − j6.207 Z o 0.2 (12 − j)

With these, we transform the voltage source in Fig. (b) to a current source. We obtain the circuit in Fig. (c). Ix Io

4Ω

Zo

-j3 Ω (c)

j5 A

Using current division, Zo 2.4 − j0.2 Ix = (I o + j5) = (15.517 − j1.207) Z o + 4 − j3 6.4 − j3.2 I x = 5 + j1.5625 = 5.238∠17.35° A Chapter 10, Solution 53. We transform the voltage source to a current source to obtain the circuit in Fig. (a). -j3 Ω

j4 Ω +

4Ω

5∠0° A

j2 Ω

2Ω

Vo

-j2 Ω

− (a) Let

j8 = 0.8 + j1.6 4 + j2 Vs = (5∠0°) Z s = (5)(0.8 + j1.6) = 4 + j8 Z s = 4 || j2 =

With these, the current source is transformed so that the circuit becomes that shown in Fig. (b). -j3 Ω

Zs

Vs

j4 Ω +

+ −

2Ω

-j2 Ω

Vo −

(b) Let

Z x = Z s − j3 = 0.8 − j1.4 V 4 + j8 = −3.0769 + j4.6154 Ix = s = Z s 0.8 − j1.4

With these, we transform the voltage source in Fig. (b) to obtain the circuit in Fig. (c). j4 Ω + Ix

Zx

2Ω

-j2 Ω

Vo −

(c)

1.6 − j2.8 = 0.8571 − j0.5714 2.8 − j1.4 Vy = I x Z y = (−3.0769 + j4.6154) ⋅ (0.8571 − j0.5714) = j5.7143

Z y = 2 || Z x =

Let

With these, we transform the current source to obtain the circuit in Fig. (d). j4 Ω

Zy

Vy

+

+ −

-j2 Ω

Vo −

(d) Using current division, Vo =

- j2 ( j5.7143) - j2 Vy = = (3.529 – j5.883) V Z y + j4 − j2 0.8571 − j0.5714 + j4 − j2

Chapter 10, Solution 54. 50 x(− j 30) = 13.24 − j 22.059 50 − j 30 We convert the current source to voltage source and obtain the circuit below. 50 //(− j 30) =

40 Ω

+ 115.91 –j31.06V

13.24 – j22.059 Ω

j20 Ω

+

-

I

134.95-j74.912 V

V -

+ -

Applying KVL gives -115.91 + j31.058 + (53.24-j2.059)I -134.95 + j74.912 = 0 or I =

− 250.86 + j105.97 = −4.7817 + j1.8055 53.24 − j 2.059

But − V + (40 + j20)I + V = 0

 →

V = Vs − (40 + j20)I

V = 115.91 − j31.05 − (40 + j20)(−4.7817 + j1.8055) = 124.06∠ − 154 o V which agrees with the result in Prob. 10.7. Chapter 10, Solution 55. (a)

To find Z th , consider the circuit in Fig. (a). j20 Ω

10 Ω Zth

-j10 Ω

(a) ( j20)(- j10) j20 − j10 = 10 − j20 = 22.36∠-63.43° Ω

Z N = Z th = 10 + j20 || (- j10) = 10 +

To find Vth , consider the circuit in Fig. (b). j20 Ω

10 Ω +

50∠30° V

+ −

-j10 Ω

Vth

− (b) Vth =

IN =

- j10 (50∠30°) = -50∠30° V j20 − j10 Vth - 50 ∠30° = = 2.236∠273.4° A Z th 22.36 ∠ - 63.43°

(b)

To find Z th , consider the circuit in Fig. (c). -j5 Ω

8Ω

Zth

j10 Ω

(c) Z N = Z th = j10 || (8 − j5) =

( j10)(8 − j5) = 10∠26° Ω j10 + 8 − j5

To obtain Vth , consider the circuit in Fig. (d). -j5 Ω Io 4∠0° A

8Ω

j10 Ω

+ Vth

− (d)

By current division, 8 32 Io = (4∠0°) = 8 + j10 − j5 8 + j5 Vth = j10 I o =

IN =

j320 = 33.92∠58° V 8 + j5

Vth 33.92 ∠58° = = 3.392∠32° A 10 ∠26° Z th

Chapter 10, Solution 56.

(a)

To find Z th , consider the circuit in Fig. (a). j4 Ω 6Ω

-j2 Ω

Zth

(a)

( j4)(- j2) = 6 − j4 j4 − j2 = 7.211∠-33.69° Ω

Z N = Z th = 6 + j4 || (- j2) = 6 +

By placing short circuit at terminals a-b, we obtain, I N = 2∠0° A Vth = Z th I th = (7.211∠ - 33.69°) (2∠0°) = 14.422∠-33.69° V

(b)

To find Z th , consider the circuit in Fig. (b). j10 Ω

30 Ω

60 Ω

-j5 Ω

(b)

30 || 60 = 20 (- j5)(20 + j10) 20 + j5 = 5.423∠-77.47° Ω

Z N = Z th = - j5 || (20 + j10) =

Zth

To find Vth and I N , we transform the voltage source and combine the 30 Ω and 60 Ω resistors. The result is shown in Fig. (c). j10 Ω

4∠45° A

20 Ω

a IN

-j5 Ω

b

(c)

20 2 (4∠45°) = (2 − j)(4∠45°) 20 + j10 5 = 3.578∠18.43° A

IN =

Vth = Z th I N = (5.423∠ - 77.47°) (3.578∠18.43°) = 19.4∠-59° V Chapter 10, Solution 57.

To find Z th , consider the circuit in Fig. (a). 5Ω

-j10 Ω

2Ω Zth

j20 Ω

(a)

( j20)(5 − j10) 5 + j10 = 18 − j12 = 21.633∠-33.7° Ω

Z N = Z th = 2 + j20 || (5 − j10) = 2 +

To find Vth , consider the circuit in Fig. (b). 5Ω

-j10 Ω

2Ω +

60∠120° V

+ −

j20 Ω

Vth

− (b)

j20 j4 (60 ∠120°) = (60∠120°) 5 − j10 + j20 1 + j2 = 107.3∠146.56° V

Vth =

IN =

Vth 107.3∠146.56° = = 4.961∠-179.7° A Z th 21.633∠ - 33.7°

Chapter 10, Solution 58.

Consider the circuit in Fig. (a) to find Z th .

8Ω

Zth

j10 Ω

-j6 Ω (a)

( j10)(8 − j6) = 5 (2 + j) 8 + j4 = 11.18∠26.56° Ω

Z th = j10 || (8 − j6) =

Consider the circuit in Fig. (b) to find Vth . Io

+

8Ω 5∠45° A

j10 Ω -j6 Ω (b)

Io =

8 − j6 4 − j3 (5∠45°) = (5∠45°) 8 − j6 + j10 4 + j2

Vth = j10 I o =

( j10)(4 − j3)(5∠45°) = 55.9∠71.56° V (2)(2 + j)

Vth

Chapter 10, Solution 59.

The frequency-domain equivalent circuit is shown in Fig. (a). Our goal is to find Vth and Z th across the terminals of the capacitor as shown in Figs. (b) and (c). 3Ω

10∠-45° V

jΩ

jΩ

a

+

+ −

3Ω

+ −

-j Ω

Vo −

Zth 5∠-60° A

b (b)

(a)

3Ω

jΩ

Zth a

+ 10∠-45° V

+ −

Vth

+ −

5∠-60° A

Vth

+

+ −

Vo

− (c)

From Fig. (b), Z th = 3 || j =

j3 3 = (1 + j3) 3 + j 10

From Fig.(c), 10∠ - 45° − Vth 5∠ - 60° − Vth =0 + 3 j 10 ∠ - 45° − 15∠30° Vth = 1 − j3 From Fig. (d),

− (d)

-j Ω b

-j V = 10∠ - 45° − 15∠30° Z th − j th Vo = 15.73∠247.9° V

Vo = Therefore,

v o = 15.73 cos(t + 247.9°) V

Chapter 10, Solution 60. (a)

To find Z th , consider the circuit in Fig. (a). 10 Ω

-j4 Ω a

j5 Ω

Zth

4Ω

b

(a) Z th = 4 || (- j4 + 10 || j5) = 4 || (- j4 + 2 + j4) Z th = 4 || 2 = 1.333 Ω To find Vth , consider the circuit in Fig. (b). 10 Ω

V1

-j4 Ω

V2 +

20∠0° V

+ −

j5 Ω

4∠0° A

4Ω

Vth −

(b)

At node 1, 20 − V1 V1 V1 − V2 = + 10 j5 - j4 (1 + j0.5) V1 − j2.5 V2 = 20 (1) At node 2,

V1 − V2 V2 = - j4 4 V1 = (1 − j) V2 + j16 (2)

4+

Substituting (2) into (1) leads to 28 − j16 = (1.5 − j3) V2 28 − j16 = 8 + j5.333 V2 = 1.5 − j3 Therefore, Vth = V2 = 9.615∠33.69° V (b)

To find Z th , consider the circuit in Fig. (c).

Zth c

d

10 Ω

-j4 Ω

j5 Ω

4Ω

(c)  j10   Z th = - j4 || (4 + 10 || j5) = - j4 ||  4 + 2 + j  - j4 Z th = - j4 || (6 + j4) = (6 + j4) = 2.667 – j4 Ω 6

To find Vth ,we will make use of the result in part (a). V2 = 8 + j5.333 = (8 3 ) (3 + j2) V1 = (1 − j) V2 + j16 = j16 + (8 3) (5 − j) Vth = V1 − V2 = 16 3 + j8 = 9.614∠56.31° V

Chapter 10, Solution 61. First, we need to find Vth and Z th across the 1 Ω resistor.

4Ω

-j3 Ω

j8 Ω

6Ω

Zth

(a) From Fig. (a), Z th = (4 − j3) || (6 + j8) =

(4 − j3)(6 + j8) = 4.4 − j0.8 10 + j5

Z th = 4.472∠-10.3° Ω

4Ω

-j16 V

+ −

-j3 Ω

j8 Ω

+ 2A

Vth − (b)

From Fig. (b), - j16 − Vth Vth +2= 4 − j3 6 + j8 3.92 − j2.56 Vth = = 20.93∠ - 43.45° 0.22 + j0.4 Vth 20.93∠ - 43.45° = 1 + Z th 5.46 ∠ - 8.43° Vo = 3.835∠ - 35.02°

Vo =

Therefore,

v o = 3.835 cos(4t – 35.02°) V

6Ω

Chapter 10, Solution 62. First, we transform the circuit to the frequency domain. 12 cos( t )  → 12∠0°, ω = 1

2H  → 1 F  → 4 1 F  → 8

jωL = j2 1 = - j4 jωC 1 = - j8 jωC

To find Z th , consider the circuit in Fig. (a). 3 Io

Io

4Ω

Vx

j2 Ω

1

Ix 2

-j4 Ω

-j8 Ω

+ −

1V

(a)

At node 1, Vx Vx 1 − Vx , + + 3Io = 4 - j4 j2 Thus,

where I o =

Vx 2 Vx 1 − Vx − = - j4 4 j2 Vx = 0.4 + j0.8

At node 2, I x + 3Io =

1 1 − Vx + - j8 j2

I x = (0.75 + j0.5) Vx − j

3 8

I x = -0.1 + j0.425

Z th =

1 = -0.5246 − j2.229 = 2.29∠ - 103.24° Ω Ix

- Vx 4

To find Vth , consider the circuit in Fig. (b).

3 Io

Io

4Ω

j2 Ω

V1

V2

1 12∠0° V

+ −

2

-j4 Ω

-j8 Ω

+ Vth −

(b) At node 1, 12 − V1 V V − V2 , = 3Io + 1 + 1 4 - j4 j2 24 = (2 + j) V1 − j2 V2

where I o = (1)

At node 2, V1 − V2 V + 3Io = 2 j2 - j8 72 = (6 + j4) V1 − j3 V2 (2) From (1) and (2),  24  2 + j - j2  V1   72  =  6 + j4 - j3  V   2   

∆ = -5 + j6 , Vth = V2 =

Thus,

∆ 2 = - j24 ∆2 = 3.073∠ - 219.8° ∆

2 (2)(3.073∠ - 219.8°) Vth = 2 + Z th 1.4754 − j2.229 6.146∠ - 219.8° Vo = = 2.3∠ - 163.3° 2.673∠ - 56.5°

Vo =

Therefore,

v o = 2.3 cos(t – 163.3°) V

12 − V1 4

Chapter 10, Solution 63. Transform the circuit to the frequency domain. 4 cos(200t + 30°)  → 4∠30°, ω = 200 10 H  → 5 µF  →

jωL = j (200)(10) = j2 kΩ 1 1 = = - j kΩ jωC j (200)(5 × 10 -6 )

Z N is found using the circuit in Fig. (a). -j kΩ

j2 kΩ

ZN

2 kΩ

(a) Z N = - j + 2 || j2 = - j + 1 + j = 1 kΩ We find I N using the circuit in Fig. (b).

-j kΩ

4∠30° A

j2 kΩ

(b) j2 || 2 = 1 + j By the current division principle, 1+ j IN = (4 ∠30°) = 5.657 ∠75° 1+ j − j Therefore, i N = 5.657 cos(200t + 75°) A

Z N = 1 kΩ

2 kΩ

IN

Chapter 10, Solution 64. Z N is obtained from the circuit in Fig. (a).

60 Ω

40 Ω

ZN

-j30 Ω

j80 Ω (a) Z N = (60 + 40) || ( j80 − j30) = 100 || j50 =

(100)( j50) 100 + j50

Z N = 20 + j40 = 44.72∠63.43° Ω To find I N , consider the circuit in Fig. (b).

60 Ω 3∠60° A

j80 Ω

(b)

For mesh 1, 100 I 1 − 60 I s = 0 I 1 = 1.8∠60° For mesh 2, ( j80 − j30) I 2 − j80 I s = 0 I 2 = 4.8∠60°

I N = I 1 − I 2 = 3∠60° A

40 Ω

I2

-j30 Ω

IN

Is

I s = 3∠60°

I1

Chapter 10, Solution 65. 5 cos(2 t )  → 5∠0°, ω = 2 4H  → 1 F  → 4 1 F  → 2

jωL = j (2)(4) = j8 1 1 = = - j2 jωC j (2)(1 / 4) 1 1 = = -j jωC j (2)(1 / 2)

To find Z N , consider the circuit in Fig. (a).

2Ω ZN -j2 Ω

-j Ω

(a) Z N = - j || (2 − j2) =

- j (2 − j2) 1 = (2 − j10) 2 − j3 13

To find I N , consider the circuit in Fig. (b).

5∠0° V

2Ω

+ −

-j2 Ω

-j Ω

IN (b)

IN =

5∠0° = j5 -j

The Norton equivalent of the circuit is shown in Fig. (c).

Io IN

ZN (c)

j8 Ω

Using current division, ZN (1 13)(2 − j10)( j5) 50 + j10 Io = IN = = Z N + j8 (1 13)(2 − j10) + j8 2 + j94 I o = 0.1176 − j0.5294 = 0542∠ - 77.47° Therefore, i o = 0.542 cos(2t – 77.47°) A

Chapter 10, Solution 66. ω = 10 0.5 H  →

jωL = j (10)(0.5) = j5 1 1 10 mF  → = = - j10 jωC j (10)(10 × 10 -3 )

-j10 Ω

Vx

+ 10 Ω

j5 Ω

Vo

2 Vo

1A

−

(a) To find Z th , consider the circuit in Fig. (a).

Vx Vx , + j5 10 − j10 V 19 Vx - 10 + j10 1+ = x  → Vx = 10 − j10 j5 21 + j2 1 + 2 Vo =

Z N = Z th =

where Vo =

10Vx 10 − j10

Vx 14.142 ∠135° = 0.67∠129.56° Ω = 1 21.095∠5.44°

To find Vth and I N , consider the circuit in Fig. (b).

12∠0° V

-j10 Ω

− +

+

+ -j2 A

10 Ω

Vo

j5 Ω

I

2 Vo

Vth −

−

(b)

where Thus,

(10 − j10 + j5) I − (10)(- j2) + j5 (2 Vo ) − 12 = 0 Vo = (10)(- j2 − I ) (10 − j105) I = -188 − j20 188 + j20 I= - 10 + j105 Vth = j5 (I + 2 Vo ) = j5 (21I + j40) = j105 I − 200 j105 (188 + j20) Vth = − 200 = -11.802 + j2.076 - 10 + j105 Vth = 11.97∠170° V IN =

Vth 11.97 ∠170° = 17.86∠40.44° A = Z th 0.67 ∠129.56°

Chapter 10, Solution 67. 10(13 − j5) 12(8 + j6) + = 11.243 + j1.079Ω 23 − j5 20 + j6 10 (8 + j6) Va = (60∠45 o ) = 13.78 + j21.44, Vb = (60∠45 o ) = 25.93 + j454.37Ω 23 − j5 20 + j6 V VTh = Va − Vb = 433.1∠ − 1.599 o V, I N = Th = 38.34∠ − 97.09 o A Z Th Z N = Z Th = 10 //(13 − j5) + 12 //(8 + j6) =

Chapter 10, Solution 68. 1H  → jωL = j10x1 = j10 1 1 1 F  → = = − j2 1 20 jω C j10 x 20 We obtain VTh using the circuit below.

Io

4Ω a

+

+

6<0o -

+ -

Vo/3

-j2 j10

Vo

4Io

j10(− j2) = − j2.5 j10 − j2 Vo = 4I o x (− j2.5) = − j10I o 1 − 6 + 4I o + Vo = 0 3

b

j10 //(− j2) =

(1) (2)

Combining (1) and (2) gives Io =

6 , 4 − j10 / 3

VTh = Vo = − j10I o =

− j60 = 11.52∠ − 50.19 o 4 − j10 / 3

v Th = 11.52 sin(10 t − 50.19 o ) To find RTh, we insert a 1-A source at terminals a-b, as shown below. Io

4Ω a + Vo/3

1 4I o + Vo = 0 3

1 + 4I o =

→

Vo Vo + − j2 j10

+ -

V Io = − o 12

-j2 j10 4Io

Vo -

1<0o

Combining the two equations leads to Vo =

1 = 1.2293 − j1.4766 0.333 + j0.4 V Z Th = o = 1.2293 − 1.477Ω 1

Chapter 10, Solution 69. This is an inverting op amp so that Vo - Z f -R = = = -jωRC Vs Zi 1 jωC When Vs = Vm and ω = 1 RC , 1 Vo = - j ⋅ ⋅ RC ⋅ Vm = - j Vm = Vm ∠ - 90° RC Therefore, v o ( t ) = Vm sin(ωt − 90°) = - Vm cos(ωt)

Chapter 10, Solution 70. This may also be regarded as an inverting amplifier. 2 cos(4 × 10 4 t )  → 2 ∠0°, ω = 4 × 10 4 1 1 10 nF  → = = - j2.5 kΩ 4 jωC j (4 × 10 )(10 × 10 -9 ) Vo - Z f = Vs Zi

where Z i = 50 kΩ and Z f = 100k || (- j2.5k ) =

- j100 kΩ . 40 − j

Vo - j2 = Vs 40 − j

Thus, If Vs = 2 ∠0° ,

Vo = Therefore,

- j4 4 ∠ - 90° = = 0.1∠ - 88.57° 40 − j 40.01∠ - 1.43°

v o ( t ) = 0.1 cos(4x104 t – 88.57°) V

Chapter 10, Solution 71.

8 cos(2t + 30 o )

→ 8∠30 o 1 1 0. 5µF  → = = − j1kΩ jωC j2x 0.5x10 − 6 At the inverting terminal, Vo − 8∠30 o Vo − 8∠30 o 8∠30 o + = − j1k 10k 2k

Vo =

 →

Vo (0.1 + j) = 8∠30(0.6 + j)

(6.9282 + j4)(0.6 + j) = 9.283∠4.747 o 0.1 + j vo(t) = 9.283cos(2t + 4.75o) V

Chapter 10, Solution 72.

4 cos(10 4 t )  → 4 ∠0°, ω = 10 4 1 1 1 nF  → = = - j100 kΩ 4 jωC j (10 )(10 -9 )

Consider the circuit as shown below. 50 kΩ

4∠0° V

+ −

Therefore,

+ −

-j100 kΩ

At the noninverting node, 4 − Vo Vo = 50 - j100 Io =

Vo

 → Vo =

Vo Io 100 kΩ

4 1 + j0.5

Vo 4 = mA = 35.78∠ - 26.56° µA 100k (100)(1 + j0.5)

i o ( t ) = 35.78 cos(104 t – 26.56°) µA

Chapter 10, Solution 73. As a voltage follower, V2 = Vo 1 1 = = -j20 kΩ 3 jωC1 j (5 × 10 )(10 × 10 -9 ) 1 1 C 2 = 20 nF  → = = -j10 kΩ 3 jωC 2 j (5 × 10 )(20 × 10 -9 )

C1 = 10 nF  →

Consider the circuit in the frequency domain as shown below.

-j20 kΩ Is 10 kΩ

20 kΩ V1

VS

+ −

V2

+ −

Io Vo

-j10 kΩ

Zin At node 1, Vs − V1 V1 − Vo V1 − Vo = + 10 - j20 20 2 Vs = (3 + j)V1 − (1 + j)Vo

(1)

At node 2,

V1 − Vo Vo − 0 = 20 - j10 V1 = (1 + j2)Vo Substituting (2) into (1) gives 2 Vs = j6Vo

or

(2)

1 Vo = -j Vs 3

2 1 V1 = (1 + j2)Vo =  − j  Vs 3 3 Is =

Vs − V1 (1 3)(1 − j) = Vs 10k 10k

Is 1− j = Vs 30k Vs 30k = = 15 (1 + j) k Is 1− j Z in = 21.21∠45° kΩ Z in =

Chapter 10, Solution 74. Zi = R1 +

1 , jωC1

Zf = R 2 +

1 jωC 2

1 Vo - Z f jωC 2  C 1   1 + jωR 2 C 2   = = Av = =   1 Vs Zi  C 2   1 + jωR 1C 1  R1 + jωC1 R2 +

Av =

At ω = 0 ,

As ω → ∞ ,

Av =

C1 C2

R2 R1

At ω =

1 , R 1 C1

 C   1 + j R 2 C 2 R 1C1   Av =  1   1+ j   C2  

At ω =

1 , R 2C2

C   1+ j  Av =  1    C 2   1 + j R 1C1 R 2 C 2 

Chapter 10, Solution 75.

ω = 2 × 10 3 → C1 = C 2 = 1 nF 

1 1 = = -j500 kΩ 3 jωC1 j (2 × 10 )(1 × 10 -9 )

Consider the circuit shown below.

100 kΩ -j500 kΩ

-j500 kΩ

V2

V1 VS

+ −

20 kΩ

+ −

100 kΩ

+ Vo

20 kΩ

−

At node 1, Vs − V1 Vo − V1 V1 − V2 = + - j500 100 - j500 Vs = (2 + j5) V1 − j5 Vo − V2

(1)

V1 − V2 V2 = - j500 100 V1 = (1 − j5) V2

(2)

At node 2,

But V2 =

Vo R3 Vo = R3 + R4 2

From (2) and (3), 1 V1 = ⋅ (1 − j5) Vo 2 Substituting (3) and (4) into (1), 1 1 Vs = ⋅ (2 + j5)(1 − j5) Vo − j5 Vo − Vo 2 2 1 Vs = ⋅ (26 − j25) Vo 2 Vo 2 = = 0.0554∠43.88° Vs 26 − j25

Chapter 10, Solution 76.

(3)

(4)

Let the voltage between the -jk Ω capacitor and the 10k Ω resistor be V1. 2∠30 o − V1 V1 − Vo V1 − Vo = + − j4k 10k 20k

 →

(1)

2∠30 o = (1 − j0.6)V1 + j0.6Vo Also,

V1 − Vo Vo = − j2k 10k

→

V1 = (1 + j5)Vo

(2)

Solving (2) into (1) yields Vo = 0.047 − j0.3088 = 0.3123∠ − 81.34 o V Chapter 10, Solution 77. Consider the circuit below. R3 2 R1 1 VS

At node 1,

+ −

V1 V1

C2 − +

C1

Vs − V1 = jωC V1 R1 Vs = (1 + jωR 1C1 ) V1

At node 2, 0 − V1 V1 − Vo = + jωC 2 (V1 − Vo ) R3 R2

 R3  V1 = (Vo − V1 )  + jωC 2 R 3  R2 

R2

+ Vo −

(1)

  1  V1 Vo = 1 +  (R 3 R 2 ) + jωC 2 R 3  From (1) and (2),   Vs R2 1 +  Vo = 1 + jωR 1C1  R 3 + jωC 2 R 2 R 3 

(2)

Vo R 2 + R 3 + jωC 2 R 2 R 3 = Vs (1 + jωR 1C 1 ) ( R 3 + jωC 2 R 2 R 3 )

Chapter 10, Solution 78. 2 sin(400t )  → 2∠0°, ω = 400 1 1 0.5 µF  → = = - j5 kΩ jωC j (400)(0.5 × 10 -6 ) 1 1 0.25 µF  → = = - j10 kΩ jωC j (400)(0.25 × 10 -6 ) Consider the circuit as shown below. 20 kΩ 10 kΩ V 1

2∠0° V

+ −

-j5 kΩ

V2

+ −

Vo 40 kΩ

-j10 kΩ

10 kΩ 20 kΩ

At node 1,

2 − V1 V V − V2 V1 − Vo = 1 + 1 + 10 - j10 - j5 20 4 = (3 + j6) V1 − j4 V2 − Vo

(1)

V1 − V2 V2 = − j5 10 V1 = (1 − j0.5) V2

(2)

At node 2,

But V2 =

20 1 Vo = Vo 20 + 40 3

(3)

From (2) and (3), 1 V1 = ⋅ (1 − j0.5) Vo 3 Substituting (3) and (4) into (1) gives

(4)

 1 4 1 4 = (3 + j6) ⋅ ⋅ (1 − j0.5) Vo − j Vo − Vo = 1 − j  Vo  3 3 6

Vo =

24 = 3.945∠9.46° 6− j

Therefore, v o ( t ) = 3.945 sin(400t + 9.46°) V Chapter 10, Solution 79. 5 cos(1000t )  → 5∠0°, ω = 1000 0.1 µF  →

1 1 = = - j10 kΩ jωC j (1000)(0.1 × 10 -6 )

0.2 µF  →

1 1 = = - j5 kΩ jωC j (1000)(0.2 × 10 -6 )

Consider the circuit shown below. 20 kΩ -j10 kΩ 10 kΩ

Vs = 5∠0° V

+ −

− +

40 kΩ V1 -j5 kΩ

Since each stage is an inverter, we apply Vo =

− +

- Zf V to each stage. Zi i

+ Vo −

Vo =

- 40 V - j15 1 (1)

and V1 =

- 20 || (- j10) Vs 10 (2)

From (1) and (2),  - j8  - (20)(-j10)   5∠0°  Vo =   10  20 − j10  Vo = 16 (2 + j) = 35.78∠26.56° Therefore,

v o ( t ) = 35.78 cos(1000t + 26.56°) V

Chapter 10, Solution 80. 4 cos(1000t − 60°)  → 4∠ - 60°, ω = 1000 0.1 µF  →

1 1 = = - j10 kΩ jωC j (1000)(0.1 × 10 -6 )

0.2 µF  →

1 1 = = - j5 kΩ jωC j (1000)(0.2 × 10 -6 )

The two stages are inverters so that  20 20  - j5   ⋅ (4∠ - 60°) + Vo =  V  50 o  10   - j10 =

-j -j 2 ⋅ ( j2) ⋅ (4∠ - 60°) + ⋅ Vo 2 2 5

(1 + j 5) Vo = 4∠ - 60° Vo = Therefore,

4∠ - 60° = 3.922 ∠ - 71.31° 1+ j 5

v o ( t ) = 3.922 cos(1000t – 71.31°) V

Chapter 10, Solution 81. The schematic is shown below. The pseudocomponent IPRINT is inserted to print the value of Io in the output. We click Analysis/Setup/AC Sweep and set Total Pts. = 1, Start Freq = 0.1592, and End Freq = 0.1592. Since we assume that w = 1. The output file includes: FREQ 1.592 E-01

IM(V_PRINT1) 1.465 E+00

IP(V_PRINT1) 7.959 E+01

Io = 1.465∠79.59o A

Thus,

Chapter 10, Solution 82. The schematic is shown below. We insert PRINT to print Vo in the output file. For AC Sweep, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we print out the output file which includes: FREQ 1.592 E-01

VM($N_0001) 7.684 E+00

VP($N_0001) 5.019 E+01

which means that

Vo = 7.684∠50.19o V

Chapter 10, Solution 83.

The schematic is shown below. The frequency is f = ω / 2π =

1000 = 159.15 2π

When the circuit is saved and simulated, we obtain from the output file FREQ 1.592E+02

VM(1) 6.611E+00

Thus,

VP(1) -1.592E+02 vo = 6.611cos(1000t – 159.2o) V

Chapter 10, Solution 84.

The schematic is shown below. We set PRINT to print Vo in the output file. In AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain the output file which includes: FREQ

VM($N_0003)

1.592 E-01

1.664 E+00

VP($N_0003) E+02 Vo = 1.664∠-146.4o V

Namely, Chapter 10, Solution 85.

The schematic is shown below. We let ω = 1 rad/s so that L=1H and C=1F.

When the circuit is saved and simulated, we obtain from the output file FREQ 1.591E-01

VM(1) 2.228E+00

VP(1) -1.675E+02

From this, we conclude that Vo = 2.228∠ − 167.5 V

-1.646

Chapter 10, Solution 86.

We insert three pseudocomponent PRINTs at nodes 1, 2, and 3 to print V1, V2, and V3, into the output file. Assume that w = 1, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After saving and simulating the circuit, we obtain the output file which includes: FREQ

VM($N_0002)

1.592 E-01

6.000 E+01

FREQ

VM($N_0003)

1.592 E-01

2.367 E+02

VP($N_0002) 3.000

E+01 VP($N_0003) E+01

-8.483

FREQ

VM($N_0001)

1.592 E-01

1.082 E+02

VP($N_0001) 1.254

E+02

Therefore, V1 = 60∠30o V V2 = 236.7∠-84.83o V V3 = 108.2∠125.4o V

Chapter 10, Solution 87.

The schematic is shown below. We insert three PRINTs at nodes 1, 2, and 3. We set Total Pts = 1, Start Freq = 0.1592, End Freq = 0.1592 in the AC Sweep box. After simulation, the output file includes: FREQ

VM($N_0004)

1.592 E-01

1.591 E+01

FREQ

VM($N_0001)

1.592 E-01

5.172 E+00

VP($N_0004) 1.696

E+02 VP($N_0001) E+02

-1.386

FREQ

VM($N_0003)

1.592 E-01

2.270 E+00

VP($N_0003) -1.524

E+02 Therefore, V1 = 15.91∠169.6o V V2 = 5.172∠-138.6o V V3 = 2.27∠-152.4o V

Chapter 10, Solution 88.

The schematic is shown below. We insert IPRINT and PRINT to print Io and Vo in the output file. Since w = 4, f = w/2π = 0.6366, we set Total Pts = 1, Start Freq = 0.6366, and End Freq = 0.6366 in the AC Sweep box. After simulation, the output file includes: FREQ

VM($N_0002)

6.366 E-01

3.496 E+01

1.261

FREQ

IM(V_PRINT2)

IP

6.366 E-01

8.912 E-01

VP($N_0002) E+01

(V_PRINT2) -8.870 E+01

Vo = 34.96∠12.6o V, Io = 0.8912∠-88.7o A

Therefore,

vo = 34.96 cos(4t + 12.6o)V,

io = 0.8912cos(4t - 88.7o )A

Chapter 10, Solution 89.

Consider the circuit below. R1

Vin

R2

2

1

R3

Vin

C

4

R4

3 − +

Iin − +

+ −

Vin

At node 1, 0 − Vin Vin − V2 = R1 R2

- Vin + V2 =

R2 V R 1 in

At node 3, V2 − Vin Vin − V4 = R3 1 jωC

(1)

- Vin + V4 =

Vin − V2 jωCR 3

(2)

From (1) and (2), - R2 V jωCR 3 R 1 in

- Vin + V4 =

Thus, I in =

R2 Vin − V4 = V R4 jωCR 3 R 1 R 4 in

Z in =

Vin jωCR 1R 3 R 4 = = jωL eq I in R2 L eq =

where

R 1R 3 R 4C R2

Chapter 10, Solution 90.

Let

Z 4 = R ||

1 R = jωC 1 + jωRC

Z3 = R +

1 1 + jωRC = jωC jωC

Consider the circuit shown below. Z3

Vi

+ −

R1

+ Z4

Vo

− R2

Vo =

Z4 R2 Vi − V Z3 + Z 4 R1 + R 2 i

R Vo R2 1 + jωC = − R 1 + jωRC R 1 + R 2 Vi + 1 + jωC jωC =

jωRC R2 − 2 jωRC + (1 + jωRC) R1 + R 2

Vo R2 jωRC = − 2 2 2 Vi 1 − ω R C + j3ωRC R 1 + R 2

For Vo and Vi to be in phase,

Vo must be purely real. This happens when Vi

1 − ω2 R 2 C 2 = 0 ω=

1 = 2πf RC f=

or

1 2πRC

At this frequency, Av =

Vo 1 R2 = − Vi 3 R 1 + R 2

Chapter 10, Solution 91.

(a)

Let

V2 = voltage at the noninverting terminal of the op amp Vo = output voltage of the op amp

Z p = 10 kΩ = R o Z s = R + jωL + As in Section 10.9,

1 jωC

Zp V2 = = Vo Z s + Z p

Ro R + R o + jωL −

j ωC

ωCR o V2 = Vo ωC (R + R o ) + j (ω2 LC − 1)

For this to be purely real, 1

ωo2 LC − 1 = 0  → ωo = fo =

1 2π LC

=

LC 1

2π (0.4 × 10 -3 )(2 × 10 -9 )

f o = 180 kHz (b)

At oscillation, Ro ωo CR o V2 = = Vo ωo C (R + R o ) R + R o This must be compensated for by Vo 80 = 1+ =5 Av = V2 20 Ro 1 = R + Ro 5

 → R = 4R o = 40 kΩ

Chapter 10, Solution 92. Let

V2 = voltage at the noninverting terminal of the op amp Vo = output voltage of the op amp

Zs = R o Z p = jωL ||

As in Section 10.9,

ωRL 1 1 = || R = 1 1 jωC ωL + jR (ω2 LC − 1) + jωC + R jωL

ωRL Zp V2 ωL + jR (ω2 LC − 1) = = ωRL Vo Z s + Z p Ro + ωL + jR (ω2 LC − 1)

V2 ωRL = Vo ωRL + ωR o L + jR o R (ω2 LC − 1)

For this to be purely real, ωo2 LC = 1  → f o =

(a)

1 2π LC

At ω = ωo , ωo RL V2 R = = Vo ωo RL + ωo R o L R + R o This must be compensated for by Vo Rf 1000k Av = = 1+ = 1+ = 11 V2 Ro 100k Hence, R 1 =  → R o = 10R = 100 kΩ R + R o 11

(b)

1

fo =

2π (10 × 10 -6 )(2 × 10 -9 ) f o = 1.125 MHz

Chapter 10, Solution 93. As shown below, the impedance of the feedback is jωL

1 jωC2

ZT =

1 jωC1

1 jωC1

 1   ||  jωL + jωC 2  

ZT

-j  -j  1  jωL +  − ωLC 2 ωC1  ωC 2  ω ZT = = -j -j j (C1 + C 2 − ω2 LC1C 2 ) + jωL + ωC1 ωC 2 In order for Z T to be real, the imaginary term must be zero; i.e. C1 + C 2 − ωo2 LC1C 2 = 0 C1 + C 2 1 ωo2 = = LC1C 2 LC T 1 fo = 2π LC T

Chapter 10, Solution 94. If we select C1 = C 2 = 20 nF CT =

Since f o =

1 2π LC T

L=

C1 C 2 C1 = = 10 nF C1 + C 2 2

,

1 1 = = 10.13 mH 2 2 (2πf ) C T (4π )(2500 × 10 6 )(10 × 10 -9 )

Xc =

1 1 = = 159 Ω ωC 2 (2π )(50 × 10 3 )(20 × 10 -9 )

We may select R i = 20 kΩ and R f ≥ R i , say R f = 20 kΩ . Thus, C1 = C 2 = 20 nF, R f = R i = 20 kΩ L = 10.13 mH Chapter 10, Solution 95.

First, we find the feedback impedance. C ZT L2

L1

 1   Z T = jωL1 ||  jωL 2 + jωC    j   jωL1  jωL 2 −  ω2 L1C (1 − ωL 2 ) ωC  ZT = = j j (ω2 C (L1 + L 2 ) − 1) jωL1 + jωL 2 − ωC

In order for Z T to be real, the imaginary term must be zero; i.e. ωo2 C (L1 + L 2 ) − 1 = 0

ωo = 2πf o =

fo =

1 C ( L1 + L 2 ) 1

2π C (L 1 + L 2 )

Chapter 10, Solution 96. (a)

Consider the feedback portion of the circuit, as shown below.

jωL

Vo

V2 =

+ −

jωL V R + jωL 1

V1

R

R

 → V1 =

Applying KCL at node 1, Vo − V1 V1 V1 = + jωL R R + jωL

 1 1  Vo − V1 = jωL V1  +  R R + jωL 

V2

jωL

R + jωL V2 jωL

(1)

 j2ωRL − ω2 L2   Vo = V1 1 + R (R + jωL)   (2)

From (1) and (2),  R + jωL  j2ωRL − ω2 L2  V 1 + Vo =  R (R + jωL)  2  jωL  Vo R 2 + jωRL + j2ωRL − ω2 L2 = jωRL V2

V2 = Vo

1 R − ω2 L2 3+ jωRL 2

V2 1 = Vo 3 + j (ωL R − R ωL )

(b)

V2 must be real, Vo

Since the ratio

ωo L R − =0 R ωo L ωo L =

R2 ωo L

ωo = 2πf o = fo =

(c)

R L

R 2π L

When ω = ωo V2 1 = Vo 3 This must be compensated for by A v = 3 . But R2 Av = 1+ =3 R1 R 2 = 2 R1

Chapter 11, Solution 1.

v( t ) = 160 cos(50t ) i( t ) = -20 sin(50t − 30°) = 2 cos(50t − 30° + 180° − 90°) i( t ) = 20 cos(50t + 60°) p( t ) = v( t ) i( t ) = (160)(20) cos(50t ) cos(50t + 60°) p( t ) = 1600 [ cos(100 t + 60°) + cos(60°) ] W p( t ) = 800 + 1600 cos(100t + 60°) W P=

1 1 Vm I m cos(θ v − θi ) = (160)(20) cos(60°) 2 2

P = 800 W

Chapter 11, Solution 2. First, transform the circuit to the frequency domain. 30 cos(500t )  → 30 ∠0° , ω = 500

0.3 H  → → 20µF 

I

jωL = j150 1 -j = = - j100 jωC (500)(20)(10 -6 )

I2

-j100 Ω

I1 30∠0° V

I1 =

+ −

j150 Ω

200 Ω

30∠0° = 0.2∠ − 90° = - j0.2 j150

i1 ( t ) = 0.2 cos(500 t − 90°) = 0.2 sin(500 t )

I2 =

30∠0° 0.3 = = 0.1342∠26.56° = 0.12 + j0.06 200 − j100 2 − j

i 2 ( t ) = 0.1342 cos(500 t + 25.56°) I = I 1 + I 2 = 0.12 − j0.14 = 0.1844 ∠ - 49.4° i( t ) = 0.1844 cos(500t − 35°)

For the voltage source, p( t ) = v( t ) i( t ) = [ 30 cos(500t ) ] × [ 0.1844 cos(500t − 35°) ] At t = 2 s ,

p = 5.532 cos(1000) cos(1000 − 35°) p = (5.532)(0.5624)(0.935) p = 2.91 W

For the inductor, p( t ) = v( t ) i( t ) = [ 30 cos(500t ) ] × [ 0.2 sin(500t ) ] At t = 2 s ,

p = 6 cos(1000) sin(1000) p = (6)(0.5624)(0.8269) p = 2.79 W

For the capacitor, Vc = I 2 (- j100) = 13.42∠ - 63.44° p( t ) = v( t ) i( t ) = [13.42 cos(500 − 63.44°) ] × [ 0.1342 cos(500t + 25.56°) At t = 2 s ,

p = 18 cos(1000 − 63.44°) cos(1000 + 26.56°) p = (18)(0.991)(0.1329) p = 2.37 W

For the resistor, VR = 200 I 2 = 26.84 ∠25.56° p( t ) = v( t ) i( t ) = [ 26.84 cos(500t + 26.56°) ] × [ 0.1342 cos(500t + 26.56°) ] At t = 2 s ,

p = 3.602 cos 2 (1000 + 25.56°) p = (3.602)(0.1329 2 p = 0.0636 W

Chapter 11, Solution 3.

ω= 2

10 cos(2t + 30°)  → 10∠30° , 1H  →

jωL = j2 1 = -j2 jωC

0.25 F  →

I

4Ω

I1

2Ω

I2 10∠30° V

+ −

j2 || (2 − j2) = I=

j2 Ω

( j2)(2 − j2) = 2 + j2 2

10 ∠30° = 1.581∠11.565° 4 + 2 + j2

I1 =

j2 I = j I = 1.581∠101.565° 2

I2 =

2 − j2 I = 2.236 ∠56.565° 2

For the source, S = V I* =

1 (10∠30°)(1.581∠ - 11.565°) 2

S = 7.905∠18.43° = 7.5 + j2.5 The average power supplied by the source = 7.5 W For the 4-Ω resistor, the average power absorbed is 1 2 1 P = I R = (1.581) 2 (4) = 5 W 2 2 For the inductor, 1 1 2 S = I 2 Z L = (2.236) 2 ( j2) = j5 2 2 The average power absorbed by the inductor = 0 W

-j2 Ω

For the 2-Ω resistor, the average power absorbed is 1 1 2 P = I 1 R = (1.581) 2 (2) = 2.5 W 2 2 For the capacitor, S=

1 1 2 I 1 Z c = (1.581) 2 (- j2) = - j2.5 2 2

The average power absorbed by the capacitor = 0 W

Chapter 11, Solution 4. 20 Ω

50 V

+ −

10 Ω

I1

-j10 Ω

I2

j5 Ω

For mesh 1, 50 = (20 − j10) I 1 + j10 I 2 5 = (2 − j) I 1 + j I 2

(1)

0 = (10 + j5 − j10) I 2 + j10 I 1 0 = (2 − j) I 2 + j2 I 1

(2)

For mesh 2,

In matrix form, 5  2 − j j  I 1   0 =  j2 2 − j I   2     ∆ = 5 − j4 ,

∆ 1 = 5 (2 − j) ,

I1 =

∆ 1 5 (2 − j) = = 1.746∠12.1° ∆ 5 − j4

I2 =

∆ 2 - j10 = = 1.562 ∠128.66° ∆ 5 - j4

For the source, S=

1 V I 1* = 43.65∠ - 12.1° 2

∆ 2 = -j10

The average power supplied = 43.65 cos(12.1°) = 42.68 W For the 20-Ω resistor, 1 2 P = I 1 R = 30.48 W 2 For the inductor and capacitor, P=0W For the 10-Ω resistor, 1 2 P = I 2 R = 12.2 W 2 Chapter 11, Solution 5. Converting the circuit into the frequency domain, we get: 1Ω

8∠–40˚

I1Ω =

+ −

2Ω

j6

8∠ − 40° = 1.6828∠ − 25.38° j6(2 − j2) 1+ j6 + 2 − j2

1.6828 2 P1Ω = 1 = 1.4159 W 2

P3H = P0.25F = 0

I 2Ω =

j6 1.6828∠ − 25.38° = 2.258 j6 + 2 − j2

2.258 2 P2Ω = 2 = 5.097 W 2

–j2

Chapter 11, Solution 6. 20 Ω

50 V

+ −

I1

10 Ω

I2

-j10 Ω

j5 Ω

For mesh 1, (4 + j2) I 1 − j2 (4 ∠60°) + 4 Vo = 0 Vo = 2 (4 ∠60° − I 2 )

(1) (2)

For mesh 2, (2 − j) I 2 − 2 (4∠60°) − 4Vo = 0 Substituting (2) into (3), (2 − j) I 2 − 8∠60° − 8 (4 ∠60° − I 2 ) = 0 I2 =

(3)

40∠60° 10 − j

Hence,  40∠60°  - j8∠60° = Vo = 2  4 ∠60° − 10 − j  10 − j  Substituting this into (1), 14 − j  j32 ∠60°  (4 + j2) I 1 = j8∠60° + = ( j8∠60°)  10 − j 10 − j  I1 =

(4∠60°)(1 + j14) = 2.498∠125.06° 21 + j8

P4 =

1 1 2 I 1 R = (2.498) 2 (4) = 12.48 W 2 2

Chapter 11, Solution 7. 20 Ω

50 V

+ −

I1

10 Ω

-j10 Ω

I2

j5 Ω

Applying KVL to the left-hand side of the circuit, 8∠20° = 4 I o + 0.1Vo Applying KCL to the right side of the circuit, V V1 8Io + 1 + =0 j5 10 − j5 10 V 10 − j5 1

But,

Vo =

Hence,

8Io +

 → V1 =

(1)

10 − j5 Vo 10

Vo 10 − j5 Vo + =0 j50 10

I o = j0.025 Vo Substituting (2) into (1), 8∠20° = 0.1 Vo (1 + j)

(2)

Vo =

80∠20° 1+ j

I1 =

Vo 10 = ∠ - 25° 10 2

P=

 1 100  1 2 (10) = 250 W I 1 R =    2  2  2

Chapter 11, Solution 8. We apply nodal analysis to the following circuit. V1 Io -j20 Ω

V2 I2

j10 Ω

6∠0° A

0.5 Io

40 Ω

At node 1, 6= At node 2,

V1 V1 − V2 V1 = j120 − V2 + j10 - j20

(1)

0 .5 I o + I o =

V2 40

V1 − V2 - j20

But,

Io =

Hence,

1.5 (V1 − V2 ) V2 = - j20 40 3V1 = (3 − j) V2

(2)

Substituting (1) into (2), j360 − 3V2 − 3V2 + j V2 = 0 V2 =

j360 360 = (-1 + j6) 6 − j 37

I2 =

V2 9 = (-1 + j6) 40 37

1 2 P = I2 R = 2

2

1 9    (40) = 43.78 W 2  37 

Chapter 11, Solution 9.  6 Vo = 1 +  Vs = (4)(2) = 8 V rms  2 P10 =

Vo2 64 = mW = 6.4 mW R 10

The current through the 2 -kΩ resistor is Vs = 1 mA 2k P2 = I 2 R = 2 mW Similarly, P6 = I 2 R = 6 mW

Chapter 11, Solution 10. No current flows through each of the resistors. Hence, for each resistor, P = 0 W.

Chapter 11, Solution 11. ω = 377 , R = 10 4 , C = 200 × 10 -9 ωRC = (377)(10 4 )(200 × 10 -9 ) = 0.754

tan -1 (ωRC) = 37.02° Z ab =

10k 1 + (0.754) 2

∠ - 37.02° = 6.375∠ - 37.02° kΩ

i( t ) = 2 sin(377 t + 22°) = 2 cos(377 t − 68°) mA I = 2 ∠ - 68° 2

S= I

2 rms

 2 × 10 -3   (6.375∠ - 37.02°) × 10 3 Z ab =   2 

S = 12.751∠ - 37.02° mVA P = S cos(37.02) = 10.181 mW

Chapter 11, Solution 12. (a)

We find Z Th using the circuit in Fig. (a). Zth 8Ω

Z Th

-j2 Ω

(a) (8)(-j2) 8 = 8 || -j2 = = (1 − j4) = 0.471 − j1.882 8 − j2 17

Z L = Z *Th = 0.471 + j1.882 Ω

We find VTh using the circuit in Fig. (b). Io

+

8Ω

-j2 Ω

Vth

4∠0° A

−

(b) Io =

- j2 (4∠0°) 8 − j2

VTh = 8 I o =

- j64 8 − j2 2

Pmax =

(b)

VTh 8RL

2

 64     68  = = 15.99 W (8)(0.471)

We obtain Z Th from the circuit in Fig. (c). 5Ω

-j3 Ω

j2 Ω 4Ω

Zth

(c) Z Th = j2 + 5 || (4 − j3) = j2 + Z L = Z *Th = 2.5 − j1.167 Ω

(5)(4 − j3) = 2.5 + j1.167 9 − j3

Chapter 11, Solution 13. (a)

We find Z Th at the load terminals using the circuit in Fig. (a). j100 Ω

80 Ω

Zth

-j40 Ω

(a) (-j40)(80 + j100) = 51.2 − j1.6 80 + j60

Z Th = -j40 || (80 + j100) = Z L = Z *Th = 51.2 + j1.6 Ω (b)

We find VTh at the load terminals using Fig. (b). Io

j100 Ω +

3∠20° A

80 Ω

-j40 Ω

Vth −

(b) Io =

80 (8)(3∠20°) (3∠20°) = 80 + j100 − j40 8 + j6

VTh = - j40 I o =

(- j40)(24∠20°) 8 + j6 2

Pmax =

VTh 8RL

2

 40   ⋅ 24  10  = = 22.5 W (8)(51.2)

From Fig.(d), we obtain VTh using the voltage division principle.

5Ω

-j3 Ω

j2 Ω 10∠30° V

+ −

4Ω

+ Vth −

(d)  4 − j3   4 − j3  10  (10∠30°) =   ∠30° VTh =    9 − j3   3 − j  3 2

Pmax =

VTh

8RL

2

 5 10   ⋅   10 3  = = 1.389 W (8)(2.5)

Chapter 11, Solution 14.

I

j24 Ω

–j10 Ω

16 Ω 40∠90º A

VTh

10 Ω j8 Ω

Z Th = − j10 +

+ ZTh

_

(10 + j24)(16 + j8) = − j10 + 8.245 + j7.7 = 8.245 − j2.3Ω 10 + j24 + 16 + j8

Z = Z∗Th = 8.245 + j2.3Ω

10 j40(16 + j8) 10 + j24 + 16 + j8 = 173.55∠65.66° = 71.53 + j158.12 V

VTh = I(16 + j8) =

2 VTh

Pmax =

I 2rms 8.245 =

2

2 8.245 = 456.6 W (2x8.245) 2

Chapter 11, Solution 15. To find Z Th , insert a 1-A current source at the load terminals as shown in Fig. (a).

1Ω

1

-j Ω

2

+ 2 Vo

jΩ

Vo

1A

− (a) At node 1, Vo Vo V2 − Vo + = 1 j -j At node 2,

1 + 2 Vo =

V2 − Vo -j

 → Vo = j V2

 → 1 = j V2 − (2 + j) Vo

Substituting (1) into (2), 1 = j V2 − (2 + j)( j) V2 = (1 − j) V2 V2 =

1 1− j

VTh =

V2 1 + j = = 0.5 + j0.5 1 2

Z L = Z *Th = 0.5 − j0.5 Ω

(1)

(2)

We now obtain VTh from Fig. (b). 1Ω

12∠0° V

-j Ω +

+

+ −

Vo

jΩ

2 Vo

Vth −

− (b) 12 − Vo Vo = 1 j - 12 Vo = 1+ j 2 Vo +

Vo − (- j × 2 Vo ) + VTh = 0 VTh = -(1 + j2)Vo =

(12)(1 + j2) 1+ j 2

Pmax =

VTh

2

8RL

12 5     2  = 90 W = (8)(0.5)

Chapter 11, Solution 16.

ω = 4,

1H

 →

jωL = j 4,

 →

1 / 20F

1 1 = = − j5 jωC j 4 x1 / 20

We find the Thevenin equivalent at the terminals of ZL. To find VTh, we use the circuit shown below. 0.5Vo

2Ω

4Ω

V1

V2 +

+ 10<0o -

+ Vo -

-j5

j4

VTh -

At node 1, V V − V2 10 − V1 = 1 + 0.25V1 + 1 2 − j5 4 At node 2, V1 − V2 V + 0.25V1 = 2 4 j4

 →

 →

5 = V1 (1 + j 0.2) − 0.25V2

0 = 0.5V1 + V2 (−0.25 + j 0.25)

Solving (1) and (2) leads to VTh = V2 = 6.1947 + j 7.0796 = 9.4072∠48.81o Chapter 11, Solution 17. We find R Th at terminals a-b following Fig. (a).

-j10 Ω

30 Ω a

b

40 Ω

j20 Ω (a)

Z Th = 30 || j20 + 40 || (- j10) =

(30)( j20) (40)(-j10) + 30 + j20 40 − j10

Z Th = 9.23 + j13.85 + 2.353 − j9.41 Z Th = 11.583 + j4.44 Ω

Z L = Z *Th = 11.583 − j4.44 Ω We obtain VTh from Fig. (b).

I1

I2

-j10 Ω

30 Ω j5 A

+ VTh − 40 Ω

j20 Ω (b)

(1)

(2)

Using current division, 30 + j20 I1 = ( j5) = -1.1 + j2.3 70 + j10

I2 =

40 − j10 ( j5) = 1.1 + j2.7 70 + j10

VTh = 30 I 2 + j10 I 1 = 10 + j70 Pmax =

VTh

2

8RL

=

5000 = 53.96 W (8)(11.583)

Chapter 11, Solution 18. We find Z Th at terminals a-b as shown in the figure below.

40 Ω

40 Ω

-j10 Ω

80 Ω a

j20 Ω

Zth

b

Z Th = j20 + 40 || 40 + 80 || (-j10) = j20 + 20 + Z Th = 21.23 + j10.154

Z L = Z *Th = 21.23 − j10.15 Ω

Chapter 11, Solution 19. At the load terminals,

Z Th = - j2 + 6 || (3 + j) = -j2 + Z Th = 2.049 − j1.561

R L = Z Th = 2.576 Ω

(6)(3 + j) 9+ j

(80)(-j10) 80 − j10

To get VTh , let Z = 6 || (3 + j) = 2.049 + j0.439 . By transforming the current sources, we obtain VTh = (4 ∠0°) Z = 8.196 + j1.756

Pmax =

VTh

2

8RL

=

70.258 = 3.409 W 20.608

Chapter 11, Solution 20. Combine j20 Ω and -j10 Ω to get j20 || -j10 = -j20 To find Z Th , insert a 1-A current source at the terminals of R L , as shown in Fig. (a). Io

40 Ω

V1

4 Io

V2

+ −

-j20 Ω

-j10 Ω

1A

(a) At the supernode, 1=

V1 V V + 1 + 2 40 - j20 - j10

40 = (1 + j2) V1 + j4 V2

Also,

V1 = V2 + 4 I o , 1.1 V1 = V2

 → V1 =

Substituting (2) into (1), V  40 = (1 + j2)  2  + j4 V2  1 .1 

(1) where I o = V2 1 .1

- V1 40

(2)

V2 =

44 1 + j6.4

Z Th =

V2 = 1.05 − j6.71 Ω 1

R L = Z Th = 6.792 Ω To find VTh , consider the circuit in Fig. (b). 40 Ω

Io

V1

4 Io

V2

+ −

+ 120∠0° V

+ −

-j20 Ω

-j10 Ω

Vth −

(b)

At the supernode, V V 120 − V1 = 1 + 2 40 - j20 - j10 120 = (1 + j2) V1 + j4 V2

Also,

V1 = V2 + 4 I o , V1 =

(3) where I o =

120 − V1 40

V2 + 12 1 .1

(4)

Substituting (4) into (3), 109.09 − j21.82 = (0.9091 + j5.818) V2 VTh = V2 =

Pmax =

109.09 − j21.82 = 18.893∠ - 92.43° 0.9091 + j5.818

VTh 8RL

2

(18.893) 2 = = 6.569 W (8)(6.792)

Chapter 11, Solution 21. We find Z Th at terminals a-b, as shown in the figure below.

100 Ω

-j10 Ω a

40 Ω

Zth

50 Ω j30 Ω

b Z Th = 50 || [ - j10 + 100 || (40 + j30) ]

where 100 || (40 + j30) =

(100)(40 + j30) = 31.707 + j14.634 140 + j30

Z Th = 50 || (31.707 + j4.634) =

(50)(31.707 + j4.634) 81.707 + j4.634

Z Th = 19.5 + j1.73

R L = Z Th = 19.58 Ω

Chapter 11, Solution 22. i (t ) = 4 sin t ,

I

2

rms

=

1

0
π

16 sin π∫

2

tdt =

0

I rms = 8 = 2.828 A

16  t sin 2t   −  π 2 4 

π 0

=

16 π ( − 0) = 8 π 2

Chapter 11, Solution 23. 15, 0 < t < 2 v( t ) =   5, 2 < t < 6 2 Vrms =

1 6

[ ∫ 15 2

2

0

]

dt + ∫2 5 2 dt = 6

550 6

Vrms = 9.574 V

Chapter 11, Solution 24.

 5, 0 < t < 1 v( t ) =  - 5, 1 < t < 2

T = 2,

2 = Vrms

1 2

[∫ 5 1

0

2

]

dt + ∫1 (-5) 2 dt = 2

25 [1 + 1] = 25 2

Vrms = 5 V

Chapter 11, Solution 25.

[

1 T 2 1 1 2 3 f ( t ) dt (−4) 2 dt + ∫ 1 0dt + ∫2 4 2 dt = ∫ ∫ 0 0 T 3 1 32 = [16 + 0 + 16] = 3 3

2 f rms =

f rms =

32 = 3.266 3

]

Chapter 11, Solution 26. 5 0< t<2 v( t ) =  10 2 < t < 4

T = 4,

2 Vrms =

1 4

[∫ 5 2

2

0

]

4 1 dt + ∫2 (10) 2 dt = [50 + 200 ] = 62.5 4

Vrms = 7.906 V

Chapter 11, Solution 27. T = 5, I

2 rms

i( t ) = t , 0 < t < 5

1 t3 1 5 2 = ∫0 t dt = ⋅ 5 3 5

5 0

=

125 = 8.333 15

I rms = 2.887 A

Chapter 11, Solution 28.

2 Vrms =

2 rms

V

1 5

[ ∫ (4t ) 2

2

dt + ∫2 0 2 dt

2 0

=

0

1 16 t 3 = ⋅ 5 3

5

]

16 (8) = 8.533 15

Vrms = 2.92 V 2 Vrms 8.533 P= = = 4.267 W R 2

Chapter 11, Solution 29.

 20 − 2t 5 < t < 15 i( t ) =  - 40 + 2t 15 < t < 25

T = 20 ,

[∫

]

2 I eff =

1 20

2 I eff =

25 1  15 2 ( 100 20 t t ) dt ( t 2 − 40 t + 400) dt  − + + ∫ ∫   5 15 5

I

2 eff

15

5

(20 − 2 t ) 2 dt + ∫15 (-40 + 2t) 2 dt 25

1 t 3  15 2  =  100 t − 10 t +  5 5  3

 25   t3 +  − 20 t 2 + 400 t  15    3

1 2 I eff = [83.33 + 83.33 ] = 33.332 5 I eff = 5.773 A 2 P = I eff R = 400 W

Chapter 11, Solution 30.

t 0
1 4

[∫ t 2

2

0

]

4  1 8 dt + ∫2 (-1) 2 dt =  + 2  = 1.1667  43

Vrms = 1.08 V

Chapter 11, Solution 31.

V

2

rms

2 1 2  1 4 1 1  2 = ∫ v(t )dt =  ∫ (2t ) dt + ∫ (−4) 2 dt  =  + 16 = 8.6667 20 2 0  1  2 3

Vrms = 2.944 V

Chapter 11, Solution 32.

I 2rms =

I

2 rms

2 1 1 (10t 2 ) 2 dt + ∫ 0 dt  ∫  1 2  0

t5 = 50 ∫0 t dt = 50 ⋅ 5 1

4

1 0

= 10

I rms = 3.162 A

Chapter 11, Solution 33.  10 0 < t <1  i( t ) =  20 − 10t 1 < t < 2  0 2
I 2rms =

1 3

[ ∫ 10 1

0

2

dt + ∫1 (20 − 10t ) 2 dt + 0 2

]

3 I 2rms = 100 + 100∫1 (4 − 4t + t 2 ) dt = 100 + (100)(1 3) = 133.33 2

I rms =

133.33 = 6.667 A 3

Chapter 11, Solution 34.

[

1 T 2 1 2 3 f ( t )dt = ∫ 0 (3t ) 2 dt + ∫ 2 6 2 dt ∫ 0 T 3 2   1  9t 3 = + 36 = 20  3 3 0  

2 f rms =

f rms = 20 = 4.472

]

Chapter 11, Solution 35.

2 Vrms =

1 6

[ ∫ 10 1

0

2

dt + ∫1 20 2 dt + ∫2 30 2 dt + ∫4 20 2 dt + ∫5 10 2 dt 6

5

4

2

]

1 2 Vrms = [100 + 400 + 1800 + 400 + 100 ] = 466.67 6 Vrms = 21.6 V

Chapter 11, Solution 36.

(a) Irms = 10 A 2

 3   → (b) V rms = 4 +    2 36 (c) = 9.055 A I rms = 64 + 2 2

(d)

2

Vrms =

Vrms = 16 +

9 = 4.528 V (checked) 2

25 16 + = 4.528 V 2 2

Chapter 11, Solution 37. i = i1 + i2 + i3 = 8 + 4 sin(t + 10 o ) + 6 cos(2t + 30 o ) I rms = I 21rms + I 2 2 rms + I 2 3rms = 64 +

16 36 + = 90 = 9.487 A 2 2

Chapter 11, Solution 38. 0.5 H  →

jωL = j (2π )(50)(0.5) = j157.08

Z = R + jX L = 30 + j157.08

S=

V

2

Z*

Apparent power = S =

=

(210) 2 30 − j157.08

(210) 2 = 275.6 VA 160

 157.08   = cos(79.19°) pf = cos θ = cos  tan -1   36  

pf = 0.1876 (lagging)

Chapter 11, Solution 39.

Z T = j4 || (12 − j8) =

( j4)(12 − j8) 12 − j4

Z T = 0.4 (3 + j11) = 4.56 ∠74.74°

pf = cos(74.74°) = 0.2631

Chapter 11, Solution 40. At node 1, 120∠30 o − V1 V1 V1 − V2 = + 20 j30 50

 →

103.92 + j60 = V1 (1.4 − j0.6667) − 0.4V2

(1)

At node 2, V1 − V2 V2 V = + 2 50 10 − j 40

 →

0 = −V1 + (6 + j1.25)V2

Solving (1) and (2) leads to

V1 = 45.045 + j66.935, V2 = 9.423 + j9.193

(2)

(a) Pj 30 Ω = 0 = P− j 40 Ω P10 Ω =

P50 Ω

1 | V1 − V2 | 2 = = 4603.1 / 100 = 46.03 W R 2

P20 Ω =

(b) I =

V 2 rms 1 | V2 | 2 = = 173.3 / 20 = 8.665 W R 2 R

1 | 120∠30 o − V1 | 2 = 3514 / 40 = 87.86 W R 2

120∠30 o − V1 = 2.944 − j 0.3467, 20

1 S = Vs I • = 142.5 − j106.3, 2

Vs = 120∠30 o = 103.92 + j 60

S =| S | = 177.8 VA

(c ) pf = 142.5/177.8 = 0.8015 (leading).

Chapter 11, Solution 41.

(a)

- j2 || ( j5 − j2) = -j2 || -j3 =

(-j2)(-j3) = -j6 j

Z T = 4 − j6 = 7.211∠ - 56.31°

pf = cos(-56.31°) = 0.5547 (leading) (b)

j2 || (4 + j) =

( j2)(4 + j) = 0.64 + j1.52 4 + j3

Z = 1 || (0.64 + j1.52 − j) =

0.64 + j0.44 = 0.4793∠21.5° 1.64 + j0.44

pf = cos(21.5°) = 0.9304 (lagging)

Chapter 11, Solution 42.

pf = 0.86 = cos θ  → θ = 30.683° Q = S sin θ  → S =

S = V I*

 → I * =

Q 5 = = 9.798 kVA sin θ sin(30.683°)

S 9.798 × 10 3 ∠30.683° = = 44.536 ∠30.683° 220 V

Peak current = 2 × 44.536 = 62.98 A Apparent power = S = 9.798 kVA

Chapter 11, Solution 43.

(a) Vrms = V 21rms + V 2 2 rms + V 2 3rms = 25 + (b) P =

9 1 + = 30 = 5.477 V 2 2

V 2 rms = 30 / 10 = 3 W R

Chapter 11, Solution 44. pf = 0.65 = cosθ

 →

θ = 49.46 o

S = S (cosθ + j sin θ ) = 50(0.65 + j 0.7599) = 32.5 + j 38 kVA Thus, Average power = 32.5 kW, Reactive power = 38 kVAR

Chapter 11, Solution 45.

(a) V 2 rms = 20 2 +

60 2 = 2200 2

I rms = 12 +

 →

Vrms = 46.9 V

0.5 2 = 1.125 = 1.061A 2

(b) P = Vrms I rms = 49.74 W

Chapter 11, Solution 46. (a)

S = V I * = (220∠30°)(0.5∠ - 60°) = 110∠ - 30° S = 95.26 − j55 VA Apparent power = 110 VA Real power = 95.26 W Reactive power = 55 VAR pf is leading because current leads voltage

(b)

S = V I * = (250∠ - 10°)(6.2 ∠25°) = 1550∠15° S = 1497.2 + j401.2 VA Apparent power =1550 VA Real power = 1497.2 W Reactive power = 401.2 VAR pf is lagging because current lags voltage

(c)

S = V I * = (120∠0°)(2.4∠15°) = 288∠15° S = 278.2 + j74.54 VA Apparent power = 288 VA Real power = 278.2 W Reactive power = 74.54 VAR pf is lagging because current lags voltage

(d)

S = V I * = (160 ∠45°)(8.5∠ - 180°) = 1360∠ - 135° S = - 961.7 − j961.7 VA Apparent power = 1360 VA Real power = - 961.7 W Reactive power = - 961.7 VAR pf is leading because current leads voltage

Chapter 11, Solution 47. (a)

V = 112 ∠10° , I = 4∠ - 50° 1 S = V I * = 224∠60° = 112 + j194 VA 2 Average power = 112 W

Reactive power =194 VAR (b)

V = 160 ∠0° , I = 25∠45° 1 S = V I * = 200∠ - 45° = 141.42 − j141.42 VA 2

Average power = 141.42 W Reactive power = - 141.42 VAR

(c)

S=

2

V Z*

=

(80) 2 = 128∠30° = 90.51 + j64 VA 50∠ - 30°

Average power = 90.51 W Reactive power = 64 VAR (d)

2

S = I Z = (100)(100∠45°) = 7.071 + j7.071 kVA Average power = 7.071 kW Reactive power = 7.071 kVAR

Chapter 11, Solution 48. (a)

S = P − jQ = 269 − j150 VA

(b)

pf = cos θ = 0.9  → θ = 25.84° Q = S sin θ  → S =

Q 2000 = = 4588.31 sin θ sin(25.84°)

P = S cos θ = 4129.48 S = 4129 − j2000 VA

(c)

Q 450 = = 0.75 S 600 pf = 0.6614

Q = S sin θ  → sin θ = θ = 48.59 ,

P = S cos θ = (600)(0.6614) = 396.86 S = 396.9 + j450 VA

(d)

S=

V

2

Z

=

(220) 2 = 1210 40

P = S cos θ  → cos θ =

P 1000 = = 0.8264 S 1210

θ = 34.26°

Q = S sin θ = 681.25 S = 1000 + j681.2 VA

Chapter 11, Solution 49.

(a)

4 sin(cos -1 (0.86)) kVA 0.86 S = 4 + j2.373 kVA

S = 4+ j

(b)

pf =

P 1.6 = 0.8 = cos θ  → sin θ = 0.6 S 2

S = 1.6 − j2 sin θ = 1.6 − j1.2 kVA (c)

S = Vrms I *rms = (208∠20°)(6.5∠50°) VA S = 1.352 ∠70° = 0.4624 + j1.2705 kVA 2

(d)

V (120) 2 14400 S= * = = Z 40 − j60 72.11∠ - 56.31° S = 199.7 ∠56.31° = 110.77 + j166.16 VA

Chapter 11, Solution 50.

(a)

S = P − jQ = 1000 − j

1000 sin(cos -1 (0.8)) 0.8

S = 1000 − j750

But,

Z = *

Vrms

S=

Vrms

2

Z* 2

S

(220) 2 = = 30.98 + j23.23 1000 − j750

Z = 30.98 − j23.23 Ω (b)

2

S = I rms Z Z=

(c)

Z = *

S I rms

2

Vrms S

=

2

1500 + j2000 = 10.42 + j13.89 Ω (12) 2 V

2

(120) 2 = = = 1.6 ∠ - 60° 2S (2)(4500 ∠60°)

Z = 1.6 ∠60° = 0.8 + j1.386 Ω

Chapter 11, Solution 51. (a)

Z T = 2 + (10 − j5) || (8 + j6)

ZT = 2 +

(10 − j5)(8 + j6) 110 + j20 = 2+ 18 + j 18 + j

Z T = 8.152 + j0.768 = 8.188∠5.382°

pf = cos(5.382°) = 0.9956 (lagging) 2

(b)

V 1 (16) 2 S = V I* = = 2 2 Z * (2)(8.188∠ - 5.382°) S = 15.63∠5.382°

P = S cos θ = 15.56 W

(c)

Q = S sin θ = 1.466 VAR

(d)

S = S = 15.63 VA

(e)

S = 15.63∠5.382° = 15.56 + j1.466 VA

Chapter 11, Solution 52. 2000 0.6 = 2000 + j1500 0 .8 S B = 3000 x 0.4 − j3000 x 0.9165 = 1200 − j2749

S A = 2000 + j

SC = 1000 + j500 S = S A + S B + SC = 4200 − j749

4200

(a)

pf =

(b)

S = Vrms I ∗rms  → I ∗rms =

4200 2 + 749 2

= 0.9845 leading.

Irms = 35.55∠–55.11˚ A.

4200 − j749 = 35.55∠ − 55.11° 120∠45°

Chapter 11, Solution 53.

S = SA + SB + SC = 4000(0.8–j0.6) + 2400(0.6+j0.8) + 1000 + j500 = 5640 + j20 = 5640∠0.2˚ I∗rms = (a)

SB S + SC S 5640∠0.2° = 66.46∠ − 29.8° + A = = 120∠30° Vrms Vrms Vrms 2

I = 2 x 66.46∠29.88° = 93.97∠29.8° A (b)

pf = cos(0.2˚) ≈ 1.0 lagging.

Chapter 11, Solution 54.

(a)

S = P − jQ = 1000 − j

1000 sin(cos -1 (0.8)) 0.8

S = 1000 − j750

But,

Z = *

Vrms

S=

Vrms

2

Z* 2

=

S

(220) 2 = 30.98 + j23.23 1000 − j750

Z = 30.98 − j23.23 Ω (b)

2

S = I rms Z Z=

(c)

Z = *

S I rms

2

Vrms S

=

1500 + j2000 = 10.42 + j13.89 Ω (12) 2

2

=

V

2

2S

=

(120) 2 = 1.6 ∠ - 60° (2)(4500 ∠60°)

Z = 1.6 ∠60° = 0.8 + j1.386 Ω

Chapter 11, Solution 55. We apply mesh analysis to the following circuit.

-j20 Ω

j10 Ω I3

+ −

40∠0° V rms

I1

20 Ω

I2

+ −

40 = (20 − j20) I1 − 20 I 2 2 = (1 − j) I1 − I 2 For mesh 2, - j50 = (20 + j10) I 2 − 20 I1 - j5 = -2 I1 + (2 + j) I 2 Putting (1) and (2) in matrix form,  2  1 − j - 1  I1   - j5 =  - 2 2 + j I      2 

50∠90° V rms

For mesh 1,

∆ = 1− j ,

∆ 1 = 4 − j3 ,

I1 =

∆ 1 4 − j3 1 = = (7 − j) = 3.535∠8.13° ∆ 1− j 2

I2 =

∆ 2 - 1 − j5 = = 2 − j3 = 3.605∠ - 56.31° ∆ 1− j

(1) (2)

∆ 2 = -1 − j5

I 3 = I1 − I 2 = (3.5 + j0.5) − (2 − j3) = 1.5 + j3.5 = 3.808∠66.8° For the 40-V source, 1  S = -V I 1* = -(40)  ⋅ (7 − j)  = - 140 + j20 VA 2  For the capacitor, S = I1

2

Z c = - j250 VA

2

R = 290 VA

2

Z L = j130 VA

For the resistor, S = I3

For the inductor, S = I2

For the j50-V source, S = V I *2 = ( j50)(2 + j3) = - 150 + j100 VA

Chapter 11, Solution 56.

- j2 || 6 =

(6)(- j2) = 0.6 − j1.8 6 − j2

3 + j4 + (-j2) || 6 = 3.6 + j2.2 The circuit is reduced to that shown below.

Io +

2∠30° A

3.6 + j2.2 Ω

5Ω

Vo −

Io =

3.6 + j2.2 (2∠30°) = 0.95∠47.08° 8.6 + j2.2

Vo = 5 I o = 4.75∠47.08°

S=

1 1 Vo I *s = ⋅ (4.75∠47.08°)(2∠ - 30°) 2 2

S = 4.75∠17.08° = 4.543 + j1.396 VA

Chapter 11, Solution 57. Consider the circuit as shown below.

4Ω

24∠0° V

+ −

Vo

-j1 Ω

V1

2Ω +

1Ω

j2 Ω

V2 −

At node o, 24 − Vo Vo Vo − V1 = + 4 1 -j

2 Vo

24 = (5 + j4) Vo − j4 V1 At node 1,

(1)

Vo − V1 V + 2 Vo = 1 -j j2 V1 = (2 − j4) Vo

(2)

Substituting (2) into (1), 24 = (5 + j4 − j8 − 16) Vo

Vo =

- 24 , 11 + j4

V1 =

(-24)(2 - j4) 11 + j4

The voltage across the dependent source is V2 = V1 + (2)(2 Vo ) = V1 + 4 Vo

V2 =

(-24)(6 − j4) - 24 ⋅ (2 − j4 + 4) = 11 + j4 11 + j4

S=

1 1 V2 I * = V2 (2 Vo* ) 2 2

S=

(-24)(6 − j4) - 24  576  (6 − j4) ⋅ = 11 + j4 11 - j4  137 

S = 25.23 − j16.82 VA

Chapter 11, Solution 58. Ix -j3 kΩ

8 mA

From the left portion of the circuit, 0.2 Io = = 0.4 mA 500 20 I o = 8 mA

4 kΩ

j1 kΩ

10 kΩ

From the right portion of the circuit, 16 4 Ix = mA (8 mA) = 7− j 4 + 10 + j − j3 S = Ix

2

R=

(16 × 10 -3 ) 2 ⋅ (10 × 10 3 ) 50

S = 51.2 mVA

Chapter 11, Solution 59. Consider the circuit below. Ix -j3 kΩ

8 mA

4+

j1 kΩ

4 kΩ

10 kΩ

Vo Vo 240 − Vo = + 50 - j20 40 + j30

88 = (0.36 + j0.38) Vo Vo =

88 = 168.13∠ - 46.55° 0.36 + j0.38

I1 =

Vo = 8.41∠43.45° - j20

I2 =

Vo = 3.363∠ - 83.42° 40 + j30

Reactive power in the inductor is 1 1 2 S = I 2 Z L = ⋅ (3.363) 2 ( j30) = j169.65 VAR 2 2 Reactive power in the capacitor is 1 1 2 S = I 1 Z c = ⋅ (8.41) 2 (- j20) = - j707.3 VAR 2 2

Chapter 11, Solution 60.

S1 = 20 + j

20 sin(cos -1 (0.8)) = 20 + j15 0.8

S 2 = 16 + j

16 sin(cos -1 (0.9)) = 16 + j7.749 0.9

S = S1 + S 2 = 36 + j22.749 = 42.585∠32.29°

S = Vo I * = 6 Vo

But

Vo =

S = 7.098 ∠ 32.29° 6

pf = cos(32.29°) = 0.8454 (lagging)

Chapter 11, Solution 61. Consider the network shown below.

I2

Io

S2 I1

+ Vo

So

S1

S3

− S 2 = 1.2 − j0.8 kVA

S3 = 4 + j

4 sin(cos -1 (0.9)) = 4 + j1.937 kVA 0.9

Let

S 4 = S 2 + S 3 = 5.2 + j1.137 kVA

But

S4 =

1 V I* 2 o 2

2 S 4 (2)(5.2 + j1.137) × 10 3 I = = = 22.74 − j104 Vo 100 ∠90° * 2

I 2 = 22.74 + j104

2 sin(cos -1 (0.707)) = 2 − j2 kVA 0.707

Similarly,

S1 = 2 − j

But

S1 =

1 Vo I 1* 2

I 1* =

2 S 1 (4 − j4) × 10 3 = = -40 − j40 Vo j100

I 1 = -40 + j40

I o = I 1 + I 2 = -17.26 + j144 = 145∠96.83° So =

1 Vo I *o 2

So =

1 ⋅ (100∠90°)(145∠ - 96.83°) VA 2

S o = 7.2 − j0.862 kVA

Chapter 11, Solution 62. Consider the circuit below

0.2 + j0.04 Ω

I

I2

0.3 + j0.15 Ω

I1 Vs

+ −

+

+

V1

V2

−

−

S 2 = 15 − j

15 sin(cos -1 (0.8)) = 15 − j11.25 0.8

S 2 = V2 I *2

But

I *2 =

S 2 15 − j11.25 = V2 120

I 2 = 0.125 + j0.09375 V1 = V2 + I 2 (0.3 + j0.15) V1 = 120 + (0.125 + j0.09375)(0.3 + j0.15) V1 = 120.02 + j0.0469

S1 = 10 + j

10 sin(cos -1 (0.9)) = 10 + j4.843 0.9

S1 = V1 I 1*

But

I 1* =

S 1 11.111∠25.84° = V1 120.02 ∠0.02°

I 1 = 0.093∠ - 25.82° = 0.0837 − j0.0405 I = I 1 + I 2 = 0.2087 + j0.053 Vs = V1 + I (0.2 + j0.04)

Vs = (120.02 + j0.0469) + (0.2087 + j0.053)(0.2 + j0.04) Vs = 120.06 + j0.0658 Vs = 120.06∠0.03° V

Chapter 11, Solution 63. Let

S = S1 + S 2 + S 3 . S1 = 12 − j

12 sin(cos -1 (0.866)) = 12 − j6.929 0.866

S 2 = 16 + j

16 sin(cos -1 (0.85)) = 16 + j9.916 0.85

S3 =

(20)(0.6) + j20 = 15 + j20 sin(cos -1 (0.6)

S = 43 + j22.987 = I *o =

1 V I *o 2

2 S 44 + j22.98 = V 110

I o = 0.4513∠ - 27.58° A

Chapter 11, Solution 64. I2 I1 8Ω + −

Is

120∠0º V

j12

Is + I2 = I1 or Is = I1 – I2 I1 = But,

120 8 + j12

= 4.615 − j6.923

2500 − j400 S S = VI ∗2  → I ∗2 = = = 20.83 − j3.333 V 120 or I 2 = 20.83 + j3.333 Is = I1 – I2 = –16.22 – j10.256 = 19.19∠–147.69˚ A.

Chapter 11, Solution 65.

C = 1 nF  →

1 -j = 4 = -j100 kΩ jωC 10 × 10 -9

At the noninverting terminal, 4∠0° − Vo Vo = 100 - j100 Vo =

4 2

v o (t) =

 → Vo =

4 1+ j

∠ - 45° 4 2

cos(10 4 t − 45°) 2

2  4 1   1  Vrms   W = ⋅ P= R  2 2   50 × 10 3 

P = 80 µW

Chapter 11, Solution 66.

As an inverter, - Zf - (2 + j4) Vo = Vs = ⋅ (4 ∠45°) Zi 4 + j3 Io =

Vo - (2 + j4)(4∠45°) mA mA = (6 - j2)(4 + j3) 6 − j2

The power absorbed by the 6-kΩ resistor is 2 2 1 1  20 × 4  -6 3 P = Io R = ⋅  × 10 × 6 × 10 2 2  40 × 5  P = 0.96 mW

Chapter 11, Solution 67.

ω = 2,

3H

10 //( − j 5) =

 →

jωL = j 6,

0.1F

 →

1 1 = = − j5 jωC j 2 x0.1

− j 50 = 2 − j4 10 − j 5

The frequency-domain version of the circuit is shown below. Z2=2-j4 Ω

Z1 =8+j6 Ω

I1

+

+

+

0.6∠20 o V

Io

Z 3 = 12Ω

Vo

-

(a) I 1 =

0.6∠20 o − 0 0.5638 + j 0.2052 = = 0.06∠ − 16.87 o 8 + j6 8 + j6

1 S = Vs I *1 = (0.3∠20 o )(0.06∠ + 16.87 o ) = 14.4 + j10.8 mVA = 18∠36.86 o mVA 2 (b) Vo = −

Z2 Vs , Z1

P=

Io =

Vo ( 2 − j 4) =− (0.6∠20 o ) = 0.0224∠99.7 o 12(8 + j 6) Z3

1 | I o | 2 R = 0.5(0.0224) 2 (12) = 2.904 mW 2

Chapter 11, Solution 68. S = SR + SL + Sc

Let

S R = PR + jQ R =

where

1 2 I R + j0 2 o

1 S L = PL + jQ L = 0 + j I o2 ωL 2 1 1 S c = Pc + jQ c = 0 − j I o2 ⋅ ωC 2 S=

Hence,

 1 2 1   I o  R + jωL −  2  ωC 

Chapter 11, Solution 69.

(a)

Given that Z = 10 + j12 tan θ =

12 10

 → θ = 50.19°

pf = cos θ = 0.6402

(b)

S=

V

2

2 Z*

=

(120) 2 = 295.12 + j354.09 (2)(10 − j12)

The average power absorbed = P = Re(S) = 295.1 W (c)

For unity power factor, θ1 = 0° , which implies that the reactive power due to the capacitor is Q c = 354.09 But

C=

Qc =

V2 1 = ωC V 2 2 Xc 2

2 Qc (2)(354.09) = 130.4 µF 2 = (2π )(60)(120) 2 ωV

Chapter 11, Solution 70.

pf = cos θ = 0.8  → sin θ = 0.6 Q = S sin θ = (880)(0.6) = 528 If the power factor is to be unity, the reactive power due to the capacitor is Q c = Q = 528 VAR But

C=

2 Vrms 1 Q= = ωC V 2 Xc 2

 → C =

2 Qc ωV2

(2)(528) = 69.45 µF (2π)(50)(220) 2

Chapter 11, Solution 71. P1 = Q1 = 150 x0.7071 = 106.065,

S 1 = 106.065 + j106.065,

Q2 = 50,

S2 =

Q2 , 0 .6

S 2 = 66.67 − j 50

S = S 1 + S 2 = 172.735 + j 56.06 = 181.6∠17.98 o ,

pf = cos17.98 o = 0.9512

Qc = P(tan θ 1 − tan θ 2 ) = 172.735(tan 17.98 o − 0) = 56.058 C=

Qc 56.058 = = 10.33 µF 2 ωV rms 2πx60 x120 2

Chapter 11, Solution 72.

(a)

θ1 = cos -1 (0.76) = 40.54° θ 2 = cos -1 (0.9) = 25.84° Q c = P (tan θ1 − tan θ 2 ) Q c = (40)[ tan(40.54°) − tan(25.84°) ] kVAR Q c = 14.84 kVAR C=

P2 = 0.8S = 0.8

Qc 14840 = = 2.734 mF 2 ω Vrms (2π )(60)(120) 2

50 = 66.67 0 .6

(b)

θ1 = 40.54° ,

θ 2 = 0°

Q c = (40)[ tan(40.54°) − 0 ] kVAR = 34.21 kVAR C=

Qc 34210 = 6.3 mF 2 ω Vrms (2π)(60)(120) 2

Chapter 11, Solution 73.

(a)

S = 10 − j15 + j22 = 10 + j7 kVA S = S = 10 2 + 7 2 = 12.21 kVA

(b)

S = V I*

 → I * =

S 10,000 + j7,000 = 240 V

I = 41.667 − j29.167 = 50.86∠ - 35° A (c)

7 θ1 = tan -1   = 35° , 10 

θ 2 = cos -1 (0.96) = 16.26°

Q c = P1 [ tan θ1 − tan θ 2 ] = 10 [ tan(35°) - tan(16.26°) ] Q c = 4.083 kVAR C=

(d)

Qc 4083 = = 188.03 µF 2 ω Vrms (2π )(60)(240) 2

S 2 = P2 + jQ 2 ,

P2 = P1 = 10 kW

Q 2 = Q1 − Q c = 7 − 4.083 = 2.917 kVAR S 2 = 10 + j2.917 kVA

But I *2 =

S 2 = V I *2 S 2 10,000 + j2917 = V 240

I 2 = 41.667 − j12.154 = 43.4∠ - 16.26° A

Chapter 11, Solution 74. (a)

θ1 = cos -1 (0.8) = 36.87° P1 24 = = 30 kVA cos θ1 0.8

S1 =

Q1 = S1 sin θ1 = (30)(0.6) = 18 kVAR S1 = 24 + j18 kVA

θ 2 = cos -1 (0.95) = 18.19° S2 =

P2 40 = = 42.105 kVA cos θ 2 0.95

Q 2 = S 2 sin θ 2 = 13.144 kVAR S 2 = 40 + j13.144 kVA

S = S1 + S 2 = 64 + j31.144 kVA

 31.144   = 25.95° θ = tan -1   64  pf = cos θ = 0.8992 (b)

θ 2 = 25.95° ,

θ1 = 0°

Q c = P [ tan θ 2 − tan θ1 ] = 64 [ tan(25.95°) − 0 ] = 31.144 kVAR C=

Qc 31,144 = = 5.74 mF 2 ω Vrms (2π )(60)(120) 2

Chapter 11, Solution 75.

(a)

2

V

S1 =

Z1*

=

(240) 2 5760 = = 517.75 − j323.59 VA 80 + j50 8 + j5

(240) 2 5760 S2 = = = 358.13 + j208.91 VA 120 − j70 12 − j7 S3 =

(240) 2 = 960 VA 60

S = S1 + S 2 + S 3 = 1835.88 − j114.68 VA (b)

 114.68   = 3.574° θ = tan -1  1835.88  pf = cos θ = 0.998

(c)

Q c = P [ tan θ 2 − tan θ1 ] = 18.35.88[ tan(3.574°) − 0 ] Q c = 114.68 VAR C=

Qc 114.68 = = 6.336 µF 2 ω Vrms (2π )(50)(240) 2

Chapter 11, Solution 76. The wattmeter reads the real power supplied by the current source. Consider the circuit below. 4Ω

12∠0° V

-j3 Ω

+ −

3∠30° +

Vo

j2 Ω

12 − Vo Vo Vo = + 4 − j3 j2 8

8Ω

3∠30° A

Vo =

S=

36.14 + j23.52 = 0.7547 + j11.322 = 11.347 ∠86.19° 2.28 − j3.04

1 1 Vo I *o = ⋅ (11.347 ∠86.19°)(3∠ - 30°) 2 2

S = 17.021∠56.19°

P = Re(S) = 9.471 W

Chapter 11, Solution 77.

The wattmeter measures the power absorbed by the parallel combination of 0.1 F and 150 Ω. 120 cos(2t )  → 120∠0° , ω= 2 4H  →

jωL = j8 1 0.1 F  → = -j5 jωC Consider the following circuit. 6Ω

120∠0° V

Z = 15 || (-j5) =

j8 Ω

I

+ −

(15)(-j5) = 1.5 − j4.5 15 − j5

I=

120 = 14.5∠ - 25.02° (6 + j8) + (1.5 − j4.5)

S=

1 1 2 1 V I * = I Z = ⋅ (14.5) 2 (1.5 − j4.5) 2 2 2

S = 157.69 − j473.06 VA

The wattmeter reads P = Re(S) = 157.69 W

Z

Chapter 11, Solution 78.

The wattmeter reads the power absorbed by the element to its right side. ω= 4 2 cos(4t )  → 2∠0° , 1H  →

jωL = j4

1 → F  12

1 = -j3 jωC

Consider the following circuit. 10 Ω

20∠0° V

I

+ −

Z = 5 + j4 + 4 || - j3 = 5 + j4 +

Z

(4)(- j3) 4 − j3

Z = 6.44 + j2.08 I=

20 = 1.207 ∠ - 7.21° 16.44 + j2.08

S=

1 2 1 I Z = ⋅ (1.207) 2 (6.44 + j2.08) 2 2

P = Re(S) = 4.691 W

Chapter 11, Solution 79.

The wattmeter reads the power supplied by the source and partly absorbed by the 40- Ω resistor.

ω = 100, j100x10x10 − 3 = j,

 →

10 mH

500µF

 →

1 1 = = − j20 jωC j100x500 x10 − 6

The frequency-domain circuit is shown below. 20 40

I

Io

j V1

V2

+1 2 Io

o

10<0

-j20

At node 1, 10 − V1 V − V2 V1 − V2 3(V1 − V2 ) V1 − V2 = 2I o + 1 + = + 40 j 20 20 j 10 = (7 − j40)V1 + (−6 + j40)V2

→

(1)

At node 2, V1 − V 2 V1 − V 2 V + = 2 j 20 − j 20

 →

0 = (20 + j )V1 − (19 + j )V 2

Solving (1) and (2) yields V1 = 1.5568 –j4.1405 I=

10 − V1 = 0.8443 + j 0.4141, 40

1 S = VI • = 4.2216 − j 2.0703 2

P = Re(S) = 4.222 W.

Chapter 11, Solution 80.

(a)

I=

V 110 = = 17.19 A Z 6.4

(2)

(b)

V 2 (110) 2 = = 1890.625 S= Z 6 .4

cos θ = pf = 0.825  → θ = 34.41° P = S cos θ = 1559.76 ≅ 1.6 kW

Chapter 11, Solution 81. kWh consumed = 4017 − 3246 = 771 kWh The electricity bill is calculated as follows : (a) Fixed charge = $12 (b) First 100 kWh at $0.16 per kWh = $16 (c) Next 200 kWh at $0.10 per kWh = $20 (d) The remaining energy (771 – 300) = 471 kWh at $0.06 per kWh = $28.26. Adding (a) to (d) gives $76.26

Chapter 11, Solution 82. (a)

P1 = 5,000,

Q1 = 0

P2 = 30,000 x0.82 = 24,600,

Q2 = 30,000 sin(cos −1 0.82) = 17,171

S = S1 + S 2 = (P1 + P2 ) + j(Q1 + Q 2 ) = 29,600 + j17,171 S =| S |= 34.22 kVA (b)

Q = 17.171 kVAR

(c )

pf =

P 29,600 = = 0.865 S 34,220

Q c = P(tan θ1 − tan θ 2 )

[

]

= 29,600 tan(cos −1 0.865) − tan(cos −1 0.9) = 2833 VAR

(d)

C=

Qc 2833 = = 130.46µ F 2 ωV rms 2πx60 x 240 2

Chapter 11, Solution 83.

1 1 (a) S = VI ∗ = (210∠60 o )(8∠ − 25 o ) = 840∠35 o 2 2 P = S cosθ = 840 cos 35 o = 688.1 W

(b) S = 840 VA (c) Q = S sin θ = 840 sin 35 o = 481.8 VAR (d) pf = P / S = cos 35 o = 0.8191 (lagging) Chapter 11, Solution 84.

(a)

Maximum demand charge = 2,400 × 30 = $72,000 Energy cost = $0.04 × 1,200 × 10 3 = $48,000 Total charge = $120,000

(b)

To obtain $120,000 from 1,200 MWh will require a flat rate of $120,000 per kWh = $0.10 per kWh 1,200 × 10 3

Chapter 11, Solution 85.  → j 2πx60 x15 x10 −3 = j 5.655 (a) 15 mH We apply mesh analysis as shown below. I1 + Ix

120<0o V -

10 Ω

In

30 Ω Iz

120<0o V

10 Ω

+ Iy

-

j5.655 Ω I2

For mesh x, (1) 120 = 10 Ix - 10 Iz For mesh y, (2) 120 = (10+j5.655) Iy - (10+j5.655) Iz For mesh z, (3) 0 = -10 Ix –(10+j5.655) Iy + (50+j5.655) Iz Solving (1) to (3) gives Ix =20, Iy =17.09-j5.142, Iz =8 Thus, I1 =Ix =20 A I2 =-Iy =-17.09+j5.142 = 17.85∠163.26 o A In =Iy - Ix =-2.091 –j5.142 = 5.907∠ − 119.5 o A (b)

S1 =

1 (120) I • x = 60 x 20 = 1200, 2

S2 =

1 (120) I • y = 1025.5 − j 308.5 2

S = S1 + S 2 = 2225.5 − j 308.5 VA

(c ) pf = P/S = 2225.5/2246.8 = 0.9905 Chapter 11, Solution 86. For maximum power transfer Z L = Z *Th  → Z i = Z Th = Z *L Z L = R + jωL = 75 + j (2π)(4.12 × 10 6 )(4 × 10 -6 ) Z L = 75 + j103.55 Ω Z i = 75 − j103.55 Ω

Chapter 11, Solution 87.

Z = R ± jX VR = I R 2

 → R =

Z = R 2 + X2 X = 2.5377 kΩ

VR 80 = = 1 .6 k Ω I 50 × 10 -3

 → X 2 = Z

2

− R 2 = (3) 2 − (1.6) 2

X  2.5377   = 57.77° θ = tan -1   = tan -1  R   1.6 

pf = cos θ = 0.5333

Chapter 11, Solution 88. (a)

S = (110)(2 ∠55°) = 220∠55° P = S cos θ = 220 cos(55°) = 126.2 W

(b)

S = S = 220 VA

Chapter 11, Solution 89. (a)

Apparent power = S = 12 kVA P = S cos θ = (12)(0.78) = 9.36 kW Q = S sin θ = 12 sin(cos -1 (0.78)) = 7.51 kVAR

S = P + jQ = 9.36 + j7.51 kVA

(b)

V S= * Z

2

V  → Z = S

2

*

=

(210) 2 (9.36 + j7.51) × 10 3

Z = 34.398 + j27.6 Ω

Chapter 11, Solution 90 Original load : P1 = 2000 kW , S1 =

cos θ1 = 0.85  → θ1 = 31.79°

P1 = 2352.94 kVA cos θ1

Q1 = S1 sin θ1 = 1239.5 kVAR

Additional load : P2 = 300 kW , S2 =

cos θ 2 = 0.8  → θ 2 = 36.87°

P2 = 375 kVA cos θ 2

Q 2 = S 2 sin θ 2 = 225 kVAR

Total load : S = S1 + S 2 = (P1 + P2 ) + j (Q1 + Q 2 ) = P + jQ P = 2000 + 300 = 2300 kW Q = 1239.5 + 225 = 1464.5 kVAR

The minimum operating pf for a 2300 kW load and not exceeding the kVA rating of the generator is P 2300 cos θ = = = 0.9775 S1 2352.94

or

θ = 12.177°

The maximum load kVAR for this condition is Q m = S1 sin θ = 2352.94 sin(12.177°) Q m = 496.313 kVAR The capacitor must supply the difference between the total load kVAR ( i.e. Q ) and the permissible generator kVAR ( i.e. Q m ). Thus, Q c = Q − Q m = 968.2 kVAR

Chapter 11, Solution 91 P = S cos θ

pf = cos θ =

P 2700 = = 0.8182 S (220)(15)

Q = S sin θ = 220(15) sin(35.09°) = 1897.3 When the power is raised to unity pf, θ1 = 0° and Q c = Q = 1897.3 C=

Qc 1897.3 = = 104 µF 2 ω Vrms (2π )(60)(220) 2

Chapter 11, Solution 92 (a)

Apparent power drawn by the motor is P 60 Sm = = = 80 kVA cos θ 0.75

Q m = S 2 − P 2 = (80) 2 − (60) 2 = 52.915 kVAR Total real power P = Pm + Pc + PL = 60 + 0 + 20 = 80 kW Total reactive power Q = Q m + Q c + Q L = 52.915 − 20 + 0 = 32.91 kVAR Total apparent power S = P 2 + Q 2 = 86.51 kVA (b)

pf =

(c)

I=

P 80 = = 0.9248 S 86.51

S 86510 = = 157.3 A V 550

Chapter 11, Solution 93

(a)

P1 = (5)(0.7457) = 3.7285 kW

S1 =

P1 3.7285 = = 4.661 kVA pf 0.8

Q1 = S1 sin(cos -1 (0.8)) = 2.796 kVAR S 1 = 3.7285 + j2.796 kVA P2 = 1.2 kW , S 2 = 1.2 + j0 kVA

Q 2 = 0 VAR

P3 = (10)(120) = 1.2 kW , S 3 = 1.2 + j0 kVA

Q 3 = 0 VAR

Q 4 = 1.6 kVAR ,

S4 =

cos θ 4 = 0.6  → sin θ 4 = 0.8

Q4 = 2 kVA sin θ 4

P4 = S 4 cos θ 4 = (2)(0.6) = 1.2 kW S 4 = 1.2 − j1.6 kVA

S = S1 + S 2 + S 3 + S 4 S = 7.3285 + j1.196 kVA Total real power = 7.3285 kW Total reactive power = 1.196 kVAR (b)

 1.196   = 9.27° θ = tan -1   7.3285 

pf = cos θ = 0.987

Chapter 11, Solution 94 cos θ1 = 0.7  → θ1 = 45.57° S1 = 1 MVA = 1000 kVA P1 = S1 cos θ1 = 700 kW Q1 = S1 sin θ1 = 714.14 kVAR For improved pf, cos θ 2 = 0.95  → θ 2 = 18.19° P2 = P1 = 700 kW S2 =

P2 700 = = 736.84 kVA cos θ 2 0.95

Q 2 = S 2 sin θ 2 = 230.08 kVAR

P1 = P2 = 700 kW θ1

θ2

Q2 S2

S1 Q1

(a)

Qc

Reactive power across the capacitor Q c = Q1 − Q 2 = 714.14 − 230.08 = 484.06 kVAR Cost of installing capacitors = $30 × 484.06 = $14,521.80

(b)

Substation capacity released = S1 − S 2 = 1000 − 736.84 = 263.16 kVA Saving in cost of substation and distribution facilities = $120 × 263.16 = $31,579.20

(c)

Yes, because (a) is greater than (b). Additional system capacity obtained by using capacitors costs only 46% as much as new substation and distribution facilities.

Chapter 11, Solution 95 (a)

Zs = R s − jXc ZL = R L + jX 2

Source impedance Load impedance

For maximum load transfer Z L = Z *s  → R s = R L , X c = X L Xc = XL or

ω=

 → 1 LC

1 = ωL ωC = 2π f

f=

(b)

1 2π LC

=

1 2π (80 × 10 -3 )(40 × 10 -9 )

Vs2 (4.6) 2 P= = = 529 mW 4 R L (4)(10)

= 2.814 kHz

(since Vs is in rms)

Chapter 11, Solution 96 ZTh

+ −

VTh

(a)

ZL

VTh = 146 V, 300 Hz Z Th = 40 + j8 Ω

Z L = Z *Th = 40 − j8 Ω

(b)

P=

VTh

2

8 R Th

(146) 2 = = 66.61 W (8)(40)

Chapter 11, Solution 97 Z T = (2)(0.1 + j) + (100 + j20) = 100.2 + j22 Ω I=

Vs 240 = Z T 100.2 + j22 2

P = I R L = 100 I

2

(100)(240) 2 = = 547.3 W (100.2) 2 + (22) 2

Chapter 12, Solution 1. (a)

If Vab = 400 , then Van =

400 3

∠ - 30° = 231∠ - 30° V

Vbn = 231∠ - 150° V Vcn = 231∠ - 270° V (b)

For the acb sequence, Vab = Van − Vbn = Vp ∠0° − Vp ∠120°

 1 3 Vab = Vp 1 + − j  = Vp 3∠ - 30° 2   2 i.e. in the acb sequence, Vab lags Van by 30°. Hence, if Vab = 400 , then Van =

400 3

∠30° = 231∠30° V

Vbn = 231∠150° V Vcn = 231∠ - 90° V

Chapter 12, Solution 2. Since phase c lags phase a by 120°, this is an acb sequence. Vbn = 160∠(30° + 120°) = 160∠150° V

Chapter 12, Solution 3. Since Vbn leads Vcn by 120°, this is an abc sequence. Van = 208∠(130° + 120°) = 208∠ 250° V

Chapter 12, Solution 4. Vbc = Vca ∠120° = 208∠140° V Vab = Vbc ∠120° = 208∠260° V

Van =

Vab 3 ∠30°

=

208∠260° 3 ∠30°

= 120∠230° V

Vbn = Van ∠ - 120° = 120∠110° V

Chapter 12, Solution 5. This is an abc phase sequence. Vab = Van 3 ∠30°

or

Van =

Vab 3 ∠30°

=

420∠0° 3 ∠30°

= 242.5∠ - 30° V

Vbn = Van ∠ - 120° = 242.5∠ - 150° V Vcn = Van ∠120° = 242.5∠90° V

Chapter 12, Solution 6. Z Y = 10 + j5 = 11.18∠26.56°

The line currents are Van 220 ∠0° = = 19.68∠ - 26.56° A Ia = Z Y 11.18∠26.56° I b = I a ∠ - 120° = 19.68∠ - 146.56° A I c = I a ∠120° = 19.68∠93.44° A

The line voltages are Vab = 200 3 ∠30° = 381∠30° V Vbc = 381∠ - 90° V Vca = 381∠ - 210° V The load voltages are VAN = I a Z Y = Van = 220∠0° V VBN = Vbn = 220∠ - 120° V VCN = Vcn = 220∠120° V

Chapter 12, Solution 7. This is a balanced Y-Y system.

440∠0° V

+ −

ZY = 6 − j8 Ω

Using the per-phase circuit shown above, 440∠0° Ia = = 44∠53.13° A 6 − j8 I b = I a ∠ - 120° = 44∠ - 66.87° A I c = I a ∠120° = 44∠173.13° A

Chapter 12, Solution 8. VL = 220 V ,

I an =

Vp ZY

=

I L = 6.918 A

Z Y = 16 + j9 Ω

VL 3 ZY

=

220 3 (16 + j9)

= 6.918∠ - 29.36°

Chapter 12, Solution 9.

Ia =

Van 120 ∠0° = 4.8∠ - 36.87° A = Z L + Z Y 20 + j15

I b = I a ∠ - 120° = 4.8∠ - 156.87° A I c = I a ∠120° = 4.8∠83.13° A As a balanced system, I n = 0 A

Chapter 12, Solution 10. Since the neutral line is present, we can solve this problem on a per-phase basis. For phase a, Ia =

Van 220 ∠0° = 6.55∠36.53° = Z A + 2 27 − j20

Ib =

Vbn 220 ∠ - 120° = = 10 ∠ - 120° ZB + 2 22

Ic =

Vcn 220 ∠120° = 16.92 ∠97.38° = ZC + 2 12 + j5

For phase b,

For phase c,

The current in the neutral line is I n = -(I a + I b + I c ) or - In = Ia + Ib + Ic - I n = (5.263 + j3.9) + (-5 − j8.66) + (-2.173 + j16.78) I n = 1.91 − j12.02 = 12.17 ∠ - 81° A

Chapter 12, Solution 11.

Van =

Vbc 3 ∠ - 90°

VBC

=

3 ∠ - 90°

=

220∠10° 3 ∠ - 90°

Van = 127 ∠100° V VAB = VBC ∠120° = 220∠130° V VAC = VBC ∠ - 120° = 220∠ - 110° V If I bB = 30 ∠60° , then I aA = 30∠180° , I AB =

I aA 3 ∠ - 30°

I cC = 30 ∠ - 60° =

30∠180° 3 ∠ - 30°

I BC = 17.32∠90° ,

= 17.32∠210°

I CA = 17.32 ∠ - 30°

I AC = -I CA = 17.32∠150° A

I BC Z = VBC Z=

VBC 220 ∠0° = = 12.7 ∠ - 80° Ω I BC 17.32 ∠90°

Chapter 12, Solution 12. Convert the delta-load to a wye-load and apply per-phase analysis.

Ia 110∠0° V

ZY =

+ −

Z∆ = 20 ∠45° Ω 3

ZY

110∠0° = 5.5∠ - 45° A 20∠45° I b = I a ∠ - 120° = 5.5∠ - 165° A

Ia =

I c = I a ∠120° = 5.5∠75° A

Chapter 12, Solution 13. First we calculate the wye equivalent of the balanced load. ZY = (1/3)Z∆ = 6+j5 Now we only need to calculate the line currents using the wye-wye circuits. 110 = 6.471∠ − 61.93° A 2 + j10 + 6 + j5 110∠ − 120° Ib = = 6.471∠178.07° A 8 + j15 110∠120° Ic = = 6.471∠58.07° A 8 + j15

Ia =

Chapter 12, Solution 14. We apply mesh analysis.

1 + j 2Ω

A

a + 100∠0 o V -

ZL ZL I3

n 100∠120 o V + c

I1

-

100∠120 o V + b

I2

B

C Z L = 12 + j12Ω

1 + j 2Ω

1 + j 2Ω For mesh 1, − 100 + 100∠120 o + I 1 (14 + j16) − (1 + j 2) I 2 − (12 + j12) I 3 = 0 or (14 + j16) I 1 − (1 + j 2) I 2 − (12 + j12) I 3 = 100 + 50 − j86.6 = 150 − j86.6 (1) For mesh 2, 100∠120 o − 100∠ − 120 o − I 1 (1 + j 2) − (12 + j12) I 3 + (14 + j16) I 2 = 0 or − (1 + j 2) I 1 + (14 + j16) I 2 − (12 + j12) I 3 = −50 − j86.6 + 50 − j86.6 = − j173.2 (2) For mesh 3, − (12 + j12) I 1 − (12 + j12) I 2 + (36 + j 36) I 3 = 0 (3) Solving (1) to (3) gives I 1 = −3.161 − j19.3,

I 2 = −10.098 − j16.749,

I aA = I 1 = 19.58∠ − 99.3 A o

I bB = I 2 − I 1 = 7.392∠159.8 o A I cC = − I 2 = 19.56∠58.91o A

I 3 = −4.4197 − j12.016

Chapter 12, Solution 15.

Convert the delta load, Z ∆ , to its equivalent wye load. Z Ye =

Z∆ = 8 − j10 3

Z p = Z Y || Z Ye =

(12 + j5)(8 − j10) = 8.076∠ - 14.68° 20 − j5

Z p = 7.812 − j2.047

Z T = Z p + Z L = 8.812 − j1.047 Z T = 8.874 ∠ - 6.78°

We now use the per-phase equivalent circuit. Vp 210 Ia = , where Vp = Zp + ZL 3 Ia =

210 3 (8.874 ∠ - 6.78°)

= 13.66 ∠6.78°

I L = I a = 13.66 A

Chapter 12, Solution 16. (a)

I CA = - I AC = 10∠(-30° + 180°) = 10∠150° This implies that I AB = 10 ∠30° I BC = 10∠ - 90° I a = I AB 3 ∠ - 30° = 17.32∠0° A

I b = 17.32∠ - 120° A I c = 17.32∠120° A (b)

Z∆ =

VAB 110 ∠0° = = 11∠ - 30° Ω I AB 10 ∠30°

Chapter 12, Solution 17. Convert the ∆-connected load to a Y-connected load and use per-phase analysis.

ZL

Van

ZY = Ia =

Ia

+ −

ZY

Z∆ = 3 + j4 3

Van 120 ∠0° = = 19.931∠ - 48.37° Z Y + Z L (3 + j4) + (1 + j0.5)

But I AB =

I a = I AB 3 ∠ - 30° 19.931∠ - 48.37° 3 ∠ - 30°

= 11.51∠ - 18.37° A

I BC = 11.51∠ - 138.4° A I CA = 11.51∠101.6° A VAB = I AB Z ∆ = (11.51∠ - 18.37°)(15∠53.13°) VAB = 172.6∠34.76° V

VBC = 172.6∠ - 85.24° V VCA = 172.6∠154.8° V

Chapter 12, Solution 18. VAB = Van 3 ∠30° = (440 ∠60°)( 3 ∠30°) = 762.1∠90° Z ∆ = 12 + j9 = 15∠36.87°

I AB =

VAB 762.1∠90° = 50.81∠53.13° A = Z ∆ 15∠36.87°

I BC = I AB ∠ - 120° = 50.81∠ - 66.87° A I CA = I AB ∠120° = 50.81∠173.13° A

Chapter 12, Solution 19. Z ∆ = 30 + j10 = 31.62 ∠18.43°

The phase currents are Vab 173∠0° I AB = = 5.47 ∠ - 18.43° A = Z ∆ 31.62 ∠18.43° I BC = I AB ∠ - 120° = 5.47 ∠ - 138.43° A

I CA = I AB ∠120° = 5.47 ∠101.57° A The line currents are I a = I AB − I CA = I AB 3 ∠ - 30° I a = 5.47 3 ∠ - 48.43° = 9.474∠ - 48.43° A

I b = I a ∠ - 120° = 9.474∠ - 168.43° A I c = I a ∠120° = 9.474∠71.57° A

Chapter 12, Solution 20. Z ∆ = 12 + j9 = 15∠36.87°

The phase currents are

210∠0° = 14∠ - 36.87° A 15∠36.87° = I AB ∠ - 120° = 14∠ - 156.87° A

I AB = I BC

I CA = I AB ∠120° = 14∠83.13° A The line currents are I a = I AB 3 ∠ - 30° = 24.25∠ - 66.87° A I b = I a ∠ - 120° = 24.25∠ - 186.87° A I c = I a ∠120° = 24.25∠53.13° A

Chapter 12, Solution 21.

(a)

I AC =

− 230∠120° − 230∠120° = = 17.96∠ − 98.66° A(rms) 10 + j8 12.806∠38.66°

230∠ − 120 230∠0° − 10 + j8 10 + j8 = 17.96∠ − 158.66° − 17.96∠ − 38.66° = −16.729 − j6.536 − 14.024 + j11.220 = −30.75 + j4.684 = 31.10∠171.34° A

I bB = I BC + I BA = I BC − I AB = (b)

Chapter 12, Solution 22. Convert the ∆-connected source to a Y-connected source. Vp 208 Van = ∠ - 30° = ∠ - 30° = 120 ∠ - 30° 3 3 Convert the ∆-connected load to a Y-connected load. Z (4 + j6)(4 − j5) Z = Z Y || ∆ = (4 + j6) || (4 − j5) = 3 8+ j

Z = 5.723 − j0.2153 ZL

Van

Ia

+ −

Ia =

Z

Van 120 ∠30° = = 15.53∠ - 28.4° A Z L + Z 7.723 − j0.2153

I b = I a ∠ - 120° = 15.53∠ - 148.4° A I c = I a ∠120° = 15.53∠91.6° A

Chapter 12, Solution 23.

(a) I AB =

V AB 208 = Z∆ 25∠60 o

I a = I AB

208 3∠ − 30 o 3∠ − 30 = = 14.411∠ − 90 o o 25∠60 o

I L =| I a |= 14.41 A

 208 3   cos 60 o = 2.596 kW (b) P = P1 + P2 = 3VL I L cosθ = 3 (208)   25 

Chapter 12, Solution 24. Convert both the source and the load to their wye equivalents. Z∆ = 20 ∠30° = 17.32 + j10 ZY = 3

Van =

Vab 3

∠ - 30° = 240.2∠0°

We now use per-phase analysis.

1+jΩ

Van

Ia =

+ −

Ia

20∠30° Ω

Van 240.2 = = 11.24∠ - 31° A (1 + j) + (17.32 + j10) 21.37 ∠31°

I b = I a ∠ - 120° = 11.24∠ - 151° A I c = I a ∠120° = 11.24∠89° A

But I AB =

I a = I AB 3 ∠ - 30° 11.24 ∠ - 31° 3 ∠ - 30°

= 6.489∠ - 1° A

I BC = I AB ∠ - 120° = 6.489∠ - 121° A I CA = I AB ∠120° = 6.489∠119° A

Chapter 12, Solution 25. Convert the delta-connected source to an equivalent wye-connected source and consider the single-phase equivalent. Ia =

where

440 ∠(10° − 30°) 3 ZY

Z Y = 3 + j2 + 10 − j8 = 13 − j6 = 14.32 ∠ - 24°.78° Ia =

440 ∠ - 20° 3 (14.32 ∠ - 24.78°)

= 17.74∠4.78° A

I b = I a ∠ - 120° = 17.74∠ - 115.22° A I c = I a ∠120° = 17.74∠124.78° A

Chapter 12, Solution 26. Transform the source to its wye equivalent. Vp Van = ∠ - 30° = 72.17 ∠ - 30° 3 Now, use the per-phase equivalent circuit. Van , Z = 24 − j15 = 28.3∠ - 32° I aA = Z

I aA =

72.17 ∠ - 30° = 2.55∠ 2° A 28.3∠ - 32°

I bB = I aA ∠ - 120° = 2.55∠ - 118° A I cC = I aA ∠120° = 2.55∠122° A

Chapter 12, Solution 27.

Ia =

Vab ∠ - 30° 3 ZY

=

220∠ - 10° 3 (20 + j15)

I a = 5.081∠ - 46.87° A I b = I a ∠ - 120° = 5.081∠ - 166.87° A I c = I a ∠120° = 5.081∠73.13° A

Chapter 12, Solution 28. Let

Vab = 400∠0° Ia =

Van ∠ - 30° 3 ZY

=

400∠ - 30° 3 (30 ∠ - 60°)

= 7.7 ∠30°

I L = I a = 7.7 A VAN = I a Z Y =

Van 3

∠ - 30° = 230.94∠ - 30°

Vp = VAN = 230.9 V

Chapter 12, Solution 29.

P = 3Vp I p cos θ ,

Vp =

VL 3

IL = Ip

,

P = 3 VL I L cos θ IL =

P 3 VL cos θ

ZY =

Vp Ip

=

=

VL 3 IL

5000 240 3 (0.6)

=

= 20.05 = I p

240 3 (20.05)

= 6.911

cos θ = 0.6  → θ = 53.13° Z Y = 6.911∠ - 53.13° (leading) Z Y = 4.15 − j5.53 Ω

S=

P 5000 = = 8333 pf 0.6

Q = S sin θ = 6667 S = 5000 − j6667 VA

Chapter 12, Solution 30. Since this a balanced system, we can replace it by a per-phase equivalent, as shown below.

+ Vp

-

ZL

S = 3S p =

S=

3V 2 p , Z*p

Vp =

VL 3

V 2L (208) 2 = = 1.4421∠45 o kVA Z * p 30∠ − 45 o

P = S cosθ = 1.02 kW

Chapter 12, Solution 31.

(a)

Pp = 6,000,

cosθ = 0.8,

Sp =

PP = 6 / 0.8 = 7.5 kVA cos θ

Q p = S P sin θ = 4.5 kVAR S = 3S p = 3(6 + j 4.5) = 18 + j13.5 kVA For delta-connected load, Vp = VL= 240 (rms). But S=

3V 2 p Z*p

 →

Z*p =

3V 2 p 3(240) 2 = , S (18 + j13.5) x10 3 6000

(b)

Pp = 3VL I L cosθ

(c )

We find C to bring the power factor to unity

 →

Qc = Q p = 4.5 kVA

IL =

 →

C=

Chapter 12, Solution 32. S = 3 VL I L ∠θ

S = S = 3 VL I L = 50 × 10 3 IL =

5000 3 (440)

= 65.61 A

3 x 240 x0.8

Z P = 6.144 + j 4.608Ω

= 18.04 A

Qc 4500 = = 207.2 µF 2 ωV rms 2πx60 x 240 2

For a Y-connected load, I p = I L = 65.61 ,

Z =

Vp Ip

=

Vp =

VL 3

=

440

= 254.03

254.03 = 3.872 65.61

Z = Z ∠θ ,

θ = cos -1 (0.6) = 53.13°

Z = (3.872)(cos θ + j sin θ) Z = (3.872)(0.6 + j0.8)

Z = 2.323 + j3.098 Ω

Chapter 12, Solution 33.

S = 3 VL I L ∠θ S = S = 3 VL I L For a Y-connected load, VL = 3 Vp IL = Ip , S = 3 Vp I p IL = Ip =

3

S 4800 = = 7.69 A 3 Vp (3)(208)

VL = 3 Vp = 3 × 208 = 360.3 V

Chapter 12, Solution 34.

Vp =

Ia =

VL 3

Vp ZY

=

220 3

200

=

3 (10 − j16)

= 6.73∠58°

I L = I p = 6.73 A S = 3 VL I L ∠θ = 3 × 220 × 6.73∠ - 58° S = 1359 − j2174.8 VA

Chapter 12, Solution 35. (a) This is a balanced three-phase system and we can use per phase equivalent circuit. The delta-connected load is converted to its wye-connected equivalent

Z '' y =

1 Z ∆ = (60 + j 30) / 3 = 20 + j10 3 IL +

230 V -

Z’y

Z y = Z ' y // Z '' y = (40 + j10) //( 20 + j10) = 13.5 + j 5.5

IL =

230 = 14.61 − j 5.953 A 13.5 + j 5.5

(b) S = Vs I * L = 3.361 + j1.368 kVA (c ) pf = P/S = 0.9261

Z’’y

Chapter 12, Solution 36.

S = 1 [0.75 + sin(cos-10.75) ] =0.75 + 0.6614 MVA

(a) (b) S = 3V p I * p

 →

I*p =

S (0.75 + j 0.6614) x10 6 = = 59.52 + j 52.49 3V p 3x 4200

PL =| I p | 2 Rl = (79.36) 2 (4) = 25.19 kW

(c) Vs = VL + I p (4 + j ) = 4.4381 − j 0.21 kV = 4.443∠ - 2.709 o kV

Chapter 12, Solution 37.

S=

P 12 = = 20 pf 0.6

S = S∠θ = 20∠θ = 12 − j16 kVA But

IL =

S = 3 VL I L ∠θ

20 × 10 3 3 × 208

S = 3 Ip

2

= 55.51 A

Zp

For a Y-connected load, I L = I p . Zp =

S

3 IL

2

(12 − j16) × 10 3 = (3)(55.51) 2

Z p = 1.298 − j1.731 Ω

Chapter 12, Solution 38.

As a balanced three-phase system, we can use the per-phase equivalent shown below. Ia =

110∠0° 110∠0° = (1 + j2) + (9 + j12) 10 + j14

Sp =

1 I 2 a

2

ZY =

1 (110) 2 ⋅ ⋅ (9 + j12) 2 (10 2 + 14 2 )

The complex power is 3 (110) 2 S = 3S p = ⋅ ⋅ (9 + j12) 2 296 S = 551.86 + j735.81 VA

Chapter 12, Solution 39.

Consider the system shown below. 5Ω

a 100∠120° c

+

−

+

− + 100∠-120°

−

100∠0°

A

5Ω b

8Ω B

I2

4Ω

-j6 Ω

I1

j3 Ω

I3

C 10 Ω

5Ω

For mesh 1, 100 = (18 − j6) I 1 − 5 I 2 − (8 − j6) I 3

(1)

100 ∠ - 120° = 20 I 2 − 5 I 1 − 10 I 3 20∠ - 120° = - I 1 + 4 I 2 − 2 I 3

(2)

For mesh 2,

For mesh 3, 0 = - (8 − j6) I 1 − 10 I 2 + (22 − j3) I 3

(3)

To eliminate I 2 , start by multiplying (1) by 2, 200 = (36 − j12) I 1 − 10 I 2 − (16 − j12) I 3

(4)

Subtracting (3) from (4), 200 = (44 − j18) I 1 − (38 − j15) I 3

(5)

Multiplying (2) by 5 4 , 25∠ - 120° = -1.25 I 1 + 5 I 2 − 2.5 I 3

(6)

Adding (1) and (6), 87.5 − j21.65 = (16.75 − j6) I 1 − (10.5 − j6) I 3

(7)

In matrix form, (5) and (7) become   44 − j18 - 38 + j15  I 1   200 87.5 − j12.65 = 16.75 − j6 - 10.5 + j6  I   3    

∆ = 192.5 − j26.25 ,

∆ 1 = 900.25 − j935.2 ,

∆ 3 = 110.3 − j1327.6

I1 =

∆ 1 1298.1∠ - 46.09° = 6.682 ∠ - 38.33° = 5.242 − j4.144 = 194.28∠ - 7.76° ∆

I3 =

∆ 3 1332.2∠ - 85.25° = = 6.857∠ - 77.49° = 1.485 − j6.694 ∆ 194.28∠ - 7.76°

We obtain I 2 from (6), 1 1 I 2 = 5∠ - 120° + I 1 + I 3 4 2 I 2 = (-2.5 − j4.33) + (1.3104 − j1.0359) + (0.7425 − j3.347) I 2 = -0.4471 − j8.713

The average power absorbed by the 8-Ω resistor is 2

2

P1 = I 1 − I 3 (8) = 3.756 + j2.551 (8) = 164.89 W The average power absorbed by the 4-Ω resistor is 2

P2 = I 3 (4) = (6.8571) 2 (4) = 188.1 W

The average power absorbed by the 10-Ω resistor is 2 2 P3 = I 2 − I 3 (10) = - 1.9321 − j2.019 (10) = 78.12 W Thus, the total real power absorbed by the load is P = P1 + P2 + P3 = 431.1 W

Chapter 12, Solution 40. Transform the delta-connected load to its wye equivalent. Z∆ ZY = = 7 + j8 3 Using the per-phase equivalent circuit above, 100 ∠0° Ia = = 8.567 ∠ - 46.75° (1 + j0.5) + (7 + j8) For a wye-connected load, I p = I a = I a = 8.567 S = 3 Ip

2

Z p = (3)(8.567) 2 (7 + j8)

P = Re(S) = (3)(8.567) 2 (7) = 1.541 kW

Chapter 12, Solution 41.

S=

P 5 kW = = 6.25 kVA pf 0.8

But IL =

S = 3 VL I L S 3 VL

=

6.25 × 10 3 3 × 400

= 9.021 A

Chapter 12, Solution 42. The load determines the power factor. 40 tan θ = = 1.333  → θ = 53.13° 30

pf = cos θ = 0.6 (leading)  7.2  S = 7.2 − j (0.8) = 7.2 − j9.6 kVA  0.6  S = 3 Ip

But Ip

2

=

2

Zp

S (7.2 − j9.6) × 10 3 = 80 = 3Zp (3)(30 − j40)

I p = 8.944 A I L = I p = 8.944 A VL =

S 3 IL

=

12 × 10 3 3 (8.944)

= 774.6 V

Chapter 12, Solution 43.

S = 3 Ip

2

Zp ,

I p = I L for Y-connected loads

S = (3)(13.66) 2 (7.812 − j2.047) S = 4.373 − j1.145 kVA

Chapter 12, Solution 44.

For a ∆-connected load, Vp = VL , IL = 3 Ip S = 3 VL I L IL =

S 3 VL

=

(12 2 + 5 2 ) × 10 3 3 (240)

= 31.273

At the source, VL' = VL + I L Z L VL' = 240∠0° + (31.273)(1 + j3) VL' = 271.273 + j93.819 VL' = 287.04 V

Also, at the source, S ' = 3VL' I *L

S ' = 3 (271.273 + j93.819)(31.273)  93.819   = 19.078 θ = tan -1   271.273  pf = cos θ = 0.9451

Chapter 12, Solution 45. S = 3 VL I L ∠θ IL =

IL =

S ∠-θ 3 VL

,

(635.6) ∠ - θ 3 × 440

S =

P 450 × 10 3 = = 635.6 kVA pf 0.708

= 834 ∠ - 45° A

At the source, VL = 440 ∠0° + I L (0.5 + j2)

VL VL VL VL

= 440 + (834 ∠ - 45°)(2.062 ∠76°) = 440 + 1719.7 ∠31° = 1914.1 + j885.7 = 2.109∠24.83° V

Chapter 12, Solution 46. For the wye-connected load, IL = Ip , VL = 3 Vp

S = 3 Vp I *p =

S=

VL

2

Z*

3 Vp

2

=

Z*

I p = Vp Z

3 VL

(110) 2 = = 121 W 100

S = 3V I =

S=

2

Z*

For the delta-connected load, Vp = VL , IL = 3 Ip ,

* p p

3

3 Vp Z*

2

=

3 VL

I p = Vp Z 2

Z*

(3)(110) 2 = 363 W 100

This shows that the delta-connected load will deliver three times more average Z power than the wye-connected load. This is also evident from Z Y = ∆ . 3 Chapter 12, Solution 47. pf = 0.8 (lagging)  → θ = cos -1 (0.8) = 36.87° S1 = 250 ∠36.87° = 200 + j150 kVA pf = 0.95 (leading)  → θ = cos -1 (0.95) = -18.19° S 2 = 300 ∠ - 18.19° = 285 − j93.65 kVA

pf = 1.0  → θ = cos -1 (1) = 0° S 3 = 450 kVA S T = S1 + S 2 + S 3 = 935 + j56.35 = 936.7 ∠3.45° kVA S T = 3 VL I L IL =

936.7 × 10 3 3 (13.8 × 10 3 )

= 39.19 A rms

pf = cos θ = cos(3.45°) = 0.9982 (lagging)

Chapter 12, Solution 48. (a) We first convert the delta load to its equivalent wye load, as shown below. A A 18-j12 Ω

ZA

40+j15 Ω

ZB ZC C

B 60 Ω

ZA =

(40 + j15)(18 − j12) = 7.577 − j1.923 118 + j 3

ZB =

60(40 + j15). = 20.52 − j 7.105 118 + j 3

ZC =

60(18 − j12) = 8.992 − j 6.3303 118 + j 3

The system becomes that shown below.

C

B

a

2+j3

A

+ 240<0o 240<120 + c

o

-

I2

ZA I1 o

240<-120 + b

ZB

ZC

2+j3 B

C

2+j3 We apply KVL to the loops. For mesh 1, − 240 + 240∠ − 120 o + I 1 (2Z l + Z A + Z B ) − I 2 ( Z B + Z l ) = 0 or (32.097 + j11.13) I 1 − (22.52 + j10.105) I 2 = 360 + j 207.85 For mesh 2, 240∠120 o − 240∠ − 120 o − I 1 ( Z B + Z l ) + I 2 (2Z l + Z B + Z C ) = 0 or

(1)

− (22.52 + j10.105) I 1 + (33.51 + j 6.775) I 2 = − j 415.69 Solving (1) and (2) gives I 1 = 23.75 − j 5.328, I 2 = 15.165 − j11.89

(2)

I aA = I 1 = 24.34∠ − 12.64 o A,

I bB = I 2 − I 1 = 10.81∠ − 142.6 o A

I cC = − I 2 = 19.27∠141.9 o A (b)

S a = (240∠0 o )(24.34∠12.64 o ) = 5841.6∠12.64 o S b = (240∠ − 120 o )(10.81∠142.6 o ) = 2594.4∠22.6 o S b = (240∠120 o )(19.27∠ − 141.9 o ) = 4624.8∠ − 21.9 o S = S a + S b + S c = 12.386 + j 0.55 kVA = 12.4∠2.54 o kVA

Chapter 12, Solution 49. (a) For the delta-connected load, Z p = 20 + j10Ω, S=

V p = VL = 220 (rms) ,

3V 2 p 3 x 220 2 = = 5808 + j 2904 = 6.943∠26.56 o kVA * (20 − j10) Z p

(b) For the wye-connected load, Z p = 20 + j10Ω, S=

V p = VL / 3 ,

3V 2 p 3 x 220 2 = = 2.164∠26.56 o kVA * 3(20 − j10) Z p

Chapter 12, Solution 50.

S = S 1 + S 2 = 8(0.6 + j 0.8) = 4.8 + j 6.4 kVA, Hence,

S 1 = 3 kVA

S 2 = S − S 1 = 1.8 + j 6.4 kVA But S 2 =

Z*p =

3V 2 p , Z*p

Vp =

VL 3

V *L 240 2 = (1.8 + j 6.4) x10 3 S2

 →

 →

S2 =

.V 2 L Z*p

Z p = 2.346 + j8.34Ω

Chapter 12, Solution 51. Apply mesh analysis to the circuit as shown below. Za

+

150∠120°

−

150∠0°

i1 Zb

− +

n −

+

150∠-120°

i2 Zc

For mesh 1, - 150 + (Z a + Z b ) I 1 − Z b I 2 = 0 150 = (18 + j) I 1 − (12 + j9) I 2

(1)

For mesh 2, - 150 ∠ - 120° + (Z b + Z c ) I 2 − Z b I 1 = 0 150 ∠ - 120° = (27 + j9) I 2 − (12 + j9) I 1 From (1) and (2),   18 + j - 12 − j9 I 1   150 150∠ - 120° =  - 12 − j9 27 + j9  I   2    

∆ = 414 − j27 ,

∆ 1 = 3780.9 + j3583.8 ,

(2)

∆ 2 = 579.9 − j1063.2

I1 =

∆ 1 5209.5∠43.47° = = 12.56 ∠47.2° ∆ 414.88∠ - 3.73°

I2 =

∆ 2 1211.1∠ - 61.39° = = 2.919 ∠ - 57.66° ∆ 414.88∠ - 3.73°

I a = I 1 = 12.56∠47.2° A I b = I 2 − I1 =

Ib =

∆ 2 − ∆ 1 - 3201 − j4647 = ∆ ∆

5642.3∠235.44° = 13.6∠239.17° A 414.88∠ - 3.73°

I c = - I 2 = 2.919∠122.34° A

Chapter 12, Solution 52.

Since the neutral line is present, we can solve this problem on a per-phase basis. Van 120 ∠120° = = 6 ∠60° Ia = 20 ∠60° Z AN Ib =

Vbn 120 ∠0° = = 4 ∠0° 30 ∠0° Z BN

Vcn 120 ∠ - 120° = = 3∠ - 150° 40 ∠30° Z CN

Ic =

Thus, - In - In - In - In

= Ia + Ib + Ic = 6 ∠60° + 4 ∠0° + 3∠ - 150° = (3 + j5.196) + (4) + (-2.598 − j1.5) = 4.405 + j3.696 = 5.75∠40°

I n = 5.75∠ 220° A

Chapter 12, Solution 53.

Vp =

250 3

Since we have the neutral line, we can use per-phase equivalent circuit for each phase. 250∠0° 1 Ia = ⋅ = 3.608∠ - 60° A 40∠60° 3 Ib =

Ic =

250∠ - 120° 3 250∠120° 3

⋅

⋅

1 = 2.406∠ - 75° A 60∠ - 45°

1 = 7.217 ∠120° A 20∠0°

- In = Ia + Ib + Ic - I n = (1.804 − j3.125) + (0.6227 − j2.324) + (-3.609 + j6.25) I n = 1.1823 − j0.801 = 1.428 ∠ - 34.12° A

Chapter 12, Solution 54. Consider the circuit shown below.

Ia

a

Vp∠0°

+ −

IAB Vp∠120° +

−

−

A

50 Ω

Vp∠-120°

j50 Ω

+

c

b

B

C -j50 Ω

VAB = Vab = 100 × 3 ∠30°

I AB =

VAB 100 3 ∠30° = = 3.464∠ 30° A 50 Z AB

I BC =

VBC 100 3 ∠ - 90° = = 3.464∠0° A 50∠ - 90° Z BC

I CA =

VCA 100 3 ∠150° = = 3.464∠60° A 50∠90° Z CA

Chapter 12, Solution 55. Consider the circuit shown below. Ia

a

220∠0°

220∠120° +

+ −

−

c

A

60 + j80

I1

−

220∠-120° + b

100 – j120 Ib

30 + j40

B

C

I2 Ic For mesh 1, 220 ∠ - 120° − 220 ∠0° + (160 − j40) I 1 − (100 − j120) I 2 = 0 11 − 11∠ - 120° = (8 − j2) I 1 − (5 − j6) I 2

(1)

For mesh 2, 220 ∠120° − 220 ∠ - 120° + (130 − j80) I 2 − (100 − j120) I 1 = 0 11∠ - 120° − 11∠120° = - (5 − j6) I 1 + (6.5 − j4) I 2

(2)

From (1) and (2), 16.5 + j9.526  8 − j2 - 5 + j6  I 1   - j19.053  =  - 5 + j6 6.5 - j4  I      2 

∆ = 55 + j15 ,

∆ 1 = 31.04 − j99.35 ,

∆ 2 = 101.55 − j203.8

I1 =

∆ 1 104.08∠ - 72.65° = 1.8257 ∠ - 87.91° = ∆ 57.01∠15.26°

I2 =

∆ 2 227.7 ∠ - 63.51° = = 3.994 ∠ - 78.77° ∆ 57.01∠15.26°

I a = I 1 = 1.8257 ∠ - 87.91°

I b = I 2 − I1 =

∆ 2 − ∆ 1 70.51 − j104.45 = = 2.211∠ - 71.23° ∆ 55 + j15

I c = - I 2 = 3.994∠101.23° SA = Ia

2

Z AN = (1.8257) 2 (60 + j80) = 199.99 + j266.7

SB = Ib

2

Z BN = (2.211) 2 (100 − j120) = 488.9 − j586.6

SC = Ic

2

Z CN = (3.994) 2 (30 + j40) = 478.6 + j638.1

S = S A + S B + S C = 1167.5 + j318.2 VA

Chapter 12, Solution 56. (a)

Consider the circuit below.

a

A

440∠0° + −

440∠120°

+

j10 Ω

I1 b

−

B

− + 440∠-120°

I2

I3

-j5 Ω

20 Ω C

c For mesh 1, 440∠ - 120° − 440∠0° + j10 (I 1 − I 3 ) = 0 I1 − I 3 =

(440)(1.5 + j0.866) = 76.21∠ - 60° j10

(1)

For mesh 2, 440∠120° − 440∠ - 120° + 20 (I 2 − I 3 ) = 0

I3 − I2 =

(440)( j1.732) = j38.1 20

For mesh 3, j10 (I 3 − I 1 ) + 20 (I 3 − I 2 ) − j5 I 3 = 0

(2)

Substituting (1) and (2) into the equation for mesh 3 gives, (440)(-1.5 + j0.866) I3 = = 152.42∠60° j5 From (1), I 1 = I 3 + 76.21∠ - 60° = 114.315 + j66 = 132∠30° From (2), I 2 = I 3 − j38.1 = 76.21 + j93.9 = 120.93∠50.94° I a = I 1 = 132∠30° A I b = I 2 − I 1 = -38.105 + j27.9 = 47.23∠143.8° A I c = - I 2 = 120.9∠230.9° A (b)

2

S AB = I 1 − I 3 ( j10) = j58.08 kVA 2

S BC = I 2 − I 3 (20) = 29.04 kVA 2

S CA = I 3 (-j5) = (152.42) 2 (-j5) = -j116.16 kVA S = S AB + S BC + S CA = 29.04 − j58.08 kVA Real power absorbed = 29.04 kW (c)

Total complex supplied by the source is S = 29.04 − j58.08 kVA

(3)

Chapter 12, Solution 57. We apply mesh analysis to the circuit shown below. Ia + Va -

80 + j 50Ω I1

-

20 + j 30Ω

-

Vc +

Vb

+

I2

60 − j 40Ω

Ib

Ic

(1) (100 + j80) I 1 − (20 + j 30) I 2 = Va − Vb = 165 + j 95.263 − (20 + j 30) I 1 + (80 − j10) I 2 = Vb − Vc = − j190.53 (2) Solving (1) and (2) gives I 1 = 1.8616 − j 0.6084, I 2 = 0.9088 − j1.722 . I a = I 1 = 1.9585∠ − 18.1o A,

I b = I 2 − I 1 = −0.528 − j1.1136 = 1.4656∠ − 130.55 o A

I c = − I 2 = 1.947∠117.8 o A Chapter 12, Solution 58.

The schematic is shown below. IPRINT is inserted in the neutral line to measure the current through the line. In the AC Sweep box, we select Total Ptss = 1, Start Freq. = 0.1592, and End Freq. = 0.1592. After simulation, the output file includes FREQ

IM(V_PRINT4)

IP(V_PRINT4)

1.592 E–01

1.078 E+01

–8.997 E+01

i.e.

In = 10.78∠–89.97° A

Chapter 12, Solution 59.

The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq = 60, and End Freq = 60. After simulation, we obtain an output file which includes

i.e.

FREQ

VM(1)

VP(1)

6.000 E+01

2.206 E+02

–3.456 E+01

FREQ

VM(2)

VP(2)

6.000 E+01

2.141 E+02

–8.149 E+01

FREQ

VM(3)

VP(3)

6.000 E+01

4.991 E+01

–5.059 E+01

VAN = 220.6∠–34.56°, VBN = 214.1∠–81.49°, VCN = 49.91∠–50.59° V

Chapter 12, Solution 60.

The schematic is shown below. IPRINT is inserted to give Io. We select Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box. Upon simulation, the output file includes

from which,

FREQ

IM(V_PRINT4)

IP(V_PRINT4)

1.592 E–01

1.421 E+00

–1.355 E+02

Io = 1.421∠–135.5° A

Chapter 12, Solution 61.

The schematic is shown below. Pseudocomponents IPRINT and PRINT are inserted to measure IaA and VBN. In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. Once the circuit is simulated, we get an output file which includes FREQ VM(2) VP(2) 1.592 E–01

2.308 E+02

–1.334 E+02

FREQ

IM(V_PRINT2)

IP(V_PRINT2)

1.592 E–01

1.115 E+01

3.699 E+01

from which IaA = 11.15∠37° A, VBN = 230.8∠–133.4° V

Chapter 12, Solution 62.

Because of the delta-connected source involved, we follow Example 12.12. In the AC Sweep box, we type Total Pts = 1, Start Freq = 60, and End Freq = 60. After simulation, the output file includes

From which

FREQ

IM(V_PRINT2)

IP(V_PRINT2)

6.000 E+01

5.960 E+00

–9.141 E+01

FREQ

IM(V_PRINT1)

IP(V_PRINT1)

6.000 E+01

7.333 E+07

1.200 E+02

Iab = 7.333x107∠120° A, IbB = 5.96∠–91.41° A

Chapter 12, Solution 63.

Let ω = 1 so that L = X/ω = 20 H, and C =

1 = 0.0333 F ωX

The schematic is shown below..

. When the file is saved and run, we obtain an output file which includes the following: FREQ 1.592E-01 FREQ 1.592E-01

IM(V_PRINT1)IP(V_PRINT1) 1.867E+01

1.589E+02

IM(V_PRINT2)IP(V_PRINT2) 1.238E+01

1.441E+02

From the output file, the required currents are: I aA = 18.67∠158.9 o A, I AC = 12.38∠144.1o A

Chapter 12, Solution 64. We follow Example 12.12. In the AC Sweep box we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation the output file includes FREQ

IM(V_PRINT1)

IP(V_PRINT1)

1.592 E–01

4.710 E+00

7.138 E+01

FREQ

IM(V_PRINT2)

IP(V_PRINT2)

1.592 E–01

6.781 E+07

–1.426 E+02

FREQ

IM(V_PRINT3)

IP(V_PRINT3)

1.592 E–01

3.898 E+00

–5.076 E+00

FREQ

IM(V_PRINT4)

IP(V_PRINT4)

1.592 E–01

3.547 E+00

6.157 E+01

FREQ

IM(V_PRINT5)

IP(V_PRINT5)

1.592 E–01

1.357 E+00

9.781 E+01

FREQ

IM(V_PRINT6)

IP(V_PRINT6)

1.592 E–01

3.831 E+00

–1.649 E+02

from this we obtain IaA = 4.71∠71.38° A, IbB = 6.781∠–142.6° A, IcC = 3.898∠–5.08° A IAB = 3.547∠61.57° A, IAC = 1.357∠97.81° A, IBC = 3.831∠–164.9° A

Chapter 12, Solution 65. Due to the delta-connected source, we follow Example 12.12. We type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. The schematic is shown below. After it is saved and simulated, we obtain an output file which includes

Thus,

FREQ

IM(V_PRINT1)

IP(V_PRINT1)

1.592 E–01

6.581 E+00

9.866 E+01

FREQ

IM(V_PRINT2)

IP(V_PRINT2)

1.592 E–01

1.140 E+01

–1.113 E+02

FREQ

IM(V_PRINT3)

IP(V_PRINT3)

1.592 E–01

6.581 E+00

3.866 E+01

IaA = 6.581∠98.66° A, IbB = 11.4∠–111.3 A, IcC = 6.581∠38.66° A

Chapter 12, Solution 66.

VL

Vp =

(b)

Because the load is unbalanced, we have an unbalanced three-phase system. Assuming an abc sequence,

3

=

208

(a)

3

= 120 V

I1 =

120 ∠0° = 2.5∠0° A 48

I2 =

120∠ - 120° = 3∠ - 120° A 40

I3 =

120∠120° = 2∠120° A 60

  3 3 - I N = I 1 + I 2 + I 3 = 2.5 + (3)  - 0.5 − j  + (2)  - 0.5 + j  2  2   

IN = j

3 = j0.866 = 0.866∠90° A 2

Hence, I1 = 2.5 A ,

(c)

I2 = 3 A ,

I3 = 2 A ,

P1 = I12 R 1 = (2.5) 2 (48) = 300 W P2 = I 22 R 2 = (3) 2 (40) = 360 W

P3 = I 32 R 3 = (2) 2 (60) = 240 W (d)

PT = P1 + P2 + P3 = 900 W

Chapter 12, Solution 67.

(a)

The power to the motor is PT = S cos θ = (260)(0.85) = 221 kW The motor power per phase is 1 Pp = PT = 73.67 kW 3 Hence, the wattmeter readings are as follows: Wa = 73.67 + 24 = 97.67 kW Wb = 73.67 + 15 = 88.67 kW Wc = 73.67 + 9 = 83.67 kW

(b)

The motor load is balanced so that I N = 0 . For the lighting loads, Ia =

24,000 = 200 A 120

Ib =

15,000 = 125 A 120

Ic =

9,000 = 75 A 120

If we let

I N = 0.866 A

I a = I a ∠0° = 200∠0° A I b = 125∠ - 120° A I c = 75∠120° A

Then, - I N = Ia + Ib + Ic   3 3 - I N = 200 + (125) - 0.5 − j  + (75) - 0.5 + j  2  2    - I N = 100 − 86.602 A

I N = 132.3 A Chapter 12, Solution 68. (a)

S = 3 VL I L = 3 (330)(8.4) = 4801 VA

(b)

P = S cos θ  → pf = cos θ = pf =

P S

4500 = 0.9372 4801.24

(c)

For a wye-connected load, I p = I L = 8.4 A

(d)

Vp =

VL 3

=

330 3

= 190.53 V

Chapter 12, Solution 69. S 1 = 1.2(0.8 + j 0.6) = 0.96 + j 0.72 MVA,

S 2 = 2(0.75 − j 0.661) = 1.5 − 1.323 MVA,

S = S 1 + S 2 + S 3 = 3.26 − j 0.603 MVA,

pf =

S3 = 0.8 MVA

3.26 P = = 0.9833 S 3.3153

Qc = P(tan old − tan new ) = 3.26[tan(cos −1 0.9833) − tan(cos −1 0.99) = 0.1379 MVA

1 x0.1379 x10 6 3 C= = 28 mF 2πx60 x6.6 2 x10 6

Chapter 12, Solution 70. PT = P1 + P2 = 1200 − 400 = 800 Q T = P2 − P1 = -400 − 1200 = -1600

tan θ =

Q T - 1600 = = -2  → θ = -63.43° PT 800

pf = cos θ = 0.4472 (leading) Zp =

VL 240 = = 40 IL 6

Z p = 40 ∠ - 63.43° Ω

Chapter 12, Solution 71.

(a)

If Vab = 208∠0° , Vbc = 208∠ - 120° , Vca = 208∠120° , I AB =

Vab 208∠0° = = 10.4 ∠0° 20 Z Ab

I BC =

Vbc 208∠ - 120° = 14.708∠ - 75° = Z BC 10 2 ∠ - 45°

I CA =

Vca 208∠120° = 16 ∠97.38° = Z CA 13∠22.62°

I aA = I AB − I CA = 10.4∠0° − 16∠97.38° I aA = 10.4 + 2.055 − j15.867 I aA = 20.171∠ - 51.87° I cC = I CA − I BC = 16∠97.83° − 14.708∠ - 75° I cC = 30.64 ∠101.03°

P1 = Vab I aA cos(θ Vab − θIaA )

P1 = (208)(20.171) cos(0° + 51.87°) = 2590 W

P2 = Vcb I cC cos(θ Vcb − θ IcC ) But

Vcb = -Vbc = 208∠60°

P2 = (208)(30.64) cos(60° − 101.03°) = 4808 W

(b)

PT = P1 + P2 = 7398.17 W

Q T = 3 (P2 − P1 ) = 3840.25 VAR S T = PT + jQ T = 7398.17 + j3840.25 VA S T = S T = 8335 VA

Chapter 12, Solution 72. From Problem 12.11, VAB = 220 ∠130° V

I aA = 30∠180° A

and

P1 = (220)(30) cos(130° − 180°) = 4242 W

VCB = -VBC = 220∠190° I cC = 30∠ - 60° P2 = (220)(30) cos(190° + 60°) = - 2257 W

Chapter 12, Solution 73. Consider the circuit as shown below.

I1 Ia 240∠-60° V

+ −

Z Z

240∠-120° V

Z

− +

I2

Ib

Ic

Z = 10 + j30 = 31.62∠71.57° Ia =

240∠ - 60° = 7.59∠ - 131.57° 31.62∠71.57°

Ib =

240 ∠ - 120° = 7.59∠ - 191.57° 31.62∠71.57°

I c Z + 240∠ - 60° − 240 ∠ - 120° = 0 Ic =

- 240 = 7.59∠108.43° 31.62∠71.57°

I 1 = I a − I c = 13.146∠ - 101.57° I 2 = I b + I c = 13.146∠138.43° P1 = Re [ V1 I 1* ] = Re [ (240∠ - 60°)(13.146 ∠101.57°) ] = 2360 W P2 = Re [ V2 I *2 ] = Re [ (240 ∠ - 120°)(13.146∠ - 138.43°) ] = - 632.8 W

Chapter 12, Solution 74. Consider the circuit shown below. Z = 60 − j30 Ω

208∠0° V

208∠-60° V

+ −

I1

− +

I2

Z

For mesh 1, 208 = 2 Z I 1 − Z I 2

For mesh 2, - 208∠ - 60° = - Z I 1 + 2 Z I 2

Z

In matrix form,    2 Z - Z  I 1  208  - 208∠ - 60° =  - Z 2 Z  I      2  ∆ = 3Z 2 ,

∆ 1 = (208)(1.5 + j0.866) Z ,

∆ 2 = (208)( j1.732) Z

I1 =

∆ 1 (208)(1.5 + j0.866) = = 1.789∠56.56° ∆ (3)(60 − j30)

I2 =

∆ 2 (208)( j1.732) = = 1.79∠116.56° ∆ (3)(60 − j30)

P1 = Re [ V1 I 1* ] = Re [ (208)(1.789∠ - 56.56°) ] = 208.98 W P2 = Re [ V2 (- I 2 ) * ] = Re [ (208∠ - 60°))(1.79∠63.44°) ] = 371.65 W Chapter 12, Solution 75. (a)

I=

V 12 = = 20 mA R 600

(b)

I=

V 120 = = 200 mA R 600

Chapter 12, Solution 76. If both appliances have the same power rating, P, P I= Vs P For the 120-V appliance, I1 = . 120 P For the 240-V appliance, I2 = . 240  P2 R  2 Power loss = I 2 R =  120 2 P R  240 2 Since

for the 120-V appliance for the 240-V appliance

1 1 , the losses in the 120-V appliance are higher. 2 > 120 240 2

Chapter 12, Solution 77. Pg = PT − Pload − Pline , But

pf = 0.85

PT = 3600 cos θ = 3600 × pf = 3060

Pg = 3060 − 2500 − (3)(80) = 320 W

Chapter 12, Solution 78.

cos θ1 =

51 = 0.85  → θ1 = 31.79° 60

Q1 = S1 sin θ1 = (60)(0.5268) = 31.61 kVAR P2 = P1 = 51 kW

cos θ 2 = 0.95  → θ 2 = 18.19° S2 =

P2 = 53.68 kVA cos θ 2

Q 2 = S 2 sin θ 2 = 16.759 kVAR

Q c = Q1 − Q 2 = 3.61 − 16.759 = 14.851 kVAR For each load, Q c1 = C=

Qc = 4.95 kVAR 3

Q c1 4950 = 67.82 µF 2 = ωV (2π )(60)(440) 2

Chapter 12, Solution 79. Consider the per-phase equivalent circuit below.

Ia

2Ω

a

A + −

Van

ZY = 12 + j5 Ω

n Ia =

N Van 255∠0° = = 17.15∠ - 19.65° A Z Y + 2 14 + j5

Thus,

I b = I a ∠ - 120° = 17.15∠ - 139.65° A I c = I a ∠120° = 17.15∠100.35° A VAN = I a Z Y = (17.15∠ - 19.65°)(13∠22.62°) = 223∠2.97° V Thus,

VBN = VAN ∠ - 120° = 223∠ - 117.63° V VCN = VAN ∠120° = 223∠122.97° V

Chapter 12, Solution 80.

S = S1 + S 2 + S 3 = 6[0.83 + j sin(cos −1 0.83)] + S 2 + 8(0.7071 − j 0.7071) S = 10.6368 − j 2.31 + S 2 kVA

(1)

But S = 3VL I L ∠θ = 3 (208)(84.6)(0.8 + j 0.6) VA = 24.383 + j18.287 kVA

(2)

From (1) and (2), S 2 = 13.746 + j 20.6 = 24.76∠56.28 kVA Thus, the unknown load is 24.76 kVA at 0.5551 pf lagging.

Chapter 12, Solution 81.

pf = 0.8 (leading)  → θ1 = -36.87° S1 = 150 ∠ - 36.87° kVA pf = 1.0  → θ 2 = 0° S 2 = 100 ∠0° kVA pf = 0.6 (lagging)  → θ3 = 53.13°

S 3 = 200∠53.13° kVA S 4 = 80 + j95 kVA

S = S1 + S 2 + S 3 + S 4 S = 420 + j165 = 451.2∠21.45° kVA

S = 3 VL I L IL =

S 3 VL

=

451.2 × 10 3 3 × 480

= 542.7 A

For the line, S L = 3 I 2L Z L = (3)(542.7) 2 (0.02 + j0.05) S L = 17.67 + j44.18 kVA At the source, S T = S + S L = 437.7 + j209.2 S T = 485.1∠25.55° kVA VT =

ST 3 IL

=

485.1 × 10 3 3 × 542.7

= 516 V

Chapter 12, Solution 82.

S 1 = 400(0.8 + j 0.6) = 320 + j 240 kVA,

S2 = 3

V 2p Z*p

For the delta-connected load, V L = V p (2400) 2 S 2 = 3x = 1053.7 + j842.93 kVA 10 − j8

S = S 1 + S 2 = 1.3737 + j1.0829 MVA Let I = I1 + I2 be the total line current. For I1, S1 = 3V p I *1 ,

I *1 =

S1 3VL

=

Vp =

VL 3

(320 + j 240) x10 3 3 (2400)

,

I 1 = 76.98 − j 57.735

For I2, convert the load to wye. I 2 = I p 3∠ − 30 o =

2400 3∠ − 30 o = 273.1 − j 289.76 10 + j8

I = I 1 + I 2 = 350 − j 347.5

Vs = VL + Vline = 2400 + I (3 + j 6) = 5.185 + j1.405 kV

 →

| Vs |= 5.372 kV

Chapter 12, Solution 83. S1 = 120 x746 x0.95(0.707 + j 0.707) = 60.135 + j 60.135 kVA,

S 2 = 80 kVA

S = S1 + S 2 = 140.135 + j 60.135 kVA

But | S |= 3VL I L

 →

IL =

|S| 3VL

=

152.49 x10 3 3 x 480

= 183.42 A

Chapter 12, Solution 84.

We first find the magnitude of the various currents. For the motor, S

IL =

3 VL

=

4000 440 3

= 5.248 A

For the capacitor, Q c 1800 = = 4.091 A IC = VL 440 For the lighting, 440 Vp = = 254 V 3 I Li =

PLi 800 = = 3.15 A Vp 254

Consider the figure below.

Ia a

IC

+ Vab

b

I1

Ib

−

-jXC

I2

Ic c

I3 ILi

In

R

n

If Van = Vp ∠0° ,

Vab = 3 Vp ∠30°

Vcn = Vp ∠120°

IC =

Vab = 4.091∠120° -j X C

I1 =

Vab = 4.091∠(θ + 30°) Z∆

where θ = cos -1 (0.72) = 43.95° I 1 = 5.249 ∠73.95° I 2 = 5.249 ∠ - 46.05°

I 3 = 5.249∠193.95° I Li =

Vcn = 3.15∠120° R

Thus,

I a = I 1 + I C = 5.249∠73.95° + 4.091∠120° I a = 8.608∠93.96° A I b = I 2 − I C = 5.249∠ - 46.05° − 4.091∠120° I b = 9.271∠ - 52.16° A

I c = I 3 + I Li = 5.249∠193.95° + 3.15∠120° I c = 6.827 ∠167.6° A I n = - I Li = 3.15∠ - 60° A

Chapter 12, Solution 85. Let

ZY = R

Vp =

VL 3

=

240 3

= 138.56 V

Vp2 27 P = Vp I p = = 9 kW = 2 R

R=

Vp2

=

P

(138.56) 2 = 2.133 Ω 9000

Z Y = 2.133 Ω

Thus,

Chapter 12, Solution 86.

Consider the circuit shown below. 1Ω a

A + −

120∠0° V rms

I1

24 – j2 Ω

1Ω n

N I2

+ −

120∠0° V rms

15 + j4 Ω

1Ω b

B

For the two meshes, 120 = (26 − j2) I 1 − I 2 120 = (17 + j4) I 2 − I 1

(1) (2)

In matrix form, 120  26 − j2 - 1  I 1  120 =  - 1 17 + j4 I 2    

∆ = 449 + j70 ,

∆ 1 = (120)(18 + j4) ,

∆ 2 = (120)(27 − j2)

I1 =

∆ 1 120 × 18.44 ∠12.53° = 4.87 ∠3.67° = ∆ 454.42 ∠8.86°

I2 =

∆ 2 120 × 27.07 ∠ - 4.24° = = 7.15∠ - 13.1° ∆ 454.42 ∠8.86°

I aA = I 1 = 4.87 ∠3.67° A I bB = - I 2 = 7.15∠166.9° A I nN = I 2 − I 1 =

I nN =

∆ 2 − ∆1 ∆

(120)(9 − j6) = 2.856∠ - 42.55° A 449 + j70

Chapter 12, Solution 87.

L = 50 mH  → jωL = j (2π)(60)(5010 -3 ) = j18.85 Consider the circuit below. 1Ω

115 V

+ −

I1

20 Ω

2Ω 15 + j18.85 Ω 115 V

+ −

I2

30 Ω

1Ω

Applying KVl to the three meshes, we obtain 23 I 1 − 2 I 2 − 20 I 3 = 115 - 2 I 1 + 33 I 2 − 30 I 3 = 115 - 20 I 1 − 30 I 2 + (65 + j18.85) I 3 = 0 In matrix form, - 20  23 - 2   I 1  115  - 2 33  I  = 115 - 30   2   - 20 - 30 65 + j18.85 I 3   0 

∆ = 12,775 + j14,232 , ∆ 2 = (115)(1825 + j471.3) ,

(1) (2) (3)

∆ 1 = (115)(1975 + j659.8) ∆ 3 = (115)(1450)

I1 =

∆ 1 115 × 2082∠18.47° = = 12.52∠ - 29.62° ∆ 19214∠48.09°

I2 =

∆ 2 115 × 1884.9 ∠14.48° = 11.33∠ - 33.61° = ∆ 19124 ∠48.09°

I n = I 2 − I1 =

∆ 2 − ∆ 1 (115)(-150 − j188.5) = = 1.448∠ - 176.6° A ∆ 12,775 + j14,231.75

S 1 = V1 I *1 = (115)(12.52∠ 29.62°) = 1252 + j711.6 VA S 2 = V2 I *2 = (115)(1.33∠33.61°) = 1085 + j721.2 VA

Chapter 13, Solution 1. For coil 1, L1 – M12 + M13 = 6 – 4 + 2 = 4 For coil 2, L2 – M21 – M23 = 8 – 4 – 5 = – 1 For coil 3, L3 + M31 – M32 = 10 + 2 – 5 = 7 LT = 4 – 1 + 7 = 10H or

LT = L1 + L2 + L3 – 2M12 – 2M23 + 2M12 LT = 6 + 8 + 10 = 10H

Chapter 13, Solution 2. L = L1 + L2 + L3 + 2M12 – 2M23 2M31 = 10 + 12 +8 + 2x6 – 2x6 –2x4 = 22H

Chapter 13, Solution 3. L1 + L2 + 2M = 250 mH

(1)

L1 + L2 – 2M = 150 mH

(2)

Adding (1) and (2), 2L1 + 2L2 = 400 mH But,

L1 = 3L2,, or 8L2 + 400,

and L2 = 50 mH

L1 = 3L2 = 150 mH From (2),

150 + 50 – 2M = 150 leads to M = 25 mH k = M/ L1 L 2 = 2.5 / 50x150 = 0.2887

Chapter 13, Solution 4. (a) For the series connection shown in Figure (a), the current I enters each coil from its dotted terminal. Therefore, the mutually induced voltages have the same sign as the self-induced voltages. Thus, Leq = L1 + L2 + 2M Is L1 I1 Vs

L2

I2

+ –

L1

L2

Leq (a) (b)

(b)

For the parallel coil, consider Figure (b). Is = I 1 + I2

and

Zeq = Vs/Is

Applying KVL to each branch gives, Vs = jωL1I1 + jωMI2

(1)

Vs = jωMI1 + jω L2I2

(2)

 Vs   jωL1  V  =  jωM  s 

or

jωM   I1  jωL 2  I 2 

∆ = –ω2L1L2 + ω2M2, ∆1 = jωVs(L2 – M), ∆2 = jωVs(L1 – M) I1 = ∆1/∆, and I2 = ∆2/∆ Is = I1 + I2 = (∆1 + ∆2)/∆ = jω(L1 + L2 – 2M)Vs/( –ω2(L1L2 – M)) Zeq = Vs/Is = jω(L1L2 – M)/[jω(L1 + L2 – 2M)] = jωLeq i.e.,

Leq = (L1L2 – M)/(L1 + L2 – 2M)

Chapter 13, Solution 5. (a) If the coils are connected in series,

L = L1 + L 2 + 2M = 25 + 60 + 2(0.5) 25x 60 = 123.7 mH (b) If they are connected in parallel, L=

L1 L 2 − M 2 25x 60 − 19.36 2 = mH = 24.31 mH L1 + L 2 − 2M 25 + 60 − 2x19.36

Chapter 13, Solution 6. V1 = (R1 + jωL1)I1 – jωMI2 V2 = –jωMI1 + (R2 + jωL2)I2

Chapter 13, Solution 7. Applying KVL to the loop, 20∠30° = I(–j6 + j8 + j12 + 10 – j4x2) = I(10 + j6) where I is the loop current. I = 20∠30°/(10 + j6) Vo = I(j12 + 10 – j4) = I(10 + j8) = 20∠30°(10 + j8)/(10 + j6) = 22∠37.66° V

Chapter 13, Solution 8. Consider the current as shown below.

j2 1Ω

10

+ –

I1

4Ω

j6

+ j4

I2

-j3

Vo

–

For mesh 1, 10 = (1 + j6)I1 + j2I2 For mesh 2,

(1)

0 = (4 + j4 – j3)I2 + j2I1 0 = j2I1 +(4 + j)I2

(2)

In matrix form,

j2 10 1 + j6  0  =  j2 4+   

  I1  j  I 2 

∆ = 2 + j25, and ∆2 = –j20 I2 = ∆2/∆ = –j20/(2 + j25) Vo = –j3I2 = –60/(2 + j25) = 2.392∠94.57°

Chapter 13, Solution 9. Consider the circuit below.

2Ω

8∠30o

+ –

I1

2Ω

j4

j4

I2

-j1

-j2V

+ –

For loop 1, 8∠30° = (2 + j4)I1 – jI2 For loop 2,

((j4 + 2 – j)I2 – jI1 + (–j2) = 0 or

Substituting (2) into (1),

(1)

I1 = (3 – j2)i2 – 2 8∠30° + (2 + j4)2 = (14 + j7)I2

I2 = (10.928 + j12)/(14 + j7) = 1.037∠21.12° Vx = 2I2 = 2.074∠21.12°

(2)

Chapter 13, Solution 10.

Consider the circuit below. jωM jωL

jωL Io

I1

Iin∠0o

I2 1/jωC

M = k L1 L 2 = L2 = L, I1 = Iin∠0°, I2 = Io Io(jωL + R + 1/(jωC)) – jωLIin – (1/(jωC))Iin = 0 Io = j Iin(ωL – 1/(ωC)) /(R + jωL + 1/(jωC)) Chapter 13, Solution 11.

Consider the circuit below. V2

R2

+–

I3 jωL1

R1

V1

+ –

I1

1/jωC

jωM jωL2 I2

For mesh 1, V1 = I1(R1 + 1/(jωC)) – I2(1/jωC)) –R1I3 For mesh 2, 0 = –I1(1/(jωC)) + (jωL1 + jωL2 + (1/(jωC)) – j2ωM)I2 – jωL1I3 + jωMI3 For mesh 3, or

–V2 = –R1I1 – jω(L1 – M)I2 + (R1 + R2 + jωL1)I3 V2 = R1I1 + jω(L1 – M)I2 – (R1 + R2 + jωL1)I3

Chapter 13, Solution 12. Let ω = 1.

j4 j2

+ 1V -

j6

•

j8

I1

j10 I2

•

Applying KVL to the loops, 1 = j8 I 1 + j 4 I 2 (1) 0 = j 4 I 1 + j18 I 2 (2) Solving (1) and (2) gives I1 = -j0.1406. Thus

Z=

1 = jLeq I1

 →

Leq =

1 = 7.111 H jI 1

We can also use the equivalent T-section for the transform to find the equivalent inductance.

Chapter 13, Solution 13.

We replace the coupled inductance with an equivalent T-section and use series and parallel combinations to calculate Z. Assuming that ω = 1, La = L1 − M = 18 − 10 = 8, Lb = L2 − M = 20 − 10 = 10, The equivalent circuit is shown below:

Lc = M = 10

12 Ω

j8 Ω

j10 Ω

2Ω

j10 Ω -j6 Ω Z

j4 Ω

Z=12 +j8 + j14//(2 + j4) = 13.195 + j11.244Ω Chapter 13, Solution 14. To obtain VTh, convert the current source to a voltage source as shown below.

j2 5Ω

j6 Ω

j8 Ω

-j3 Ω

2Ω

a j10 V

+ –

+ VTh

I

–

b Note that the two coils are connected series aiding. ωL = ωL1 + ωL2 – 2ωM jωL = j6 + j8 – j4 = j10 Thus,

–j10 + (5 + j10 – j3 + 2)I + 8 = 0 I = (– 8 + j10)/ (7 + j7)

But,

–j10 + (5 + j6)I – j2I + VTh = 0

8V

+ –

VTh = j10 – (5 + j4)I = j10 – (5 + j4)(–8 + j10)/(7 + j7) VTh = 5.349∠34.11° To obtain ZTh, we set all the sources to zero and insert a 1-A current source at the terminals a–b as shown below.

j2 5Ω

j6 Ω

I1

a

j8 Ω

-j3 Ω

+ Vo

1A

2Ω

I2

–

b Clearly, we now have only a super mesh to analyze. (5 + j6)I1 – j2I2 + (2 + j8 – j3)I2 – j2I1 = 0 (5 + j4)I1 + (2 + j3)I2 = 0

(1)

But,

I2 – I1 = 1 or I2 = I1 – 1

(2)

Substituting (2) into (1),

(5 + j4)I1 +(2 + j3)(1 + I1) = 0 I1 = –(2 + j3)/(7 + j7)

Now,

((5 + j6)I1 – j2I1 + Vo = 0 Vo = –(5 + j4)I1 = (5 + j4)(2 + j3)/(7 + j7) = (–2 + j23)/(7 + j7) = 2.332∠50° ZTh = Vo/1 = 2.332∠50° ohms

Chapter 13, Solution 15. To obtain IN, short-circuit a–b as shown in Figure (a). 20 Ω

j20 Ω

20 Ω

a

j5 + –

I1

j20 Ω j5

j10 Ω

j10 Ω

60∠30o

1

+ –

IN I1

I2

I2

b (a)

For mesh 1,

a

b (b)

60∠30° = (20 + j10)I1 + j5I2 – j10I2 or For mesh 2,

12∠30° = (4 + j2)I1 – jI2

(1)

0 = (j20 + j10)I2 – j5I1 – j10I1 or

I1 = 2I2

(2)

12∠30° = (8 + j3)I2

Substituting (2) into (1),

IN = I2 = 12∠30°/(8 + j3) = 1.404∠9.44° A To find ZN, we set all the sources to zero and insert a 1-volt voltage source at terminals a– b as shown in Figure (b). For mesh 1,

1 = I1(j10 + j20 – j5x2) + j5I2 1 = j20I1 + j5I2

For mesh 2,

(3)

0 = (20 + j10)I2 + j5I1 – j10I1 = (4 + j2)I2 – jI1 or

Substituting (4) into (3),

I2 = jI1/(4 + j2) 1 = j20I1 + j(j5)I1/(4 + j2) = (–1 + j20.5)I1 I1 = 1/(–1 + j20.5)

ZN = 1/I1 = (–1 + j20.5) ohms

(4)

Chapter 13, Solution 16.

To find IN, we short-circuit a-b. 8Ω

jΩ

-j2 Ω

a

• • j4 Ω

+ 80∠0 V o

j6 Ω

I2

IN

I1

b

− 80 + (8 − j 2 + j 4) I 1 − jI 2 = 0  → j 6 I 2 − jI 1 = 0  → I1 = 6I 2

(8 + j 2) I 1 − jI 2 = 80

(1) (2)

Solving (1) and (2) leads to 80 IN = I2 = = 1.584 − j 0.362 = 1.6246∠ − 12.91o A 48 + j11 To find ZN, insert a 1-A current source at terminals a-b. Transforming the current source to voltage source gives the circuit below. 8Ω

jΩ

-j2 Ω

2Ω a

• • j4 Ω

+ j6 Ω

I1

2V

I2

b

0 = (8 + j 2) I 1 − jI 2

 →

I1 =

jI 2 8 + j2

(3)

2 + (2 + j 6) I 2 − jI 1 = 0 (4) Solving (3) and (4) leads to I2 = -0.1055 +j0.2975, Vab=-j6I2 = 1.7853 +0.6332 ZN =

Vab = 1.894∠19.53o Ω 1

Chapter 13, Solution 17.

Z = -j6 // Zo

where

Z o = j20 +

Z=

144 = 0.5213 + j15.7 j30 − j2 + j5 + 4

− j6 xZ o = 0.1989 − j9.7Ω − j6 + Z o

Chapter 13, Solution 18.

Let ω = 1.

L1 = 5, L2 = 20, M = k L1 L2 = 0.5 x10 = 5

We replace the transformer by its equivalent T-section. La = L1 − (− M ) = 5 + 5 = 10,

Lb = L1 + M = 20 + 5 = 25,

Lc = − M = −5

We find ZTh using the circuit below. -j4

j10

j25

j2

-j5 ZTh

4+j6

Z Th = j 27 + (4 + j ) //( j 6) = j 27 +

j 6(4 + j ) = 2.215 + j 29.12Ω 4 + j7

We find VTh by looking at the circuit below.

-j4

j10

j25

j2

+ -j5 + VTh

o

120<0

4+j6

-

-

VTh =

4+ j (120) = 61.37∠ − 46.22 o V 4 + j + j6

Chapter 13, Solution 19.

Let ω = 1.

La = L1 − (− M ) = 40 + 25 = 65 H

Lb = L2 + M = 30 + 25 = 55 H,

L C = − M = −25

Thus, the T-section is as shown below. j65 Ω

j55 Ω

-j25 Ω

Chapter 13, Solution 20.

Transform the current source to a voltage source as shown below. k=0.5 4Ω

j10

8Ω

j10 I3

+ –

j12

I1

-j5

I2

20∠0o

+ –

k = M/ L1 L 2 or M = k L1 L 2 ωM = k ωL1ωL 2 = 0.5(10) = 5 For mesh 1,

j12 = (4 + j10 – j5)I1 + j5I2 + j5I2 = (4 + j5)I1 + j10I2

For mesh 2,

0 = 20 + (8 + j10 – j5)I2 + j5I1 + j5I1 –20 = +j10I1 + (8 + j5)I2

From (1) and (2),

(1)

 j12 4 + j5 + j10   I1   20  =  + j10 8 + j5 I   2    

∆ = 107 + j60, ∆1 = –60 –j296, ∆2 = 40 – j100 I1 = ∆1/∆ = 2.462∠72.18° A I2 = ∆2/∆ = 0.878∠–97.48° A I3 = I1 – I2 = 3.329∠74.89° A i1 = 2.462 cos(1000t + 72.18°) A i2 = 0.878 cos(1000t – 97.48°) A

(2)

At t = 2 ms, 1000t = 2 rad = 114.6° i1 = 0.9736cos(114.6° + 143.09°) = –2.445 i2 = 2.53cos(114.6° + 153.61°) = –0.8391 The total energy stored in the coupled coils is w = 0.5L1i12 + 0.5L2i22 – Mi1i2 Since ωL1 = 10 and ω = 1000, L1 = L2 = 10 mH, M = 0.5L1 = 5mH w = 0.5(10)(–2.445)2 + 0.5(10)(–0.8391)2 – 5(–2.445)(–0.8391) w = 43.67 mJ Chapter 13, Solution 21.

For mesh 1, 36∠30° = (7 + j6)I1 – (2 + j)I2 For mesh 2,

(1)

0 = (6 + j3 – j4)I2 – 2I1 jI1 = –(2 + j)I1 + (6 – j)I2

Placing (1) and (2) into matrix form,

(2)

36∠30°  7 + j6 − 2 − j  I1   0  = − 2 − j 6 − j  I   2    

∆ = 48 + j35 = 59.41∠36.1°, ∆1 = (6 – j)36∠30° = 219∠20.54° ∆2 = (2 + j)36∠30° = 80.5∠56.56°, I1 = ∆1/∆ = 3.69∠–15.56°, I2 = ∆2/∆ = 1.355∠20.46° Power absorbed fy the 4-ohm resistor, = 0.5(I2)24 = 2(1.355)2 = 3.672 watts

Chapter 13, Solution 22. With more complex mutually coupled circuits, it may be easier to show the effects of the coupling as sources in terms of currents that enter or leave the dot side of the coil. Figure 13.85 then becomes, -j50 Io I3 j20Ic + −

j40

j10Ib

j60

+ −

Ia

j30Ic

− + − +

Ix

j30Ib

− + j20Ia

50∠0° V + −

j80

I1

I2

100 Ω

Ib − +

j10Ia

Note the following, Ia = I 1 – I3 Ib = I2 – I1 Ic = I 3 – I2 and

Io = I 3

Now all we need to do is to write the mesh equations and to solve for Io. Loop # 1, -50 + j20(I3 – I2) j 40(I1 – I3) + j10(I2 – I1) – j30(I3 – I2) + j80(I1 – I2) – j10(I1 – I3) = 0 j100I1 – j60I2 – j40I3 = 50 Multiplying everything by (1/j10) yields 10I1 – 6I2 – 4I3 = - j5

(1)

Loop # 2, j10(I1 – I3) + j80(I2–I1) + j30(I3–I2) – j30(I2 – I1) + j60(I2 – I3) – j20(I1 – I3) + 100I2 = 0 -j60I1 + (100 + j80)I2 – j20I3 = 0

(2)

Loop # 3, -j50I3 +j20(I1 –I3) +j60(I3 –I2) +j30(I2 –I1) –j10(I2 –I1) +j40(I3 –I1) –j20(I3 –I2) = 0 -j40I1 – j20I2 + j10I3 = 0 Multiplying by (1/j10) yields,

-4I1 – 2I2 + I3 = 0

(3)

Multiplying (2) by (1/j20) yields -3I1 + (4 – j5)I2 – I3 = 0 Multiplying (3) by (1/4) yields

(4)

-I1 – 0.5I2 – 0.25I3 = 0

(5)

Multiplying (4) by (-1/3) yields I1 – ((4/3) – j(5/3))I2 + (1/3)I3 = -j0.5 (7) Multiplying [(6)+(5)] by 12 yields

(-22 + j20)I2 + 7I3 = 0

(8)

Multiplying [(5)+(7)] by 20 yields

-22I2 – 3I3 = -j10

(9)

(8) leads to I2 = -7I3/(-22 + j20) = 0.2355∠42.3o = (0.17418+j0.15849)I3 (9) leads to I3 = (j10 – 22I2)/3, substituting (1) into this equation produces, I3 = j3.333 + (-1.2273 – j1.1623)I3 I3 = Io = 1.3040∠63o amp.

or

Chapter 13, Solution 23. ω = 10 0.5 H converts to jωL1 = j5 ohms 1 H converts to jωL2 = j10 ohms 0.2 H converts to jωM = j2 ohms 25 mF converts to 1/(jωC) = 1/(10x25x10-3) = –j4 ohms The frequency-domain equivalent circuit is shown below.

j2 j5

12∠0°

+ −

I1

j10

–j4

I2

5Ω

(10)

For mesh 1,

12 = (j5 – j4)I1 + j2I2 – (–j4)I2 –j2 = I1 + 6I2

For mesh 2,

0 = (5 + j10)I2 + j2I1 –(–j4)I1 0 = (5 + j10)I2 + j6I1

From (1),

(1)

(2)

I1 = –j12 – 6I2

Substituting this into (2) produces, I2 = 72/(–5 + j26) = 2.7194∠–100.89° I1 = –j12 – 6 I2 = –j12 – 163.17∠–100.89 = 5.068∠52.54° Hence,

i1 = 5.068cos(10t + 52.54°) A, i2 = 2.719cos(10t – 100.89°) A. 10t = 10x15x10-3 0.15 rad = 8.59°

At t = 15 ms,

i1 = 5.068cos(61.13°) = 2.446 i2 = 2.719cos(–92.3°) = –0.1089 w = 0.5(5)(2.446)2 + 0.5(1)(–0.1089)2 – (0.2)(2.446)(–0.1089) = 15.02 J

Chapter 13, Solution 24.

(a)

k = M/ L1 L 2 = 1/ 4 x 2 = 0.3535

(b)

ω = 4 1/4 F leads to 1/(jωC) = –j/(4x0.25) = –j 1||(–j) = –j/(1 – j) = 0.5(1 – j) 1 H produces jωM = j4 4 H produces j16 2 H becomes j8

j4 2Ω j8 12∠0°

+ −

I1

I2

0.5(1–j)

j16

12 = (2 + j16)I1 + j4I2 or

6 = (1 + j8)I1 + j2I2

0 = (j8 + 0.5 – j0.5)I2 + j4I1 or I1 = (0.5 + j7.5)I2/(–j4)

(1) (2)

Substituting (2) into (1), 24 = (–11.5 – j51.5)I2 or I2 = –24/(11.5 + j51.5) = –0.455∠–77.41° Vo = I2(0.5)(1 – j) = 0.3217∠57.59° vo = 321.7cos(4t + 57.6°) mV (c)

From (2),

At t = 2s,

I1 = (0.5 + j7.5)I2/(–j4) = 0.855∠–81.21°

i1 = 0.885cos(4t – 81.21°) A, i2 = –0.455cos(4t – 77.41°) A 4t = 8 rad = 98.37° i1 = 0.885cos(98.37° – 81.21°) = 0.8169 i2 = –0.455cos(98.37° – 77.41°) = –0.4249 w = 0.5L1i12 + 0.5L2i22 + Mi1i2

= 0.5(4)(0.8169)2 + 0.5(2)(–.4249)2 + (1)(0.1869)(–0.4249) = 1.168 J

Chapter 13, Solution 25.

m = k L1 L 2 = 0.5 H We transform the circuit to frequency domain as shown below. 12sin2t converts to 12∠0°, ω = 2 0.5 F converts to 1/(jωC) = –j 2 H becomes jωL = j4 j1 Io 4 Ω

1Ω

a

2Ω

–j1

12∠0°

+ −

j2

j2

j4

10 Ω b Applying the concept of reflected impedance, Zab = (2 – j)||(1 + j2 + (1)2/(j2 + 3 + j4)) = (2 – j)||(1 + j2 + (3/45) – j6/45) = (2 – j)||(1 + j2 + (3/45) – j6/45) = (2 – j)||(1.0667 + j1.8667) =(2 – j)(1.0667 + j1.8667)/(3.0667 + j0.8667) = 1.5085∠17.9° ohms Io = 12∠0°/(Zab + 4) = 12/(5.4355 + j0.4636) = 2.2∠–4.88° io = 2.2sin(2t – 4.88°) A

Chapter 13, Solution 26.

M = k L1L 2 ωM = k ωL1ωL 2 = 0.6 20x 40 = 17 The frequency-domain equivalent circuit is shown below. j17

50 Ω

200∠60°

–j30

I1

+ −

Io

j20

j40

I2

For mesh 1, 200∠60° = (50 – j30 + j20)I1 + j17I2 = (50 – j10)I1 + j17I2

10 Ω

(1)

For mesh 2, 0 = (10 + j40)I2 + j17I1 In matrix form,

(2)

j17   I1  200∠60° 50 − j10 =   0 10 + j40 I 2     j17

∆ = 900 + j100, ∆1 = 2000∠60°(1 + j4) = 8246.2∠136°, ∆2 = 3400∠–30° I2 = ∆2/∆ = 3.755∠–36.34° Io = I2 = 3.755∠–36.34° A Switching the dot on the winding on the right only reverses the direction of Io. This can be seen by looking at the resulting value of ∆2 which now becomes 3400∠150°. Thus, Io = 3.755∠143.66° A

Chapter 13, Solution 27. Zin = –j4 + j5 + 9/(12 + j6) = 0.6 + j.07 = 0.922∠49.4° I1 = 12∠0°/0.922∠49.4° = 13∠–49.4° A Chapter 13, Solution 28. We find ZTh by replacing the 20-ohm load with a unit source as shown below. j10 Ω

8Ω

-jX

• • j12 Ω

j15 Ω

I2

+ 1V -

I1

For mesh 1,

0 = (8 − jX + j12) I 1 − j10 I 2

For mesh 2,

1 + j15I 2 − j10 I 1 = 0

(1)

 →

I 1 = 1.5I 2 − 0.1 j

(2)

Substituting (2) into (1) leads to − 1.2 + j 0.8 + 0.1X I2 = 12 + j8 − j1.5 X

Z Th =

| Z Th |= 20 =

1 12 + j8 − j1.5 X = − I 2 1.2 − j 0.8 − 0.1X

12 2 + (8 − 1.5 X ) 2 (1.2 − 0.1X ) + 0.8 2

2

 →

Solving the quadratic equation yields X = 6.425

0 = 1.75 X 2 + 72 X − 624

Chapter 13, Solution 29.

30 mH becomes jωL = j30x10-3x103 = j30 50 mH becomes j50 Let X = ωM Using the concept of reflected impedance, Zin = 10 + j30 + X2/(20 + j50) I1 = V/Zin = 165/(10 + j30 + X2/(20 + j50)) p = 0.5|I1|2(10) = 320 leads to |I1|2 = 64 or |I1| = 8 8 = |165(20 + j50)/(X2 + (10 + j30)(20 + j50))| = |165(20 + j50)/(X2 – 1300 + j1100)| 64 = 27225(400 + 2500)/((X2 – 1300)2 + 1,210,000)

or

(X2 – 1300)2 + 1,210,000 = 1,233,633 X = 33.86 or 38.13 If X = 38.127 = ωM M = 38.127 mH k = M/ L1 L 2 = 38.127/ 30x50 = 0.984 j38.127

10 Ω

165∠0°

+ −

I1

j30

j50

I2

20 Ω

165 = (10 + j30)I1 – j38.127I2

(1)

0 = (20 + j50)I2 – j38.127I1

(2) 165  10 + j30 − j38.127  I1   0  =  − j38.127 20 + j50   I      2 

In matrix form,

∆ = 154 + j1100 = 1110.73∠82.03°, ∆1 = 888.5∠68.2°, ∆2 = j6291 I1 = ∆1/∆ = 8∠–13.81°, I2 = ∆2/∆ = 5.664∠7.97° i1 = 8cos(1000t – 13.83°), i2 = 5.664cos(1000t + 7.97°) At t = 1.5 ms, 1000t = 1.5 rad = 85.94° i1 = 8cos(85.94° – 13.83°) = 2.457 i2 = 5.664cos(85.94° + 7.97°) = –0.3862 w = 0.5L1i12 + 0.5L2i22 + Mi1i2 = 0.5(30)(2.547)2 + 0.5(50)(–0.3862)2 – 38.127(2.547)(–0.3862) = 130.51 mJ Chapter 13, Solution 30.

(a)

Zin = j40 + 25 + j30 + (10)2/(8 + j20 – j6) = 25 + j70 + 100/(8 + j14) = (28.08 + j64.62) ohms

(b)

jωLa = j30 – j10 = j20, jωLb = j20 – j10 = j10, jωLc = j10

Thus the Thevenin Equivalent of the linear transformer is shown below. j40

25 Ω

j20

j10

j10

8Ω –j6

Zin

Zin = j40 + 25 + j20 + j10||(8 + j4) = 25 + j60 + j10(8 + j4)/(8 + j14) = (28.08 + j64.62) ohms

Chapter 13, Solution 31. (a)

La = L1 – M = 10 H Lb = L2 – M = 15 H Lc = M = 5 H

(b)

L1L2 – M2 = 300 – 25 = 275 LA = (L1L2 – M2)/(L1 – M) = 275/15 = 18.33 H LB = (L1L2 – M2)/(L1 – M) = 275/10 = 27.5 H LC = (L1L2 – M2)/M = 275/5 = 55 H

Chapter 13, Solution 32. We first find Zin for the second stage using the concept of reflected impedance.

Lb

LB

R

Zin’ Zin’ = jωLb + ω2Mb2/(R + jωLb) = (jωLbR - ω2Lb2 + ω2Mb2)/(R + jωLb)

(1)

For the first stage, we have the circuit below.

La

LA

Zin’

Zin Zin = jωLa + ω2Ma2/(jωLa + Zin) = (–ω2La2 + ω2Ma2 + jωLaZin)/( jωLa + Zin)

(2)

Substituting (1) into (2) gives, ( jωL b R − ω 2 L2b + ω 2 M 2b ) − ω L + ω M + jωL a R + jω L b = 2 2 jωL b R − ω L b + ω 2 M 2b jωL a + R + jω L b 2

=

2 a

2

2 a

–Rω2La2 + ω2Ma2R – jω3LbLa + jω3LbMa2 + jωLa(jωLbR – ω2Lb2 + ω2Mb2) jωRLa –ω2LaLb + jωLbR – ω2La2 + ω2Mb2 ω2R(La2 + LaLb – Ma2) + jω3(La2Lb + LaLb2 – LaMb2 – LbMa2)

Zin =

ω2(LaLb +Lb2 – Mb2) – jωR(La +Lb)

Chapter 13, Solution 33. Zin = 10 + j12 + (15)2/(20 + j 40 – j5) = 10 + j12 + 225/(20 + j35) = 10 + j12 + 225(20 – j35)/(400 + 1225) = (12.769 + j7.154) ohms Chapter 13, Solution 34. Insert a 1-V voltage source at the input as shown below. j6 Ω

1Ω

•

+

j12 Ω

o

1<0 V -

8Ω •

j10 Ω

I1

j4 Ω I2

-j2 Ω

For loop 1, 1 = (1 + j10) I 1 − j 4 I 2

(1)

For loop 2, 0 = (8 + j 4 + j10 − j 2) I 2 + j 2 I 1 − j 6 I 1

 →

0 = − jI 1 + (2 + j 3) I 2

(2)

Solving (1) and (2) leads to I1=0.019 –j0.1068

Z=

1 = 1.6154 + j 9.077 = 9.219∠79.91o Ω I1

Alternatively, an easier way to obtain Z is to replace the transformer with its equivalent T circuit and use series/parallel impedance combinations. This leads to exactly the same result. Chapter 13, Solution 35.

For mesh 1,

16 = (10 + j 4) I 1 + j 2 I 2

(1)

For mesh 2,

0 = j 2 I 1 + (30 + j 26) I 2 − j12 I 3

(2)

For mesh 3,

0 = − j12 I 2 + (5 + j11) I 3

(3)

We may use MATLAB to solve (1) to (3) and obtain I 1 = 1.3736 − j 0.5385 = 1.4754∠ − 21.41o A I 2 = −0.0547 − j 0.0549 = 0.0775∠ − 134.85 o A I 3 = −0.0268 − j 0.0721 = 0.077∠ − 110.41o A Chapter 13, Solution 36.

Following the two rules in section 13.5, we obtain the following: (a)

V2/V1 = –n,

I2/I1 = –1/n

(b)

V2/V1 = –n,

I2/I1 = –1/n

(c)

V2/V1 = n,

I2/I1 = 1/n

(d)

V2/V1 = n,

I2/I1 = –1/n

(n = V2/V1)

Chapter 13, Solution 37.

(a) n =

V2 2400 = =5 V1 480

(b) S1 = I 1V1 = S 2 = I 2V2 = 50,000 (c ) I 2 =

 →

I1 =

50,000 = 104.17 A 480

50,000 = 20.83 A 2400

Chapter 13, Solution 38.

Zin = Zp + ZL/n2, n = v2/v1 = 230/2300 = 0.1 v2 = 230 V, s2 = v2I2* I2* = s2/v2 = 17.391∠–53.13° or I2 = 17.391∠53.13° A ZL = v2/I2 = 230∠0°/17.391∠53.13° = 13.235∠–53.13° Zin = 2∠10° + 1323.5∠–53.13° = 1.97 + j0.3473 + 794.1 – j1058.8 Zin = 1.324∠–53.05° kohms Chapter 13, Solution 39.

Referred to the high-voltage side, ZL = (1200/240)2(0.8∠10°) = 20∠10° Zin = 60∠–30° + 20∠10° = 76.4122∠–20.31° I1 = 1200/Zin = 1200/76.4122∠–20.31° = 15.7∠20.31° A Since S = I1v1 = I2v2, I2 = I1v1/v2 = (1200/240)( 15.7∠20.31°) = 78.5∠20.31° A

Chapter 13, Solution 40.

n=

N2 500 1 = = , N1 2000 4

P=

V 2 60 2 = = 300 W R 12

n=

V2 V1

→

V2 = nV1 =

1 (240) = 60 V 4

Chapter 13, Solution 41. We reflect the 2-ohm resistor to the primary side. Zin = 10 + 2/n2,

n = –1/3

Since both I1 and I2 enter the dotted terminals,

Zin = 10 + 18 = 28 ohms

I1 = 14∠0°/28 = 0.5 A and I2 = I1/n = 0.5/(–1/3) = –1.5 A

Chapter 13, Solution 42. 10 Ω + • V1

+ 120<0o V -

I1

-j50 Ω

1:4

-

• + V2

+ 20 Ω

-

Vo -

I2

Applying mesh analysis, 120 = 10I1 + V1

(1)

0 = (20 − j50)I 2 + V2

(2)

At the terminals of the transformer, V2 =n=4 V1

→

I2 1 1 =− =− I1 n 4

V2 = 4V1

 →

(3)

I1 = −4I 2

(4)

Substituting (3) and (4) into (1) gives 120 = −40I 2 + 0.25V2

(5)

Solving (2) and (5) yields I 2 = −2.4756 − j0.6877

Vo = −20I 2 = 51.39∠15.52 o V Chapter 13, Solution 43.

Transform the two current sources to voltage sources, as shown below. 10 Ω + 20 V

+ –

Using mesh analysis,

I1

12 Ω

1:4 v1

−

+ v2

−

I2

12V

+ –

–20 + 10I1 + v1 = 0 20 = v1 + 10I1

12 + 12I2 – v2 = 0 or 12 = v2 – 12I2 At the transformer terminal, v2 = nv1 = 4v1 I1 = nI2 = 4I2

(1) (2) (3) (4)

Substituting (3) and (4) into (1) and (2), we get,

Solving (5) and (6) gives

20 = v1 + 40I2

(5)

12 = 4v1 – 12I2

(6)

v1 = 4.186 V and v2 = 4v = 16.744 V

Chapter 13, Solution 44.

We can apply the superposition theorem. Let i1 = i1’ + i1” and i2 = i2’ + i2” where the single prime is due to the DC source and the double prime is due to the AC source. Since we are looking for the steady-state values of i1 and i2, i1’ = i2’ = 0. For the AC source, consider the circuit below. R

1:n +

i1”

+

v1

v2

−

−

v2/v1 = –n,

+ –

i2”

Vn∠0°

I2”/I1” = –1/n

But v2 = vm, v1 = –vm/n or I1” = vm/(Rn) I2” = –I1”/n = –vm/(Rn2) Hence,

i1(t) = (vm/Rn)cosωt A, and i2(t) = (–vm/(n2R))cosωt A

Chapter 13, Solution 45. 48 Ω

4∠–90˚

ZL = 8 −

+ −

j = 8 − j4 , n = 1/3 ωC

Z

Z=

ZL

= 9 Z L = 72 − j36 n2 4∠ − 90° 4∠ − 90° I= = = 0.03193∠ − 73.3° 48 + 72 − j36 125.28∠ − 16.7°

We now have some choices, we can go ahead and calculate the current in the second loop and calculate the power delivered to the 8-ohm resistor directly or we can merely say that the power delivered to the equivalent resistor in the primary side must be the same as the power delivered to the 8-ohm resistor. Therefore, P8Ω =

I2 72 = 0.5098x10 − 3 72 = 36.71 mW 2

The student is encouraged to calculate the current in the secondary and calculate the power delivered to the 8-ohm resistor to verify that the above is correct. Chapter 13, Solution 46.

(a)

Reflecting the secondary circuit to the primary, we have the circuit shown below. Zin

16∠60°

+ −

I1

+ −

10∠30°/(–n) = –5∠30°

Zin = 10 + j16 + (1/4)(12 – j8) = 13 + j14 –16∠60° + ZinI1 – 5∠30° = 0 or I1 = (16∠60° + 5∠30°)/(13 + j14) Hence, (b)

I1 = 1.072∠5.88° A, and I2 = –0.5I1 = 0.536∠185.88° A

Switching a dot will not effect Zin but will effect I1 and I2. I1 = (16∠60° – 5∠30°)/(13 + j14) = 0.625 ∠25 A and I2 = 0.5I1 = 0.3125∠25° A

Chapter 13, Solution 47.

0.02 F becomes 1/(jωC) = 1/(j5x0.02) = –j10 We apply mesh analysis to the circuit shown below. –j10

I3 10 Ω

3:1 +

10∠0°

+ –

I1

v1

−

+ v2

−

+ I2

vo

2Ω

−

For mesh 1,

10 = 10I1 – 10I3 + v1

(1)

For mesh 2,

v2 = 2I2 = vo

(2)

0 = (10 – j10)I3 – 10I1 + v2 – v1

(3)

v2 = nv1 = v1/3

(4)

I1 = nI2 = I2/3

(5)

From (2) and (4),

v1 = 6I2

(6)

Substituting this into (1),

10 = 10I1 – 10I3

(7)

For mesh 3, At the terminals,

Substituting (4) and (6) into (3) yields From (5), (7), and (8)

0 = –10I1 – 4I2 + 10(1 – j)I3 0   I1   0  − 0.333  1  10 6 − 10  I 2  = 10  − 10 10 − j10  I 3   0  −4

(8)

I2 =

∆2 100 − j100 = 1.482∠32.9° = ∆ − 20 − j93.33 vo = 2I2 = 2.963∠32.9° V

(a)

Switching the dot on the secondary side effects only equations (4) and (5).

From (2) and (9),

v2 = –v1/3

(9)

I1 = –I2/3

(10)

v1 = –6I2

Substituting this into (1), 10 = 10I1 – 10I3 – 6I2 = (23 – j5)I1

(11)

Substituting (9) and (10) into (3), From (10) to (12), we get

0 = –10I1 + 4I2 + 10(1 – j)I3

(12)

0.333 0   I1   0   1  10 −6 − 10  I 2  = 10  − 10 4 10 − j10  I 3   0 

I2 =

∆2 100 − j100 = 1.482∠–147.1° = ∆ − 20 + j93.33 vo = 2I2 = 2.963∠–147.1° V

Chapter 13, Solution 48.

We apply mesh analysis.

8Ω

10 Ω

2:1 +

+ •

+ o

100∠0 V -

I1

V1

•

V2 -

Ix

j6 Ω I2

-j4 Ω

100 = (8 − j 4) I 1 − j 4 I 2 + V1

(1)

0 = (10 + j 2) I 2 − j 4 I 1 + V 2

(2)

V2 1 =n= 2 V1

(3)

But

I2 1 = − = −2 I1 n

 →

V1 = 2V2

 →

I 1 = −0.5 I 2

(4)

Substituting (3) and (4) into (1) and (2), we obtain 100 = (−4 − j 2) I 2 + 2V2 0 = (10 + j 4) I 2 +V2

(1)a (2)a

Solving (1)a and (2)a leads to I2 = -3.5503 +j1.4793

I x = I 1 + I 2 = 0.5 I 2 = 1.923∠157.4 o A Chapter 13, Solution 49.

ω = 2,

1 F 20

 →

1 = − j10 jω C -j10

Ix

2Ω

I1

1:3

I2

1 + 12<0o V -

2 + • V1 -

+ V2 •

-

6Ω

At node 1, 12 − V1 V1 − V2 = + I1  → 12 = 2 I 1 + V1 (1 + j 0.2) − j 0.2V2 2 − j10 At node 2, V − V2 V2 I2 + 1 =  → 0 = 6 I 2 + j 0.6V1 − (1 + j 0.6)V2 − j10 6

At the terminals of the transformer, V2 = −3V1 ,

(1)

(2)

1 I 2 = − I1 3

Substituting these in (1) and (2), 12 = −6 I 2 + V1 (1 + j 0.8),

0 = 6 I 2 + V1 (3 + j 2.4)

Adding these gives V1=1.829 –j1.463 and Ix =

V1 − V2 4V1 = = 0.937∠51.34 o − j10 − j10 i x = 0.937 cos(2t + 51.34 o ) A

Chapter 13, Solution 50.

The value of Zin is not effected by the location of the dots since n2 is involved. Zin’ = (6 – j10)/(n’)2, n’ = 1/4 Zin’ = 16(6 – j10) = 96 – j160 Zin = 8 + j12 + (Zin’ + 24)/n2, n = 5 Zin = 8 + j12 + (120 – j160)/25 = 8 + j12 + 4.8 – j6.4 Zin = (12.8 + j5.6) ohms Chapter 13, Solution 51.

Let Z3 = 36 +j18, where Z3 is reflected to the middle circuit. ZR’ = ZL/n2 = (12 + j2)/4 = 3 + j0.5 Zin = 5 – j2 + ZR’ = (8 – j1.5) ohms I1 = 24∠0°/ZTh = 24∠0°/(8 – j1.5) = 24∠0°/8.14∠–10.62° = 8.95∠10.62° A

Chapter 13, Solution 52. For maximum power transfer, 40 = ZL/n2 = 10/n2 or n2 = 10/40 which yields n = 1/2 = 0.5 I = 120/(40 + 40) = 3/2 p = I2R = (9/4)x40 = 90 watts. Chapter 13, Solution 53. (a)

The Thevenin equivalent to the left of the transformer is shown below. 8Ω

20 V

+ −

The reflected load impedance is ZL’ = ZL/n2 = 200/n2. 8 = 200/n2 produces n = 5.

For maximum power transfer, (b)

If n = 10, ZL’ = 200/10 = 2 and I = 20/(8 + 2) = 2 p = I2ZL’ = (2)2(2) = 8 watts.

Chapter 13, Solution 54. (a)

ZTh

VS

+ −

I1

ZL/n2

For maximum power transfer, ZTh = ZL/n2, or n2 = ZL/ZTh = 8/128 n = 0.25 (b)

I1 = VTh/(ZTh + ZL/n2) = 10/(128 + 128) = 39.06 mA

(c)

v2 = I2ZL = 156.24x8 mV = 1.25 V But

v2 = nv1 therefore v1 = v2/n = 4(1.25) = 5 V

Chapter 13, Solution 55. We reflect Zs to the primary side. ZR = (500 – j200)/n2 = 5 – j2, Zin = Zp + ZR = 3 + j4 + 5 – j2 = 8 + j2 I1 = 120∠0°/(8 + j2) = 14.552∠–14.04° Zp

Vp

+ −

1:n

I1

I2

Zs

Since both currents enter the dotted terminals as shown above, I2 = –(1/n)I1 = –1.4552∠–14.04° = 1.4552∠166° S2 = |I2|2Zs = (1.4552)(500 – j200) P2 = Re(S2) = (1.4552)2(500) = 1054 watts

Chapter 13, Solution 56. We apply mesh analysis to the circuit as shown below. 2Ω

1:2 +

+

v1 46V

+ −

I1

v2

−

−

I2

10 Ω

5Ω

For mesh 1,

46 = 7I1 – 5I2 + v1

(1)

For mesh 2,

v2 = 15I2 – 5I1

(2)

At the terminals of the transformer, v2 = nv1 = 2v1

(3)

I1 = nI2 = 2I2

(4)

Substituting (3) and (4) into (1) and (2),

Combining (5) and (6),

46 = 9I2 + v1

(5)

v1 = 2.5I2

(6)

46 = 11.5I2 or I2 = 4

P10 = 0.5I22(10) = 80 watts.

Chapter 13, Solution 57. (a)

ZL = j3||(12 – j6) = j3(12 – j6)/(12 – j3) = (12 + j54)/17 Reflecting this to the primary side gives Zin = 2 + ZL/n2 = 2 + (3 + j13.5)/17 = 2.3168∠20.04° I1 = vs/Zin = 60∠90°/2.3168∠20.04° = 25.9∠69.96° A(rms) I2 = I1/n = 12.95∠69.96° A(rms)

(b)

60∠90° = 2I1 + v1 or v1 = j60 –2I1 = j60 – 51.8∠69.96° v1 = 21.06∠147.44° V(rms) v2 = nv1 = 42.12∠147.44° V(rms) vo = v2 = 42.12∠147.44° V(rms)

(c)

S = vsI1* = (60∠90°)(25.9∠–69.96°) = 1554∠20.04° VA

Chapter 13, Solution 58. Consider the circuit below. 20 Ω

I3 20 Ω

1:5 +

80∠0°

+ –

I1

+

v1 −

v2 −

+ I2

vo

100 Ω

−

For mesh1,

80 = 20I1 – 20I3 + v1

(1)

For mesh 2,

v2 = 100I2

(2)

For mesh 3,

0 = 40I3 – 20I1 which leads to I1 = 2I3

(3)

At the transformer terminals, v2 = –nv1 = –5v1

(4)

I1 = –nI2 = –5I2 From (2) and (4),

(5)

–5v1 = 100I2 or v1 = –20I2

(6)

Substituting (3), (5), and (6) into (1), 4

= I1 – I2 – I 3 = I1 – (I1/(–5)) – I1/2 = (7/10)I1

I1 = 40/7, I2 = –8/7, I3 = 20/7 p20(the one between 1 and 3) = 0.5(20)(I1 – I3)2 = 10(20/7)2 = 81.63 watts p20(at the top of the circuit) = 0.5(20)I32 = 81.63 watts p100 = 0.5(100)I22 = 65.31 watts

Chapter 13, Solution 59. We apply nodal analysis to the circuit below. 2Ω

I3 8Ω

v1 I1 2 : 1 +

20∠0°

+ –

v1 −

I2 + v2 −

v2

4Ω

20 = 8I1 + V1

(1)

V1 = 2I3 + V2

(2)

V2 = 4I2

(3)

At the transformer terminals,

v2 = 0.5v1

(4)

I1 = 0.5I2

(5)

Solving (1) to (5) gives I1 = 0.833 A, I2 = 1.667 A, I3 = 3.333 A V1 = 13.33 V, V2 = 6.667 V. P8Ω = 0.5(8)|(20 – V1)/8|2 = 2.778 W P2Ω = 0.5(2)I32 = 11.11 W, P4Ω = 0.5V22/4 = 5.556 W Chapter 13, Solution 60. (a)

Transferring the 40-ohm load to the middle circuit, ZL’ = 40/(n’)2 = 10 ohms where n’ = 2 10||(5 + 10) = 6 ohms We transfer this to the primary side. Zin = 4 + 6/n2 = 4 + 96 = 100 ohms, where n = 0.25 I1 = 120/100 = 1.2 A and I2 = I1/n = 4.8 A 4Ω

1:4

I1 +

120∠0°

+ –

v1 −

I2

5Ω

I2 ’

+ v2

10 Ω

10 Ω

−

Using current division, I2’ = (10/25)I2 = 1.92 and I3 = I2’/n’ = 0.96 A (b)

p = 0.5(I3)2(40) = 18.432 watts

Chapter 13, Solution 61. We reflect the 160-ohm load to the middle circuit. ZR = ZL/n2 = 160/(4/3)2 = 90 ohms, where n = 4/3 2Ω

1:5

I1 +

24∠0°

+ –

v1 −

Io

14 Ω

Io ’

+ vo

60 Ω

90 Ω

−

14 + 60||90 = 14 + 36 = 50 ohms We reflect this to the primary side. ZR’ = ZL’/(n’)2 = 50/52 = 2 ohms when n’ = 5 I1 = 24/(2 + 2) = 6A 24 = 2I1 + v1 or v1 = 24 – 2I1 = 12 V vo = –nv1 = –60 V, Io = –I1 /n1 = –6/5 = –1.2 Io‘ = [60/(60 + 90)]Io = –0.48A I2 = –Io’/n = 0.48/(4/3) = 0.36 A Chapter 13, Solution 62. (a)

Reflect the load to the middle circuit. ZL’ = 8 – j20 + (18 + j45)/32 = 10 – j15 We now reflect this to the primary circuit so that Zin = 6 + j4 + (10 – j15)/n2 = 7.6 + j1.6 = 7.767∠11.89°, where n = 5/2 = 2.5 I1 = 40/Zin = 40/7.767∠11.89° = 5.15∠–11.89° S = 0.5vsI1* = (20∠0°)(5.15∠11.89°) = 103∠11.89° VA

(b)

I2 = –I1/n,

n = 2.5

I3 = –I2/n’,

n = 3

I3 = I1/(nn’) = 5.15∠–11.89°/(2.5x3) = 0.6867∠–11.89° p = 0.5|I2|2(18) = 9(0.6867)2 = 4.244 watts Chapter 13, Solution 63. Reflecting the (9 + j18)-ohm load to the middle circuit gives, Zin’ = 7 – j6 + (9 + j18)/(n’)2 = 7 – j6 + 1 + j12 = 8 + j4 when n’ = 3 Reflecting this to the primary side, Zin = 1 + Zin’/n2 = 1 + 2 – j = 3 – j, where n = 2 I1 = 12∠0°/(3 – j) = 12/3.162∠–18.43° = 3.795∠18.43A I2 = I1/n = 1.8975∠18.43° A I3 = –I2/n2 = 632.5∠161.57° mA Chapter 13, Solution 64. We find ZTh at the terminals of Z by considering the circuit below. 10 Ω

1:n +• V1

• + V2

–

+

–

I1

I2 20 Ω

1V –

For mesh 1,

30 I 1 + 20 I 2 + V1 = 0

(1)

For mesh 2,

20 I 1 + 20 I 2 + V2 = 1

(2)

At the terminals,

V2 = nV1 ,

I2 = −

I1 n

Substituting these in (1) and (2) leads to (20 − 30n) I 2 + V1 = 0,

20(1 − n) I 2 + nV1 = 1

Solving these gives

I2 =

1 30n − 40n + 20

 →

2

Z Th =

1 = 30n 2 − 40n + 20 = 7.5 I2

Solving the quadratic equation yields n=0.5 or 0.8333 Chapter 13, Solution 65.

40 Ω

10 Ω

I1

-

I2

+•

+ 200 V (rms)

50 Ω

I2

•

1 +

1:2

+

V1 -

V2 -

V3 -

1:3

I3 •

2 + V4 -

20 Ω

•

At node 1, 200 − V1 V1 − V4 = + I1 10 40

 →

200 = 1.25V1 − 0.25V4 + 10 I 1

(1)

At node 2, V1 − V4 V4 = + I3 40 20

 →

V1 = 3V4 + 40 I 3

(2)

At the terminals of the first transformer, V2 = −2 V1

 →

I2 = −1 / 2 I1

V2 = −2V1

 →

(3)

I 1 = −2 I 2

(4)

For the middle loop, − V2 + 50 I 2 + V3 = 0

 →

V3 = V2 − 50 I 2

(5)

At the terminals of the second transformer, V4 =3 V3 I3 = −1 / 3 I2

 →

V4 = 3V3

 →

(6)

I 2 = −3 I 3

(7)

We have seven equations and seven unknowns. Combining (1) and (2) leads to 200 = 3.5V4 + 10 I 1 + 50 I 3 But from (4) and (7), I 1 = −2 I 2 = −2(−3I 3 ) = 6 I 3 . Hence 200 = 3.5V4 + 110 I 3

(8)

From (5), (6), (3), and (7), V4 = 3(V2 − 50 I 2 ) = 3V2 − 150 I 2 = −6V1 + 450 I 3

Substituting for V1 in (2) gives

V4 = −6(3V4 + 40 I 3 ) + 450 I 3

 →

I3 =

19 V4 210

(9)

Substituting (9) into (8) yields 200 = 13.452V4

 →

V4 = 14.87 V 24 P= = 11.05 W 20

Chapter 13, Solution 66.

v1 = 420 V

(1)

v2 = 120I2

(2)

v1/v2 = 1/4 or v2 = 4v1

(3)

I1/I2 = 4 or I1 = 4 I2

(4)

Combining (2) and (4), v2 = 120[(1/4)I1] = 30 I1 4v1 = 30I1 4(420) = 1680 = 30I1 or I1 = 56 A Chapter 13, Solution 67.

(a)

V1 N 1 + N 2 1 = = V2 N2 0 .4

(b)

S 2 = I 2V2 = 5,000

(c )

S 2 = S1 = I 1V1 = 5,000

 →

 →

V2 = 0.4V1 = 0.4 x 400 = 160 V

I2 =

 →

5000 = 31.25 A 160 I2 =

5000 = 12.5 A 400

Chapter 13, Solution 68.

This is a step-up transformer.

I2 + N1

2 – j6

I1

10 + j40 +

20∠30°

v2

N2 −

v1

+ −

−

For the primary circuit,

20∠30° = (2 – j6)I1 + v1

(1)

For the secondary circuit,

v2 = (10 + j40)I2

(2)

At the autotransformer terminals, v1/v2 = N1/(N1 + N2) = 200/280 = 5/7,

Also,

thus v2 = 7v1/5

(3)

I1/I2 = 7/5 or I2 = 5I1/7

(4)

Substituting (3) and (4) into (2),

v1 = (10 + j40)25I1/49

Substituting that into (1) gives

20∠30° = (7.102 + j14.408)I1

I1 = 20∠30°/16.063∠63.76° = 1.245∠–33.76° A I2 = 5I1/7 = 0.8893∠–33.76° A Io = I1 – I2 = [(5/7) – 1]I1 = –2I1/7 = 0.3557∠146.2° A p = |I2|2R = (0.8893)2(10) = 7.51 watts

Chapter 13, Solution 69.

We can find the Thevenin equivalent. I2

+ N2 I1 120∠0°

v2

j125 Ω

75 Ω

+

VTh

N1

v1

+ −

+

−

−

− I1 = I2 = 0 As a step up transformer,

v1/v2 = N1/(N1 + N2) = 600/800 = 3/4

v2 = 4v1/3 = 4(120)/3 = 160∠0° rms = VTh. To find ZTh, connect a 1-V source at the secondary terminals. We now have a step-down transformer. + v1

j125 Ω

75 Ω I2

I1

1∠0° V

+ v2

+ −

−

− v1 = 1V, v2 =I2(75 + j125) But

v1/v2 = (N1 + N2)/N1 = 800/200 which leads to v1 = 4v2 = 1 and v2 = 0.25 I1/I2 = 200/800 = 1/4 which leads to I2 = 4I1

Hence

0.25 = 4I1(75 + j125) or I1 = 1/[16(75 + j125) ZTh = 1/I1 = 16(75 + j125) Therefore, ZL = ZTh* = (1.2 – j2) kΩ Since VTh is rms, p = (|VTh|/2)2/RL = (80)2/1200 = 5.333 watts

Chapter 13, Solution 70.

This is a step-down transformer. 30 + j12

I1 + I2

v1 120∠0°

+ −

+ v2

−

−

20 – j40

I1/I2 = N2/(N1 + N2) = 200/1200 = 1/6, or I1 = I2/6

(1)

v1/v2 = (N2 + N2)/N2 = 6, or v1 = 6v2

(2)

For the primary loop,

120 = (30 + j12)I1 + v1

(3)

For the secondary loop,

v2 = (20 – j40)I2

(4)

Substituting (1) and (2) into (3), 120 = (30 + j12)( I2/6) + 6v2 and substituting (4) into this yields 120 = (49 – j38)I2 or I2 = 1.935∠37.79° p = |I2|2(20) = 74.9 watts.

Chapter 13, Solution 71. Zin = V1/I1 But

V1I1 = V2I2, or V2 = I2ZL and I1/I2 = N2/(N1 + N2) V1 = V2I2/I1 = ZL(I2/I1)I2 = ZL(I2/I1)2I1

Hence

V1/I1 = ZL[(N1 + N2)/N2] 2 Zin = [1 + (N1/N2)] 2ZL Chapter 13, Solution 72. (a)

Consider just one phase at a time. 1:n

a

A B

b C c n = VL/ 3VLp = 7200 /(12470 3 ) = 1/3 (b)

The load carried by each transformer is 60/3 = 20 MVA. Hence

ILp = 20 MVA/12.47 k = 1604 A ILs = 20 MVA/7.2 k = 2778 A

(c)

The current in incoming line a, b, c is 3I Lp = 3x1603.85 = 2778 A Current in each outgoing line A, B, C is 2778/(n 3 ) = 4812 A

20MVA Load

Chapter 13, Solution 73. (a)

This is a three-phase ∆-Y transformer.

(b)

VLs = nvLp/ 3 = 450/(3 3 ) = 86.6 V, where n = 1/3 As a Y-Y system, we can use per phase equivalent circuit. Ia = Van/ZY = 86.6∠0°/(8 – j6) = 8.66∠36.87° Ic = Ia∠120° = 8.66∠156.87° A ILp = n 3 ILs I1 = (1/3) 3 (8.66∠36.87°) = 5∠36.87° I2 = I1∠–120° = 5∠–83.13° A

(c)

p = 3|Ia|2(8) = 3(8.66)2(8) = 1.8 kw.

Chapter 13, Solution 74. (a)

This is a ∆-∆ connection.

(b)

The easy way is to consider just one phase. 1:n = 4:1 or n = 1/4 n = V2/V1 which leads to V2 = nV1 = 0.25(2400) = 600 i.e. VLp = 2400 V and VLs = 600 V S = p/cosθ = 120/0.8 kVA = 150 kVA pL = p/3 = 120/3 = 40 kw 4:1 IL

VLp Ipp

ILs Ips

VLs

But

pLs = VpsIps

For the ∆-load,

IL =

Hence,

Ips = 40,000/600 = 66.67 A ILs =

(c)

3 Ips =

3 Ip and VL = Vp

3 x66.67 = 115.48 A

Similarly, for the primary side ppp = VppIpp = pps or Ipp = 40,000/2400 = 16.667 A and

(d)

ILp =

3 Ip = 28.87 A

Since S = 150 kVA therefore Sp = S/3 = 50 kVA

Chapter 13, Solution 75.

(a)

n = VLs/( 3 VLp) 4500/(900 3 ) = 2.887

(b)

S =

3 VLsILs or ILs = 120,000/(900 3 ) = 76.98 A

ILs = ILp/(n 3 ) = 76.98/(2.887 3 ) = 15.395 A

Chapter 13, Solution 76.

(a)

At the load,

VL = 240 V = VAB VAN = VL/ 3 = 138.56 V

Since S =

3 VLIL then IL = 60,000/(240 3 ) = 144.34 A

1:n

0.05 Ω

2640V

j0.1 Ω

A 240V

j0.1 Ω

0.05 Ω

B 0.05 Ω

(b)

j0.1 Ω

C

Balanced Load 60kVA 0.85pf leading

Let VAN = |VAN|∠0° = 138.56∠0° cosθ = pf = 0.85 or θ = 31.79° IAA’ = IL∠θ = 144.34∠31.79° VA’N’ = ZIAA’ + VAN = 138.56∠0° + (0.05 + j0.1)(144.34∠31.79°) = 138.03∠6.69° VLs = VA’N’

(c)

3 = 137.8

3 = 238.7 V

For Y-∆ connections, n =

3 VLs/Vps =

3 x238.7/2640 = 0.1569

fLp = nILs/ 3 = 0.1569x144.34/ 3 = 13.05 A Chapter 13, Solution 77. (a)

This is a single phase transformer.

V1 = 13.2 kV, V2 = 120 V

n = V2/V1 = 120/13,200 = 1/110, therefore n = 110 (b)

P = VI or I = P/V = 100/120 = 0.8333 A I1 = nI2 = 0.8333/110 = 7.576 mA

Chapter 13, Solution 78. The schematic is shown below. k = M / L1 L 2 = 1 / 6 x 3 = 0.2357 In the AC Sweep box, set Total Pts = 1, Start Freq = 0.1592 and End Freq = 0.1592. After simulation, the output file includes

From this,

FREQ

IM(V_PRINT1)

IP(V_PRINT1)

1.592 E–01

4.253 E+00

–8.526 E+00

FREQ

IM(V_PRINT2)

IP(V_PRINT2)

1.592 E–01

1.564 E+00

2.749 E+01

I1 = 4.253∠–8.53° A, I2 = 1.564∠27.49° A

The power absorbed by the 4-ohm resistor = 0.5|I|2R = 0.5(1.564)2x4 = 4.892 watts

Chapter 13, Solution 79. The schematic is shown below. k1 = 15 / 5000 = 0.2121, k2 = 10 / 8000 = 0.1118 In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After the circuit is saved and simulated, the output includes FREQ

IM(V_PRINT1)

IP(V_PRINT1)

1.592 E–01

4.068 E–01

–7.786 E+01

FREQ

IM(V_PRINT2)

IP(V_PRINT2)

1.592 E–01

1.306 E+00

–6.801 E+01

FREQ

IM(V_PRINT3)

IP(V_PRINT3)

1.592 E–01

1.336 E+00

–5.492 E+01

Thus, I1 = 1.306∠–68.01° A, I2 = 406.8∠–77.86° mA, I3 = 1.336∠–54.92° A

Chapter 13, Solution 80. The schematic is shown below. k1 = 10 / 40 x80 = 0.1768, k2 = 20 / 40 x 60 = 0.482 k3 = 30 / 80x 60 = 0.433 In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After the simulation, we obtain the output file which includes

i.e.

FREQ

IM(V_PRINT1)

IP(V_PRINT1)

1.592 E–01

1.304 E+00

6.292 E+01

Io = 1.304∠62.92° A

Chapter 13, Solution 81. The schematic is shown below. k1 = 2 / 4x8 = 0.3535, k2 = 1 / 2 x8 = 0.25 In the AC Sweep box, we let Total Pts = 1, Start Freq = 100, and End Freq = 100. After simulation, the output file includes FREQ 1.000 E+02

IM(V_PRINT1) 1.0448 E–01

IP(V_PRINT1) 1.396 E+01

FREQ 1.000 E+02

IM(V_PRINT2) 2.954 E–02

IP(V_PRINT2) –1.438 E+02

FREQ 1.000 E+02

IM(V_PRINT3) 2.088 E–01

IP(V_PRINT3) 2.440 E+01

i.e.

I1 = 104.5∠13.96° mA, I2 = 29.54∠–143.8° mA, I3 = 208.8∠24.4° mA.

Chapter 13, Solution 82. The schematic is shown below. In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain the output file which includes FREQ 1.592 E–01

IM(V_PRINT1) 1.955 E+01

IP(V_PRINT1) 8.332 E+01

FREQ 1.592 E–01

IM(V_PRINT2) 6.847 E+01

IP(V_PRINT2) 4.640 E+01

FREQ 1.592 E–01

IM(V_PRINT3) 4.434 E–01

IP(V_PRINT3) –9.260 E+01

i.e.

V1 = 19.55∠83.32° V, V2 = 68.47∠46.4° V, Io = 443.4∠–92.6° mA.

Chapter 13, Solution 83. The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, the output file includes FREQ 1.592 E–01

IM(V_PRINT1) 1.080 E+00

IP(V_PRINT1) 3.391 E+01

FREQ 1.592 E–01

VM($N_0001) 1.514 E+01

VP($N_0001) –3.421 E+01

i.e.

iX = 1.08∠33.91° A, Vx = 15.14∠–34.21° V.

Chapter 13, Solution 84. The schematic is shown below. We set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, the output file includes FREQ 1.592 E–01

IM(V_PRINT1) 4.028 E+00

IP(V_PRINT1) –5.238 E+01

FREQ 1.592 E–01

IM(V_PRINT2) 2.019 E+00

IP(V_PRINT2) –5.211 E+01

FREQ 1.592 E–01

IM(V_PRINT3) 1.338 E+00

IP(V_PRINT3) –5.220 E+01

i.e.

I1 = 4.028∠–52.38° A, I2 = 2.019∠–52.11° A, I3 = 1.338∠–52.2° A.

Chapter 13, Solution 85.

Z1

VS

+ −

ZL/n2

For maximum power transfer, Z1 = ZL/n2 or n2 = ZL/Z1 = 8/7200 = 1/900 n = 1/30 = N2/N1. Thus N2 = N1/30 = 3000/30 = 100 turns. Chapter 13, Solution 86. n = N2/N1 = 48/2400 = 1/50 ZTh = ZL/n2 = 3/(1/50)2 = 7.5 kΩ

Chapter 13, Solution 87. ZTh = ZL/n2 or n =

Z L / Z Th = 75 / 300 = 0.5

Chapter 13, Solution 88. n = V2/V1 = I1/I2 or I2 = I1/n = 2.5/0.1 = 25 A p = IV = 25x12.6 = 315 watts Chapter 13, Solution 89. n = V2/V1 = 120/240 = 0.5 S = I1V1 or I1 = S/V1 = 10x103/240 = 41.67 A S = I2V2 or I2 = S/V2 = 104/120 = 83.33 A Chapter 13, Solution 90. (a)

n = V2/V1 = 240/2400 = 0.1

(b)

n = N2/N1 or N2 = nN1 = 0.1(250) = 25 turns

(c)

S = I1V1 or I1 = S/V1 = 4x103/2400 = 1.6667 A S = I2V2 or I2 = S/V2 = 4x104/240 = 16.667 A

Chapter 13, Solution 91. (a)

The kVA rating is S = VI = 25,000x75 = 1875 kVA

(b)

Since S1 = S2 = V2I2 and I2 = 1875x103/240 = 7812 A

Chapter 13, Solution 92. (a)

V2/V1 = N2/N1 = n, V2 = (N2/N1)V1 = (28/1200)4800 = 112 V

(b)

I2 = V2/R = 112/10 = 11.2 A and I1 = nI2, n = 28/1200 I1 = (28/1200)11.2 = 261.3 mA

(c)

p = |I2|2R = (11.2)2(10) = 1254 watts.

Chapter 13, Solution 93. (a) For an input of 110 V, the primary winding must be connected in parallel, with series-aiding on the secondary. The coils must be series-opposing to give 12 V. Thus the connections are shown below.

110 V 12 V

(b) To get 220 V on the primary side, the coils are connected in series, with seriesaiding on the secondary side. The coils must be connected series-aiding to give 50 V. Thus, the connections are shown below.

220 V 50 V

Chapter 13, Solution 94. V2/V1 = 110/440 = 1/4 = I1/I2 There are four ways of hooking up the transformer as an auto-transformer. However it is clear that there are only two outcomes.

V1

V1

V1

V2

V1

V2

(1)

(2)

V2 (3)

V2 (4)

(1) and (2) produce the same results and (3) and (4) also produce the same results. Therefore, we will only consider Figure (1) and (3). (a)

For Figure (3), V1/V2 = 550/V2 = (440 – 110)/440 = 330/440 Thus,

(b)

V2 = 550x440/330 = 733.4 V (not the desired result)

For Figure (1), V1/V2 = 550/V2 = (440 + 110)/440 = 550/440 Thus,

V2 = 550x440/550 = 440 V (the desired result)

Chapter 13, Solution 95. (a)

n = Vs/Vp = 120/7200 = 1/60

(b)

Is = 10x120/144 = 1200/144 S = VpIp = VsIs Ip = VsIs/Vp = (1/60)x1200/144 = 139 mA

Chapter 14, Solution 1.

H (ω) =

Vo R jωRC = = Vi R + 1 jωC 1 + jωRC

H (ω) =

jω ω 0 , 1 + jω ω 0

H = H (ω) =

where ω 0 =

ω ω0 1 + (ω ω0 ) 2

1 RC

φ = ∠H (ω) =

ω π − tan -1   2  ω0 

This is a highpass filter. The frequency response is the same as that for P.P.14.1 except that ω0 = 1 RC . Thus, the sketches of H and φ are shown below. H 1 0.7071

0

ω0 = 1/RC

ω

φ

90° 45° 0

ω0 = 1/RC

ω

Chapter 14, Solution 2.

H (ω) =

R 1 1 = = , R + jωL 1 + jωL R 1 + jω ω 0

H = H (ω) =

where ω0 =

R L

ω φ = ∠H (ω) = - tan -1    ω0 

1 1 + (ω ω0 ) 2

The frequency response is identical to the response in Example 14.1 except that ω0 = R L . Hence the response is shown below. H 1 0.7071

ω0 = R/L

0 φ

ω

ω0 = R/L

0°

ω

-45° -90°

Chapter 14, Solution 3. (a)

The Thevenin impedance across the second capacitor where Vo is taken is R Z Th = R + R || 1 sC = R + 1 + sRC VTh =

Vi 1 sC Vi = R + 1 sC 1 + sRC

ZTh

VTh

+ −

1 sC

+ Vo −

Vo =

(b)

Vi 1 sC ⋅ VTh = Z Th + 1 sC (1 + sRC)(1 + sCZ Th )

H (s) =

Vo 1 1 = = Vi (1 + sCZ Th )(1 + sRC) (1 + sRC)(1 + sRC + sRC (1 + sRC))

H (s) =

1 s R C + 3sRC + 1 2

2

2

RC = (40 × 10 3 )(2 × 10 -6 ) = 80 × 10 -3 = 0.08 There are no zeros and the poles are at - 0.383 s1 = = - 4.787 RC

s2 =

- 2.617 = - 32.712 RC

Chapter 14, Solution 4.

(a)

R ||

1 R = jωC 1 + jωRC

R Vo R 1 + jωRC H (ω) = = = R Vi R + jωL (1 + jωRC) jωL + 1 + jωRC

(b)

H (ω) =

R - ω RLC + R + jωL

H (ω) =

jωC (R + jωL) R + jωL = R + jωL + 1 jωC 1 + jωC (R + jωL)

H (ω) =

- ω 2 LC + jωRC 1 − ω 2 LC + jωRC

2

Chapter 14, Solution 5.

(a)

(b)

H (ω) =

Vo 1 jωC = Vi R + jωL + 1 jωC

H (ω) =

1 1 + jωRC − ω 2 LC

R ||

1 R = jωC 1 + jωRC

H (ω) =

Vo jωL (1 + jωRC) jωL = = Vi jωL + R (1 + jωRC) R + jωL (1 + jωRC)

jωL − ω 2 RLC H (ω) = R + jωL − ω 2 RLC

Chapter 14, Solution 6. (a)

(b)

Using current division, Io R H (ω) = = I i R + jωL + 1 jωC H (ω) =

jω (20)(0.25) jωRC = 2 1 + jωRC − ω LC 1 + jω(20)(0.25) − ω2 (10)(0.25)

H (ω) =

jω5 1 + jω5 − 2.5 ω 2

We apply nodal analysis to the circuit below. Io

Vx

Is

R

1/jωC

0.5 Vx + − jωL

Is =

Vx Vx − 0.5Vx + R jωL + 1 jωC

But

Io =

0.5 Vx jωL + 1 jωC

 → Vx = 2 I o ( jωL + 1 jωC)

Is 1 0 .5 = + Vx R jωL + 1 jωC Is 1 1 = + 2 I o ( jωL + 1 jωC) R 2 ( jωL + 1 jωC) I s 2 ( jωL + 1 jωC) = +1 Io R Io 1 jωRC = = I s 1 + 2 ( jωL + 1 jωC) R jωRC + 2 (1 − ω 2 LC) jω H (ω) = jω + 2 (1 − ω2 0.25)

H (ω) =

H (ω) =

jω 2 + jω − 0.5 ω 2

Chapter 14, Solution 7. (a)

0.05 = 20 log10 H

2.5 × 10 -3 = log10 H H = 10 2.5×10 = 1.005773 -3

(b)

- 6.2 = 20 log10 H - 0.31 = log10 H H = 10 -0.31 = 0.4898

(c)

104.7 = 20 log10 H 5.235 = log10 H H = 10 5.235 = 1.718 × 10 5

Chapter 14, Solution 8.

(a)

(b)

(c)

H = 0.05 H dB = 20 log10 0.05 = - 26.02 ,

φ = 0°

H = 125 H dB = 20 log10 125 = 41.94 ,

φ = 0°

H(1) =

j10 = 4.472∠63.43° 2+ j

H dB = 20 log10 4.472 = 13.01 , (d)

H(1) =

φ = 63.43°

3 6 + = 3.9 − j1.7 = 4.254∠ - 23.55° 1+ j 2 + j

H dB = 20 log10 4.254 = 12.577 ,

φ = - 23.55°

Chapter 14, Solution 9. H (ω) =

1 (1 + jω)(1 + jω 10)

H dB = -20 log10 1 + jω − 20 log10 1 + jω / 10 φ = - tan -1 (ω) − tan -1 (ω / 10) The magnitude and phase plots are shown below.

HdB 0.1

1

10

ω

100 20 log 10

-20

1 1 + jω / 10 20 log10

-40

1 1 + jω

φ 0.1 -45°

1

10

ω

100 arg

1 1 + jω / 10

-90° -135° -180°

arg

1 1 + jω

Chapter 14, Solution 10.

H( jω) =

50 = jω(5 + jω)

10 jω   1 jω1 +  5  

HdB 40 20 log1

20 10 0.1 -20

1

100

  1 20 log  jω  1+ 5 

 1   20 log   jω 

-40

φ 0.1

     

ω

1 10

-45°

ω

100 arg

1 1 + jω / 5

-90° arg

-135°

1 jω

-180° Chapter 14, Solution 11.

H (ω) =

5 (1 + jω 10) jω (1 + jω 2)

H dB = 20 log10 5 + 20 log10 1 + jω 10 − 20 log10 jω − 20 log10 1 + jω 2 φ = -90° + tan -1 ω 10 − tan -1 ω 2

The magnitude and phase plots are shown below.

HdB 40 34 20 14 0.1 -20

1

10

100

ω

1

10

100

ω

-40

φ 90° 45° 0.1 -45° -90°

Chapter 14, Solution 12.

T ( w) =

0.1(1 + jω ) , jω (1 + jω / 10)

20 log 0.1 = −20

The plots are shown below. |T|

(db)

20

ω

0 0.1

1

10

100

-20 -40

arg T

90o

ω

0 0.1 -90o

1

10

100

Chapter 14, Solution 13.

G (ω) =

(1 10)(1 + jω) 1 + jω = 2 ( jω) (10 + jω) ( jω) 2 (1 + jω 10)

G dB = -20 + 20 log10 1 + jω − 40 log10 jω − 20 log10 1 + jω 10 φ = -180° + tan -1ω − tan -1 ω 10 The magnitude and phase plots are shown below. GdB 40 20 0.1 -20

1

10

100

ω

1

10

100

ω

-40

φ 90° 0.1 -90° -180°

Chapter 14, Solution 14. 50 25

H (ω) =

1 + jω  jω10  jω  2  +    jω1 + 25  5   

H dB = 20 log10 2 + 20 log10 1 + jω − 20 log10 jω − 20 log10 1 + jω2 5 + ( jω 5) 2  ω10 25   φ = -90° + tan -1 ω − tan -1  1 − ω2 5 

The magnitude and phase plots are shown below. HdB 40 26 20 6 0.1 -20

1

10

100

ω

1

10

100

ω

-40 φ 90° 0.1 -90° -180°

Chapter 14, Solution 15. 40 (1 + jω) 2 (1 + jω) = (2 + jω)(10 + jω) (1 + jω 2)(1 + jω 10)

H (ω) =

H dB = 20 log10 2 + 20 log10 1 + jω − 20 log10 1 + jω 2 − 20 log10 1 + jω 10 φ = tan -1 ω − tan -1 ω 2 − tan -1 ω 10 The magnitude and phase plots are shown below. HdB 40 20 6 0.1 -20

1

10

100

ω

1

10

100

ω

-40 φ 90° 45° 0.1 -45° -90°

Chapter 14, Solution 16.

G (ω) =

jω jω   100(1 + jω)1 +   10 

2

GdB 20 0.1

20 log jω

1

10

100

− 40 log

-20 -40

ω

jω 10

20 log(1/100)

-60

φ 90°

arg(jω) ω

0.1 -90° -180°

1 arg

10

100 arg

1 jω   1 +   10 

2

1 1 + jω

Chapter 14, Solution 17.

G (ω) =

(1 4) jω (1 + jω)(1 + jω 2) 2

G dB = -20log10 4 + 20 log10 jω − 20 log10 1 + jω − 40 log10 1 + jω 2 φ = -90° - tan -1ω − 2 tan -1 ω 2 The magnitude and phase plots are shown below.

GdB 20 0.1 -12 -20 -40

1

10

100 ω

φ 90° 0.1

1

10

100

ω

-90° -180° Chapter 14, Solution 18. 4 (1 + jω 2) 2 G (ω) = 50 jω (1 + jω 5)(1 + jω 10)

G dB = 20 log10 4 50 + 40 log10 1 + jω 2 − 20 log10 jω − 20 log10 1 + jω 5 − 20 log10 1 + jω 10 where 20 log10 4 50 = -21.94 φ = -90° + 2 tan -1 ω 2 − tan -1 ω 5 − tan -1 ω 10 The magnitude and phase plots are shown below.

GdB 20 0.1 -20

1

10

100

ω

1

10

100

ω

-40 -60 φ 180° 90° 0.1 -90°

Chapter 14, Solution 19.

H (ω) =

jω 100 (1 + jω 10 − ω2 100)

H dB = 20 log10 jω − 20 log10 100 − 20 log10 1 + jω 10 − ω2 100  ω 10   φ = 90° − tan -1  1 − ω2 100 

The magnitude and phase plots are shown below.

HdB 40 20 0.1 -20

1

10

100

ω

1

10

100

ω

-40 -60 φ 90° 0.1 -90° -180°

Chapter 14, Solution 20. 10 (1 + jω − ω2 ) (1 + jω)(1 + jω 10)

N(ω) =

N dB = 20 − 20 log10 1 + jω − 20 log10 1 + jω 10 + 20 log10 1 + jω − ω2  ω   − tan -1 ω − tan -1 ω 10 φ = tan -1  1 − ω2  The magnitude and phase plots are shown below.

NdB 40 20 0.1 -20

1

10

100

ω

1

10

100

ω

-40 φ 180° 90° 0.1 -90°

Chapter 14, Solution 21.

T(ω) =

jω (1 + jω) 100 (1 + jω 10)(1 + jω 10 − ω2 100)

TdB = 20 log10 jω + 20 log10 1 + jω − 20 log10 100

− 20 log10 1 + jω 10 − 20 log10 1 + jω 10 − ω2 100  ω 10   φ = 90° + tan -1 ω − tan -1 ω 10 − tan -1  1 − ω2 100 

The magnitude and phase plots are shown below. TdB 20 0.1 -20

1

10

100

ω

1

10

100

ω

-40 -60 φ 180° 90° 0.1 -90° -180°

Chapter 14, Solution 22. 20 = 20 log10 k

 → k = 10

A zero of slope + 20 dB / dec at ω = 2  → 1 + jω 2 A pole of slope - 20 dB / dec at ω = 20  →

1 1 + jω 20

A pole of slope - 20 dB / dec at ω = 100  → Hence,

H (ω) =

1 1 + jω 100

10 (1 + jω 2) (1 + jω 20)(1 + jω 100)

10 4 ( 2 + jω) H (ω) = ( 20 + jω)(100 + jω)

Chapter 14, Solution 23. A zero of slope + 20 dB / dec at the origin

 →

A pole of slope - 20 dB / dec at ω = 1  →

1 1 + jω 1

A pole of slope - 40 dB / dec at ω = 10  → Hence,

H (ω) =

jω (1 + jω)(1 + jω 10) 2

H (ω) =

100 jω (1 + jω)(10 + jω) 2

jω

1 (1 + jω 10) 2

Chapter 14, Solution 24. The phase plot is decomposed as shown below. φ 90° arg (1 + jω / 10)

45° 0.1 -45° -90°

1 arg ( jω)

10

100

1000

ω   1  arg  1 + jω / 100 

G (ω) =

k ′ (1 + jω 10) k ′ (10)(10 + jω) = jω (1 + jω 100) jω (100 + jω)

where k ′ is a constant since arg k ′ = 0 .

G (ω) =

Hence,

k (10 + jω) , jω (100 + jω)

where k = 10k ′ is constant

Chapter 14, Solution 25.

ω0 =

1 LC

=

1 (40 × 10 -3 )(1 × 10 -6 )

= 5 krad / s

Z(ω0 ) = R = 2 kΩ  ω0 4   Z(ω0 4) = R + j  L − ω0 C   4

  5 × 10 3 4  ⋅ 40 × 10 -3 − Z(ω0 4) = 2000 + j  (5 × 10 3 )(1 × 10 -6 )   4 Z(ω0 4) = 2000 + j (50 − 4000 5) Z(ω0 4) = 2 − j0.75 kΩ  ω0 2   Z(ω0 2) = R + j  L − ω0 C   2

 (5 × 10 3 )  2  (40 × 10 -3 ) − Z(ω0 2) = 2000 + j  2 (5 × 10 3 )(1 × 10 -6 )   Z(ω0 4) = 2000 + j (100 − 2000 5) Z(ω0 2) = 2 − j0.3 kΩ  1   Z(2ω0 ) = R + j  2ω0 L − 2ω0 C  

  1  Z(2ω0 ) = 2000 + j  (2)(5 × 10 3 )(40 × 10 -3 ) − (2)(5 × 10 3 )(1 × 10 -6 )   Z(2ω0 ) = 2 + j0.3 kΩ  1   Z(4ω0 ) = R + j  4ω0 L − 4ω0 C  

  1  Z(4ω0 ) = 2000 + j  (4)(5 × 10 3 )(40 × 10 -3 ) − 3 -6 (4)(5 × 10 )(1 × 10 )   Z(4ω0 ) = 2 + j0.75 kΩ

Chapter 14, Solution 26.

(a)

fo =

(b)

B=

(c )

Q=

1 2π LC

=

1 2π 5 x10 −9 x10 x10 −3

R 100 = = 10 krad/s L 10 x10 −3

ωo L R

=

L 10 6 10 x10 −3 = = 14.142 3 50 0.1x10 LC R 1

Chapter 14, Solution 27.

At resonance,

Z = R = 10 Ω ,

B=

= 22.51 kHz

R L

ω0 =

and

Q=

1 LC

ω 0 ω0 L = B R

Hence, L=

RQ (10)(80) = = 16 H ω0 50

C=

1 1 = = 25 µF 2 ω0 L (50) 2 (16)

B=

R 10 = = 0.625 rad / s L 16

Therefore, R = 10 Ω ,

L = 16 H ,

C = 25 µF ,

B = 0.625 rad / s

Chapter 14, Solution 28.

Let R = 10 Ω . L=

R 10 = = 0.5 H B 20

C=

1 1 = = 2 µF 2 ω0 L (1000) 2 (0.5)

Q=

ω0 1000 = = 50 B 20

Therefore, if R = 10 Ω then C = 2 µF , L = 0.5 H ,

Q = 50

Chapter 14, Solution 29. jω

Z

Z = jω +

1/jω

1

1 jω + j ω 1 + jω

jω

 1  ω 2 + jω Z = j ω −  +  ω  1 + ω2 Since v( t ) and i( t ) are in phase, 1 ω Im(Z) = 0 = ω − + ω 1 + ω2 ω4 + ω2 − 1 = 0 ω2 =

-1 ± 1+ 4 = 0.618 2

ω = 0.7861 rad / s

Chapter 14, Solution 30. Select R = 10 Ω . L=

R 10 = = 0.05 H = 5 mH ω0 Q (10)(20)

C=

1 1 = = 0.2 F 2 ω0 L (100)(0.05)

B=

1 1 = = 0.5 rad / s RC (10)(0.2)

Therefore, if R = 10 Ω then L = 5 mH , C = 0.2 F ,

B = 0.5 rad / s

Chapter 14, Solution 31.

X L = ωL

B=

→

L=

XL ω

R ωR 2πx10 x10 6 x 5.6 x10 3 = = = 8.796 x10 6 rad/s 3 L XL 40 x10

Chapter 14, Solution 32. Since Q > 10 ,

ω1 = ω0 −

B=

B , 2

ω 2 = ω0 +

B 2

ω0 6 × 10 6 = = 50 krad / s Q 120

ω1 = 6 − 0.025 = 5.975 × 10 6 rad / s ω2 = 6 + 0.025 = 6.025 × 10 6 rad / s

Chapter 14, Solution 33.

Q = ωo RC

Q=

→

R ωo L

→

C=

L=

Q 80 = = 56.84 pF 2πf o R 2πx5.6x10 6 x 40x10 3

R 40 x10 3 = = 14.21 µH 2πf o Q 2πx 5.6 x10 6 x80

Chapter 14, Solution 34.

(a)

ωo =

1 LC

1

= 8x10

−3

x 60x10

−6

= 1.443 krad/s

1 1 = = 3.33 rad/s RC 5x10 3 x 60x10 − 6

(b)

B=

(c)

Q = ωo RC = 1.443x10 3 x 5x10 3 x 60x10 −6 = 432.9

Chapter 14, Solution 35.

At resonance, 1 R

Y=

 → R =

1 1 = = 40 Ω Y 25 × 10 -3

→ C = Q = ω0 RC 

ω0 =

B=

1 LC

 → L =

Q 80 = = 10 µF ω0 R (200 × 10 3 )(40) 1 1 = = 2.5 µH 2 10 ω0 C (4 × 10 )(10 × 10 -6 )

ω0 200 × 10 3 = = 2.5 krad / s Q 80

ω1 = ω0 −

B = 200 − 2.5 = 197.5 krad / s 2

ω1 = ω0 +

B = 200 + 2.5 = 202.5 krad / s 2

Chapter 14, Solution 36.

ω0 =

1 LC

= 5000 rad / s

Y(ω0 ) =

1 R

 → Z(ω0 ) = R = 2 kΩ

Y(ω0 4) =

 ω0 4  1  = 0.5 − j18.75 kS + j C− ω0 L  R  4

Z(ω0 4) =

1 = 1.4212 + j53.3 Ω 0.0005 − j0.01875

Y(ω0 2) =

 ω0 2  1  = 0.5 − j7.5 kS + j C− ω0 L  R  2

Z(ω0 2) =

Y(2ω0 ) =

1 = 8.85 + j132.74 Ω 0.0005 − j0.0075

 1  1  = 0.5 + j7.5 kS + j  2ω0 L − 2ω0 C  R 

Z(2ω0 ) = 8.85 − j132.74 Ω Y(4ω0 ) =

 1  1  = 0.5 + j18.75 kS + j  4ω0 L − 4ω0 C  R 

Z(4ω0 ) = 1.4212 − j53.3 Ω

Chapter 14, Solution 37. 1  L  )  + jωLR  R + j(ωL − 1 C ωC     = Z = jωL //( R + )= 1 1 2 jωC ) R+ + jω L R 2 + ( ωL − jω C ωC jωL(R +

1 ) jω C

1  L  ωL −  ωC  C =0 1 2 2 R + ( ωL − ) ωC

ωLR 2 + Im(Z) =

 →

Thus, ω=

1 LC + R 2 C 2

Chapter 14, Solution 38.

Y

1 R − jωL + jωC = jωC + 2 R + jωL R + ω2 L2

At resonance, Im(Y) = 0 , i.e.

ω 2 ( R 2 C 2 + LC) = 1

ω0 C −

ω0 L =0 R 2 + ω02 L2

R 2 + ω02 L2 =

ω0 =

L C

1 R2 − = LC L2

 50  1   -3 -6 − (40 × 10 )(10 × 10 )  40 × 10 -3 

2

ω0 = 4841 rad / s

Chapter 14, Solution 39.

(a)

B = ω 2 − ω1 = 2π(f 2 − f1 ) = 2π(90 − 86) x10 3 = 8πkrad / s ωo = B=

1 (ω1 + ω 2 ) = 2π(88) x10 3 = 176π 2

1 RC

 → 1

C=

(b)

ωo =

(c )

ωo = 176π = 552.9krad / s

(d)

B = 8π = 25.13krad / s

(e)

Q=

LC

 →

1 1 = = 19.89nF BR 8πx10 3 x 2x10 3

L=

1 ω2 o C

=

1 (176π) 2 x19.89x10 − 9

= 164.4H

ωo 176π = 22 = B 8π

Chapter 14, Solution 40. (a)

L = 5 + 10 = 15 mH

ω0 =

1 LC

1

= 15x10

−3

x 20x10

−6

= 1.8257 k rad/sec

Q = ω0 RC = 1.8257 x10 3 x 25x10 3 x 20x10 −6 = 912.8 1 1 = = 2 rad 3 RC 25x10 20x10 −6

B= (b)

To increase B by 100% means that B’ = 4. C′ =

Since C′ =

1 1 = = 10 µF RB′ 25x10 3 x 4

C1C 2 = 10µF and C1 = 20 µF, we then obtain C2 = 20 µF. C1 + C 2

Therefore, to increase the bandwidth, we merely add another 20 µF in series with the first one. Chapter 14, Solution 41.

(a)

This is a series RLC circuit.

R = 2+ 6 = 8Ω, ω0 =

(b)

1 LC

=

1 0.4

L =1H,

C = 0.4 F

= 1.5811 rad / s

Q=

ω 0 L 1.5811 = = 0.1976 R 8

B=

R = 8 rad / s L

This is a parallel RLC circuit. 3 µF and 6 µF  →

C = 2 µF ,

(3)(6) = 2 µF 3+ 6

R = 2 kΩ ,

L = 20 mH

ω0 =

1 LC

=

1 (2 × 10 -6 )(20 × 10 -3 )

= 5 krad / s

Q=

R 2 × 10 3 = 20 = ω0 L (5 × 10 3 )(20 × 10 -3 )

B=

1 1 = = 250 krad / s 3 RC (2 × 10 )(2 × 10 -6 )

Chapter 14, Solution 42.

(a)

Z in = (1 jωC) || (R + jωL)

Z in =

R + jωL jωC R + jωL +

1 jωC

=

R + jωL 1 − ω2 LC + jωRC

(R + jωL)(1 − ω2 LC − jωRC) Z in = (1 − ω2 LC) 2 + ω2 R 2 C 2 At resonance, Im(Z in ) = 0 , i.e. 0 = ωL(1 − ω2 LC) − ωR 2 C

ω2 LC = L − R 2 C ω0 = (b)

L − R 2C = LC

1 R2 − C L

Z in = jωL || (R + 1 jωC)

Z in =

jωL (R + 1 jωC) jωL (1 + jωRC) = R + jωL + 1 jωC (1 − ω2 LC) + jωRC

Z in =

(-ω2 RLC + jωL) [(1 − ω2 LC) − jωRC] (1 − ω2 LC) 2 + ω2 R 2 C 2

At resonance, Im(Z in ) = 0 , i.e. 0 = ωL (1 − ω2 LC) + ω3 R 2 C 2 L

ω2 (LC − R 2 C 2 ) = 1 ω0 =

1 LC − R 2 C 2

Z in = R || ( jωL + 1 jωC)

(c)

R ( jωL + 1 jωC) R (1 − ω2 LC) = R + jωL + 1 jωC (1 − ω 2 LC) + jωRC R (1 − ω2 LC)[(1 − ω2 LC) − jωRC] Z in = (1 − ω2 LC) 2 + ω2 R 2 C 2

Z in =

At resonance, Im(Z in ) = 0 , i.e. 0 = R (1 − ω2 LC) ωRC 1 − ω2 LC = 0 ω0 =

1 LC

Chapter 14, Solution 43.

Consider the circuit below. 1/jωC Zin

(a)

R1

jωL

Z in = (R 1 || jωL) || (R 2 + 1 jωC)  R 1 jωL   1   ||  R 2 +  Z in =  jωC   R 1 + jωL  

R2

jωR 1 L  1   ⋅ R 2 + R 1 + jωL  jωC  Z in = jR 1ωL 1 R2 + + jωC R 1 + jωL Z in =

jωR 1 L (1 + jωR 2 C) (R 1 + jωL)(1 + jωR 2 C) − ω2 LCR 1

Z in =

- ω2 R 1 R 2 LC + jωR 1 L R 1 − ω2 LCR 1 − ω2 LCR 2 + jω (L + R 1 R 2 C)

(-ω2 R 1 R 2 LC + jωR 1 L)[R 1 − ω2 LCR 1 − ω2 LCR 2 − jω (L + R 1 R 2 C)] Z in = (R 1 − ω2 LCR 1 − ω2 LCR 2 ) 2 + ω2 (L + R 1 R 2 C) 2 At resonance, Im(Z in ) = 0 , i.e. 0 = ω3 R 1 R 2 LC (L + R 1 R 2 C) + ωR 1 L (R 1 − ω2 LCR 1 − ω2 LCR 2 ) 0 = ω3 R 12 R 22 LC 2 + R 12 ωL − ω3 R 12 L2 C 0 = ω2 R 22 C 2 + 1 − ω2 LC ω2 (LC − R 22 C 2 ) = 1 ω0 =

1 LC − R 22 C 2

ω0 =

1

(0.02)(9 × 10 -6 ) − (0.1) 2 (9 × 10 -6 ) 2 ω0 = 2.357 krad / s (b)

At ω = ω0 = 2.357 krad / s , jωL = j(2.357 × 10 3 )(20 × 10 -3 ) = j47.14 R 1 || jωL =

R2 +

j47.14 = 0.9996 + j0.0212 1 + j47.14

1 1 = 0.1 + = 0.1 − j47.14 jωC j (2.357 × 10 3 )(9 × 10 -6 )

Z in (ω0 ) = (R 1 || jωL) || (R 2 + 1 jωC)

(0.9996 + j0.0212)(0.1 − j47.14) (0.9996 + j0.0212) + (0.1 − j47.14)

Z in (ω0 ) =

Z in (ω0 ) = 1 Ω

Chapter 14, Solution 44. We find the input impedance of the circuit shown below.

1

Z

jω(2/3)

1/jω 1/jωC

1 3 1 = + jω + , Z jω2 1 + 1 jωC

ω=1

C 2 + jC 1 jC = -j1.5 + j + = -j0.5 + Z 1+ C2 1 + jC

v( t ) and i( t ) are in phase when Z is purely real, i.e. 0 = -0.5 +

C 1 + C2

1 C2 1 = 2 = Z 1+ C 2

 → (C − 1) 2 = 1

 → Z = 2 Ω

V = Z I = (2)(10) = 20 v( t ) = 20 sin( t ) V ,

i.e.

Vo = 20 V

or

C = 1F

Chapter 14, Solution 45.

(a)

jω , 1 + jω

1 || jω =

1 ||

1 jω 1 1 = = jω 1 + 1 jω 1 + jω

Transform the current source gives the circuit below.

jω I 1 + jω

jω 1 + jω

+ −

1

1 1 + jω

+ Vo −

1 jω 1 + jω Vo = ⋅ I 1 jω 1 + j ω 1+ + 1 + jω 1 + jω

(b)

H (ω) =

Vo jω = I 2 (1 + jω) 2

H (1) =

1 2 (1 + j) 2

H (1) =

1 2 ( 2)2

= 0.25

Chapter 14, Solution 46. (a) This is an RLC series circuit. ωo =

1 LC

→

(b) (c )

C=

1 ω2 o L

=

1 (2πx15x10 3 ) 2 x10 x10 −3

Z = R, I = V/Z = 120/20 = 6 A Q=

ωo L 2πx15x10 3 x10 x10 −3 = = 15π = 47.12 R 20

= 11.26nF

Chapter 14, Solution 47.

H (ω) =

Vo R 1 = = Vi R + jωL 1 + jωL R

H(0) = 1 and H(∞) = 0 showing that this circuit is a lowpass filter. 1 , i.e. At the corner frequency, H(ωc ) = 2 1

=

2

1

ωc L   1+   R 

2

 → 1 =

ωc L R

or

ωc =

R L

Hence, ωc =

R = 2πf c L

fc =

1 R 1 10 × 10 3 = 796 kHz ⋅ = ⋅ 2π L 2π 2 × 10 -3

Chapter 14, Solution 48.

R ||

H (ω) =

1 jωC

jωL + R ||

1 jωC

R jωC R + 1 jωC H (ω) = R jωC jωL + R + 1 jωC H (ω) =

R R + jωL − ω 2 RLC

H(0) = 1 and H(∞) = 0 showing that this circuit is a lowpass filter.

Chapter 14, Solution 49.

At dc, H(0) =

H(ω) =

Hence,

2 2

1 2

4 = 2. 2

H(0) =

2 2

4

=

4 + 100ωc2

4 + 100ωc2 = 8  → ωc = 0.2

H(2) =

4 2 = 2 + j20 1 + j10

H(2) =

2 101

= 0.199

In dB, 20 log10 H(2) = - 14.023 arg H(2) = -tan -110 = - 84.3°

Chapter 14, Solution 50. H (ω) =

Vo jωL = Vi R + jωL

H(0) = 0 and H(∞) = 1 showing that this circuit is a highpass filter. H (ωc ) =

or

fc =

1 2

ωc =

=

1  R   1+   ωc L 

2

 → 1 =

R = 2πf c L

1 R 1 200 ⋅ = ⋅ = 318.3 Hz 2π L 2π 0.1

R ωc L

Chapter 14, Solution 51.

H ′(ω) =

jωRC jω = 1 + jωRC jω + 1 RC

(from Eq. 14.52)

This has a unity passband gain, i.e. H(∞) = 1 . 1 = ωc = 50 RC H ^ (ω) = 10 H ′(ω) =

H (ω) =

j10ω 50 + jω

j10ω 50 + jω

Chapter 14, Problem 52.

Design an RL lowpass filter that uses a 40-mH coil and has a cut-off frequency of 5 kHz. Chapter 14, Solution 53.

ωc =

R = 2πf c L

R = 2πf c L = (2π)(10 5 )(40 × 10 -3 ) = 25.13 kΩ

Chapter 14, Solution 54. ω1 = 2πf 1 = 20π × 10 3 ω2 = 2πf 2 = 22π × 10 3 B = ω2 − ω1 = 2π × 10 3 ω0 =

ω2 + ω1 = 21π × 10 3 2

Q=

ω0 21π = 11.5 = B 2π

ω0 =

1

 → L =

LC

1 ω02 C

L=

1 = 2.872 H (21π × 10 ) (80 × 10 -12 )

B=

R L

3 2

 → R = BL

R = (2π × 10 3 )(2.872) = 18.045 kΩ

Chapter 14, Solution 55.

ωc = 2πf c =

1 RC

→

R=

1 1 = = 265.3kΩ 3 2πf c C 2πx 2x10 x300x10 −12

Chapter 14, Solution 56.

ωo =

1 LC

=

1 (25 × 10 )(0.4 × 10 − 6 ) −3

B=

R 10 = = 0.4 krad / s L 25 × 10 -3

Q=

10 = 25 0.4

= 10 krad / s

ω1 = ωo − B 2 = 10 − 0.2 = 9.8 krad / s

or

f1 =

9.8 = 1.56 kHz 2π

ω2 = ωo + B 2 = 10 + 0.2 = 10.2 krad / s

or

f2 =

10.2 = 1.62 kHz 2π

Therefore, 1.56 kHz < f < 1.62 kHz

Chapter 14, Solution 57.

(a)

From Eq 14.54, R R sRC L H (s) = = = 2 R 1 1 1 + sRC + s LC s2 + s + R + sL + L LC sC s

Since B =

R and ω0 = L

H (s) = (b)

1 LC

,

sB s + sB + ω02 2

From Eq. 14.56, H (s) =

H (s) =

sL +

1 sC

1 R + sL + sC

=

s2 + s2 + s

1 LC

R 1 + L LC

s 2 + ω02 s 2 + sB + ω02

Chapter 14, Solution 58.

(a)

Consider the circuit below. I

Vs

+ −

R

I1

1/sC +

1/sC

R

Vo

− 1  1 R +  1  1 sC  sC  Z(s) = R + ||  R +  = R + 2 sC  sC  R+ sC

Z(s) = R +

1 + sRC sC (2 + sRC)

1 + 3sRC + s 2 R 2 C 2 Z(s) = sC (2 + sRC) I=

Vs Z

I1 =

Vs 1 sC I= 2 sC + R Z (2 + sRC)

Vo = I 1 R = H (s) =

R Vs sC (2 + sRC) ⋅ 2 + sRC 1 + 3sRC + s 2 R 2 C 2

Vo sRC = Vs 1 + 3sRC + s 2 R 2 C 2

 3 s  1 RC H (s) =  3 1 3 2 s + s+ 2 2  RC R C Thus, ω02 = B= (b)

1 R C2 2

or

     ω0 =

1 = 1 rad / s RC

3 = 3 rad / s RC

Similarly,

Z(s) = sL + R || (R + sL) = sL +

R (R + sL) 2R + sL

R 2 + 3sRL + s 2 L2 Z(s) = 2R + sL I=

Vs , Z

Vo = I 1 ⋅ sL =

I1 =

R Vs R I= 2R + sL Z (2R + sL)

sLR Vs 2R + sL ⋅ 2 2R + sL R + 3sRL + s 2 L2

1  3R   s Vo sRL 3 L  H (s) = = = 3R R2 Vs R 2 + 3sRL + s 2 L2 2 s + s+ 2 L L

Thus, ω0 = B=

R = 1 rad / s L

3R = 3 rad / s L

Chapter 14, Solution 59. 1

ω0 =

(b)

R 2 × 10 3 B= = = 2 × 10 4 L 0 .1 Q=

LC

=

1

(a)

(0.1)(40 × 10 -12 )

= 0.5 × 10 6 rad / s

ω0 0.5 × 10 6 = = 250 B 2 × 10 4

As a high Q circuit, B ω1 = ω0 − = 10 4 (50 − 1) = 490 krad / s 2 ω 2 = ω0 + (c)

B = 10 4 (50 + 1) = 510 krad / s 2

As seen in part (b),

Q = 250

Chapter 14, Solution 60. Consider the circuit below. Ro + 1/sC Vi

+ −

R

Vo sL

−  1  R (sL + 1 sC) Z(s) = R || sL +  =  sC  R + sL + 1 sC Z(s) =

R (1 + s 2 LC) 1 + sRC + s 2 LC

Vo R (1 + s 2 LC) Z = = H= Vi Z + R o R o + sRR o C + s 2 LCR o + R + s 2 LCR R (1 + s 2 LC) Z in = R o + Z = R o + 1 + sRC + s 2 LC

R o + sRR o C + s 2 LCR o + R + s 2 LCR Z in = 1 + sRC + s 2 LC

s = jω Z in =

R o + jωRR o C − ω2 LCR o + R − ω2 LCR 1 − ω2 LC + jωRC

Z in =

(R o + R − ω2 LCR o − ω2 LCR + jωRR o C)(1 − ω 2 LC − jωRC) (1 − ω2 LC) 2 + (ωRC) 2

Im(Z in ) = 0 implies that

- ωRC [R o + R − ω2 LCR o − ω2 LCR ] + ωRR o C (1 − ω2 LC) = 0

R o + R − ω2 LCR o − ω2 LCR − R o + ω2 LCR o = 0

ω2 LCR = R 1

ω0 =

H=

LC

1

=

(1 × 10 )(4 × 10 -6 ) -3

R (1 − ω2 LC) R o + jωRR o C + R − ω2 LCR o − ω2 LCR

H max = H(0) =

or

H max

R Ro + R

 1  R  2 − LC  R ω  = H(∞) = lim = ω→ ∞ R o + R RR o C +j − LC (R + R o ) R + R o 2 ω ω

At ω1 and ω2 , H = R 2 (R o + R )

1 2 1 2 0=

= 15.811 krad / s

=

=

=

1 2

H mzx

R (1 − ω2 LC) R o + R − ω 2 LC (R o + R ) + jωRR o C

(R o + R )(1 − ω2 LC) (ωRR o C) 2 + (R o + R − ω2 LC(R o + R )) 2 10 (1 − ω2 ⋅ 4 × 10 -9 ) (96 × 10 -6 ω) 2 + (10 − ω2 ⋅ 4 × 10 -8 ) 2 10 (1 − ω2 ⋅ 4 × 10 -9 ) (96 × 10 -6 ω) 2 + (10 − ω2 ⋅ 4 × 10 -8 ) 2

−

1 2

(10 − ω2 ⋅ 4 × 10 -8 )( 2 ) − (96 × 10 -6 ω) 2 + (10 − ω2 ⋅ 4 × 10 -8 ) 2 = 0 (2)(10 − ω2 ⋅ 4 × 10 -8 ) 2 = (96 × 10 -6 ω) 2 + (10 − ω2 ⋅ 4 × 10 -8 ) 2 (96 × 10 -6 ω) 2 − (10 − ω2 ⋅ 4 × 10 -8 ) 2 = 0

1.6 × 10 -15 ω4 − 8.092 × 10 -7 ω2 + 100 = 0

ω4 − 5.058 × 10 8 + 6.25 × 1016 = 0  2.9109 × 10 8 ω2 =   2.1471 × 10 8

Hence, ω1 = 14.653 krad / s ω2 = 17.061 krad / s B = ω2 − ω1 = 17.061 − 14.653 = 2.408 krad / s

Chapter 14, Solution 61.

(a)

V+ =

1 jωC V, R + 1 jωC i

V− = Vo

Since V+ = V− , 1 V = Vo 1 + jωRC i H (ω) =

(b)

V+ =

Vo 1 = Vi 1 + jωRC

R V, R + 1 jωC i

Since V+ = V− , jωRC V = Vo 1 + jωRC i H (ω) =

Vo jωRC = Vi 1 + jωRC

V− = Vo

Chapter 14, Solution 62. This is a highpass filter.

(a)

H (ω) =

jωRC 1 = 1 + jωRC 1 − j ωRC

H (ω) =

1 , 1 − j ωc ω

H (ω) =

1 1 = 1 − j f c f 1 − j1000 f

H (f = 200 Hz) =

Vo =

(b)

1 − j5

Vo 1 = 1 − j5 Vi

Vo 1 = 1 − j0.5 Vi

120 mV

= 107.3 mV

1 − j0.5

Vo 1 = 1 − j0.1 Vi

H (f = 10 kHz) =

Vo =

1 = 2π (1000) RC

= 23.53 mV

H (f = 2 kHz) =

Vo =

(c)

120 mV

ωc =

120 mV 1 − j0.1

= 119.4 mV

Chapter 14, Solution 63.

For an active highpass filter, H (s) = −

sC i R f 1 + sC i R i

(1)

But H(s) = −

10s 1 + s / 10

(2)

Comparing (1) and (2) leads to: C i R f = 10

 →

Rf =

10 = 10MΩ Ci

C i R i = 0.1

→

Ri =

0.1 = 100kΩ Ci

Chapter 14, Solution 64.

Z f = R f ||

Rf 1 = jωC f 1 + jωR f C f

Zi = R i +

1 + jωR i C i 1 = jωC i jωC i

Hence, H (ω) =

Vo - Z f - jωR f C i = = Vi Zi (1 + jωR f C f )(1 + jωR i C i )

This is a bandpass filter. H(ω) is similar to the product of the transfer function of a lowpass filter and a highpass filter. Chapter 14, Solution 65.

V+ =

R jωRC Vi = V R + 1 jωC 1 + jωRC i

V− =

Ri V Ri + Rf o

Since V+ = V− ,

Ri jωRC Vo = V Ri + Rf 1 + jωRC i H (ω) =

Vo  R f   jωRC    = 1 + Vi  R i   1 + jωRC 

It is evident that as ω → ∞ , the gain is 1 +

Rf 1 and that the corner frequency is . Ri RC

Chapter 14, Solution 66. (a)

Proof

(b)

When R 1 R 4 = R 2 R 3 , H (s) =

(c)

R4 s ⋅ R 3 + R 4 s + 1 R 2C

When R 3 → ∞ , H (s) =

- 1 R 1C s + 1 R 2C

Chapter 14, Solution 67. DC gain =

Rf 1 = Ri 4

 → R i = 4R f

Corner frequency = ωc =

1 = 2π (500) rad / s R f Cf

If we select R f = 20 kΩ , then R i = 80 kΩ and C=

1 = 15.915 nF (2π)(500)(20 × 10 3 )

Therefore, if R f = 20 kΩ , then R i = 80 kΩ and C = 15.915 nF

Chapter 14, Solution 68.

High frequency gain = 5 =

Corner frequency = ωc =

Rf Ri

 → R f = 5R i

1 = 2π (200) rad / s R i Ci

If we select R i = 20 kΩ , then R f = 100 kΩ and C=

1 = 39.8 nF (2π)(200)(20 × 10 3 )

Therefore, if R i = 20 kΩ , then R f = 100 kΩ and C = 39.8 nF

Chapter 14, Solution 69. This is a highpass filter with f c = 2 kHz. 1 ωc = 2πf c = RC RC =

1 1 = 2πf c 4π × 103

10 8 Hz may be regarded as high frequency. Hence the high-frequency gain is − R f − 10 = or R f = 2 .5 R R 4 If we let R = 10 kΩ , then R f = 25 kΩ , and C =

1 = 7.96 nF . 4000π × 10 4

Chapter 14, Solution 70.

(a)

H (s) =

Vo (s) Y1 Y2 = Vi (s) Y1 Y2 + Y4 (Y1 + Y2 + Y3 )

where Y1 =

H (s) =

(b)

1 1 = G 1 , Y2 = = G 2 , Y3 = sC1 , Y4 = sC 2 . R1 R2

G 1G 2 G 1 G 2 + sC 2 (G 1 + G 2 + sC1 )

G 1G 2 H(∞) = 0 = 1, G 1G 2 showing that this circuit is a lowpass filter. H ( 0) =

Chapter 14, Solution 71.

R = 50 Ω , L = 40 mH , C = 1 µF L′ =

Km Km L  → 1 = ⋅ (40 × 10 -3 ) Kf Kf

25K f = K m C′ =

C KmKf

10 6 K f =

(1)  → 1 =

1 Km

Substituting (1) into (2), 1 10 6 K f = 25K f K f = 0.2 × 10 -3 K m = 25K f = 5 × 10 -3

10 -6 KmKf

(2)

Chapter 14, Solution 72.

L′C′ =

LC K f2

 → K f2 =

LC L ′C′

(4 × 10 -3 )(20 × 10 -6 ) K = = 4 × 10 -8 (1)(2) 2 f

K f = 2 × 10 -4

L′ L 2 = K C′ C m

 → K 2m =

L′ C ⋅ C′ L

(1)(20 × 10 -6 ) K = = 2.5 × 10 -3 (2)(4 × 10 -3 ) 2 m

K m = 5 × 10 -2

Chapter 14, Solution 73. R ′ = K m R = (12)(800 × 10 3 ) = 9.6 MΩ L′ =

Km 800 L= (40 × 10 -6 ) = 32 µF Kf 1000

C′ =

C 300 × 10 -9 = = 0.375 pF K m K f (800)(1000)

Chapter 14, Solution 74. R '1 = K m R 1 = 3x100 = 300Ω R ' 2 = K m R 2 = 10 x100 = 1 kΩ

L' =

Km 10 2 L= (2) = 200 µH Kf 10 6

1 C = 10 = 1 nF C' = K m K f 108

Chapter 14, Solution 75. R ' = K m R = 20 x10 = 200 Ω L' =

Km 10 L= (4) = 400 µH Kf 10 5

C' =

C 1 = = 1 µF K m K f 10x10 5

Chapter 14, Solution 76.

R ' = K m R = 50 x10 3

L' =

Km L = 10 µH Kf

C' = 40 pF =

C KmKf

→

R=

50 x10 3 10 3

L = 10 x10 −6 x

→

= 50 Ω

10 6 10 3

= 10 mH

C = 40 x10 −12 x10 3 x10 6 = 40 mF

→

Chapter 14, Solution 77.

L and C are needed before scaling.

B=

R L

ω0 = (a)

 → L = 1 LC

R 10 = =2H B 5

 → C =

1 1 = = 312.5 µF 2 ω0 L (1600)(2)

L′ = K m L = (600)(2) = 1200 H C 3.125 × 10 -4 C′ = = = 0.5208 µF Km 600

(b)

L′ =

L 2 = 3 = 2 mH K f 10

C 3.125 × 10 -4 = = 312.5 nF C′ = Kf 10 3

(c)

L′ =

Km (400)(2) = 8 mH L= 10 5 Kf

C′ =

C 3.125 × 10 -4 = = 7.81 pF KmKf (400)(10 5 )

Chapter 14, Solution 78. R ′ = K m R = (1000)(1) = 1 kΩ Km 10 3 L′ = L = 4 (1) = 0.1 H 10 Kf

C′ =

C 1 = = 0.1 µF 3 K m K f (10 )(10 4 )

The new circuit is shown below. 1 kΩ + I

1 kΩ

0.1 H

0.1 µF

1 kΩ

Vx

−

Chapter 14, Solution 79.

(a)

Insert a 1-V source at the input terminals. Ro Io

1V

R

V1

V2

1/sC

+ −

+ −

+ sL

3Vo

Vo −

There is a supernode. 1 − V1 V2 = R sL + 1 sC But

V1 = V2 + 3Vo

Also,

Vo =

V2 = V1 − 3Vo = Vo =

 → V2 = V1 − 3Vo

sL V sL + 1 sC 2

Combining (2) and (3)

(1)

 →

Vo V2 = sL sL + 1 sC

(3)

sL + 1 sC Vo sL

s 2 LC V 1 + 4s 2 LC 1

Substituting (3) and (4) into (1) gives 1 − V1 Vo sC = = V R sL 1 + 4s 2 LC 1 sRC 1 + 4s 2 LC + sRC 1 = V1 + V = V1 1 + 4s 2 LC 1 1 + 4s 2 LC V1 =

(2)

1 + 4s 2 LC 1 + 4s 2 LC + sRC

(4)

Io =

1 − V1 sRC = R R (1 + 4s 2 LC + sRC)

Z in =

1 1 + sRC + 4s 2 LC = sC Io

Z in = 4sL + R +

1 sC

(5)

When R = 5 , L = 2 , C = 0.1 , Z in (s) = 8s + 5 +

10 s

At resonance, Im(Z in ) = 0 = 4ωL − or (b)

ω0 =

1 2 LC

=

1 ωC

1 2 (0.1)(2)

= 1.118 rad / s

After scaling, R′  → K m R 4Ω  → 40 Ω 5Ω  → 50 Ω L′ =

Km 10 L= ( 2 ) = 0 .2 H Kf 100

C′ =

C 0.1 = = 10 -4 K m K f (10)(100)

From (5), Z in (s) = 0.8s + 50 +

ω0 =

1 2 LC

=

10 4 s

1 2 (0.2)(10 -4 )

= 111.8 rad / s

Chapter 14, Solution 80. (a)

R ′ = K m R = (200)(2) = 400 Ω L′ =

K m L (200)(1) = = 20 mH Kf 10 4

C′ =

C 0.5 = = 0.25 µF K m K f (200)(10 4 )

The new circuit is shown below. 20 mH a Ix 0.25 µF

400 Ω

0.5 Ix

b

(b)

Insert a 1-A source at the terminals a-b. a

sL

V1

V2

Ix 1A

1/(sC)

R

0.5 Ix

b

At node 1, 1 = sCV1 +

V1 − V2 sL

At node 2, V1 − V2 V2 + 0 .5 I x = sL R But, I x = sC V1 .

(1)

V1 − V2 V2 + 0.5sC V1 = sL R

(2)

Solving (1) and (2), sL + R V1 = 2 s LC + 0.5sCR + 1 Z Th =

V1 sL + R = 2 1 s LC + 0.5sCR + 1

At ω = 10 4 , Z Th

( j10 4 )(20 × 10 -3 ) + 400 = ( j10 4 ) 2 (20 × 10 -3 )(0.25 × 10 -6 ) + 0.5( j10 4 )(0.25 × 10 -6 )(400) + 1

Z Th =

400 + j200 = 600 − j200 0.5 + j0.5

Z Th = 632.5∠ - 18.435° ohms

Chapter 14, Solution 81. (a)

1 (G + jωC)(R + jωL) + 1 1 = G + jωC + = R + jω L R + jω L Z

which leads to

Z=

jωL + R 2

− ω LC + jω(RC + LG) + GR + 1

ω R + C LC Z(ω) = R G  GR + 1  − ω2 + jω +  + LC L C j

(1)

We compare this with the given impedance: Z(ω) =

1000( jω + 1) 2

− ω + 2 jω + 1 + 2500

(2)

Comparing (1) and (2) shows that

1 = 1000 C

 →

R G + =2 L C

 →

C = 1 mF,

R/L = 1

 →

R=L

G = C = 1 mS

GR + 1 10 −3 R + 1 2501 = = LC 10 −3 R

→

R = 0 .4 = L

Thus, R = 0.4Ω, L = 0.4 H, C = 1 mF, G = 1 mS (b) By frequency-scaling, Kf =1000. R’ = 0.4 Ω, G’ = 1 mS L' =

L 0.4 = = 0.4mH , K f 10 3

C' =

C 10 −3 = = 1µF K f 10 − 3

Chapter 14, Solution 82.

C′ =

C KmKf

Kf =

ω′c 200 = = 200 ω 1

Km =

C 1 1 1 ⋅ = -6 ⋅ = 5000 C′ K f 10 200

R ′ = K m R = 5 kΩ,

thus,

R ′f = 2R i = 10 kΩ

Chapter 14, Solution 83.

1µF

 →

1 10 −6 C' = C= = 0.1 pF K mKf 100 x10 5

5µF

 →

C' = 0.5 pF

10 kΩ

→

R ' = K m R = 100x10 kΩ = 1 MΩ

20 kΩ

→

R ' = 2 MΩ

Chapter 14, Solution 84. The schematic is shown below. A voltage marker is inserted to measure vo. In the AC sweep box, we select Total Points = 50, Start Frequency = 1, and End Frequency = 1000. After saving and simulation, we obtain the magnitude and phase plots in the probe menu as shown below.

Chapter 14, Solution 85. We let I s = 1∠0 o A so that Vo / I s = Vo . The schematic is shown below. The circuit is simulated for 100 < f < 10 kHz.

Chapter 14, Solution 86.

The schematic is shown below. A current marker is inserted to measure I. We set Total Points = 101, start Frequency = 1, and End Frequency = 10 kHz in the AC sweep box. After simulation, the magnitude and phase plots are obtained in the Probe menu as shown below.

Chapter 14, Solution 87.

The schematic is shown below. In the AC Sweep box, we set Total Points = 50, Start Frequency = 1, and End Frequency = 100. After simulation, we obtain the magnitude response as shown below. It is evident from the response that the circuit represents a high-pass filter.

Chapter 14, Solution 88.

The schematic is shown below. We insert a voltage marker to measure Vo. In the AC Sweep box, we set Total Points = 101, Start Frequency = 1, and End Frequency = 100. After simulation, we obtain the magnitude and phase plots of Vo as shown below.

Chapter 14, Solution 89.

The schematic is shown below. In the AC Sweep box, we type Total Points = 101, Start Frequency = 100, and End Frequency = 1 k. After simulation, the magnitude plot of the response Vo is obtained as shown below.

Chapter 14, Solution 90.

The schematic is shown below. In the AC Sweep box, we set Total Points = 1001, Start Frequency = 1, and End Frequency = 100k. After simulation, we obtain the magnitude plot of the response as shown below. The response shows that the circuit is a high-pass filter.

Chapter 14, Solution 91.

The schematic is shown below. In the AC Sweep box, we set Total Points = 1001, Start Frequency = 1, and End Frequency = 100k. After simulation, we obtain the magnitude plot of the response as shown below. The response shows that the circuit is a high-pass filter.

Chapter 14, Solution 92.

The schematic is shown below. We type Total Points = 101, Start Frequency = 1, and End Frequency = 100 in the AC Sweep box. After simulating the circuit, the magnitude plot of the frequency response is shown below.

Chapter 14, Solution 93.

L C R

1 1 R2 f0 = − 2π LC L2 R 400 10 7 = = , L 240 × 10 -6 6

Since

R 1 << L LC

1 1 1016 = = LC (240 × 10 -6 )(120 × 10 -12 ) 288

f0 ≅

1 2π LC

=

10 8 24π 2

= 938 kHz

R 1 << . L LC The result remains the same. If R is reduced to 40 Ω,

Chapter 14, Solution 94.

ωc =

1 RC

We make R and C as small as possible. To achieve this, we connect 1.8 k Ω and 3.3 k Ω in parallel so that 1.8x 3.3 R= = 1.164 kΩ 1.8 + 3.3 We place the 10-pF and 30-pF capacitors in series so that C = (10x30)/40 = 7.5 pF Hence, 1 1 ωc = = = 114.55x10 6 rad/s RC 1.164x10 3 x 7.5x10 −12 Chapter 14, Solution 95.

(a)

f0 =

1 2π LC

When C = 360 pF , f0 =

1 2π (240 × 10 -6 )(360 × 10 -12 )

= 0.541 MHz

When C = 40 pF , f0 =

1 2π (240 × 10 -6 )(40 × 10 -12 )

Therefore, the frequency range is 0.541 MHz < f 0 < 1.624 MHz

= 1.624 MHz

Q=

(b)

2πfL R

At f 0 = 0.541 MHz , Q=

(2π )(0.541 × 10 6 )(240 × 10 -6 ) = 67.98 12

At f 0 = 1.624 MHz , (2π )(1.624 × 10 6 )(240 × 10 -6 ) = 204.1 Q= 12

Chapter 14, Solution 96. Ri

Vi

L

V1

Vo +

+ −

C1

C2

RL

Vo −

Z2 Z1 = R L ||

Z2 =

Z1

RL 1 = sC 2 1 + sR 2 C 2

1 1  sL + R L + s 2 R L C 2 L   || (sL + Z1 ) = ||  sC1 sC1  1 + sR L C 2 

1 sL + R L + s 2 R L C 2 L ⋅ sC1 1 + sR L C 2 Z2 = sL + R L + s 2 R L C 2 L 1 + sC1 1 + sR L C 2 sL + R L + s 2 R L LC 2 Z2 = 1 + sR L C 2 + s 2 LC1 + sR L C1 + s 3 R L LC1C 2

V1 =

Z2 V Z2 + R i i

Vo =

Z1 Z2 Z1 V1 = ⋅ V Z1 + sL Z 2 + R 2 Z1 + sL i

Vo Z2 Z1 = ⋅ Vi Z 2 + R 2 Z1 + sL

where Z2 = Z2 + R 2

sL + R L + s 2 R L LC 2 sL + R L + s 2 R L LC 2 + R i + sR i R L C 2 + s 2 R i LC1 + sR i R L C1 + s 3 R i R L LC1C 2 and

Z1 RL = Z1 + sL R L + sL + s 2 R L LC 2

Therefore, Vo = Vi

R L (sL + R L + s 2 R L LC 2 ) (sL + R L + s 2 R L LC 2 + R i + sR i R L C 2 + s 2 R i LC 1 + sR i R L C 1 + s 3 R i R L LC 1 C 2 )( R L + sL + s 2 R L LC 2 ) where s = jω .

Chapter 14, Solution 97. Ri

Vi

L

V1

Vo +

+ −

C1

C2

RL

Vo −

Z2

Z1

 1  sL (R L + 1 sC 2 ) = Z = sL ||  R L + , sC 2  R L + sL + 1 sC 2 

V1 =

Z V Z + R i + 1 sC1 i

Vo =

RL RL Z V1 = ⋅ V R L + 1 sC 2 R L + 1 sC 2 Z + R i + 1 sC1 i

H (ω) =

Vo RL sL (R L + 1 sC 2 ) = ⋅ Vi R L + 1 sC 2 sL (R L + 1 sC 2 ) + (R i + 1 sC1 )(R L + sL + 1 sC 2 )

H (ω) =

s 3 LR L C 1C 2 (sR i C 1 + 1)(s 2 LC 2 + sR L C 2 + 1) + s 2 LC 1 (sR L C 2 + 1)

where s = jω .

Chapter 14, Solution 98. B = ω2 − ω1 = 2π (f 2 − f 1 ) = 2π (454 − 432) = 44π

ω0 = 2πf 0 = QB = (20)(44π )

f0 =

s = jω

(20)(44π) = (20)(22) = 440 Hz 2π

Chapter 14, Solution 99.

Xc =

C=

1 1 = ωC 2πf C

1 1 10 -9 = = 2πf X c (2π )(2 × 10 6 )(5 × 10 3 ) 20π

X L = ωL = 2πf L

XL 300 3 × 10 -4 = = L= 2πf (2π )(2 × 10 6 ) 4π f0 =

B=

1 2π LC

=

1 3 × 10 -4 10 -9 ⋅ 2π 4π 20π

= 1.826 MHz

 4π  R  = 4.188 × 10 6 rad / s = (100)   3 × 10 -4  L

Chapter 14, Solution 100.

ωc = 2πf c =

R=

1 RC

1 1 = = 15.91 Ω 2πf c C (2π )(20 × 10 3 )(0.5 × 10 -6 )

Chapter 14, Solution 101.

ωc = 2πf c =

R=

1 RC

1 1 = = 1.061 kΩ 2πf c C (2π )(15)(10 × 10 -6 )

Chapter 14, Solution 102.

(a)

When R s = 0 and R L = ∞ , we have a low-pass filter. ωc = 2πf c =

fc =

1 RC

1 1 = = 994.7 Hz 2πRC (2π)(4 × 10 3 )(40 × 10 -9 )

(b)

We obtain R Th across the capacitor. R Th = R L || (R + R s ) R Th = 5 || (4 + 1) = 2.5 kΩ fc =

1 1 = 2πR Th C (2π )(2.5 × 10 3 )(40 × 10 -9 )

f c = 1.59 kHz

Chapter 14, Solution 103. H (ω) =

H (ω) =

H (ω) =

Vo R2 , = Vi R 2 + R 1 || 1 sC

s = jω

R (R + 1 sC) R2 = 2 1 R (1 sC) R 2 + R 1 (1 sC) R2 + 1 R 1 + 1 sC R 2 (1 + sCR 1 ) R 1 + sCR 2

Chapter 14, Solution 104.

The schematic is shown below. We click Analysis/Setup/AC Sweep and enter Total Points = 1001, Start Frequency = 100, and End Frequency = 100 k. After simulation, we obtain the magnitude plot of the response as shown.

Chapter 15, Solution 1. e at + e - at 2 1 1 1  s L [ cosh(at ) ] =  + = 2  2  s − a s + a  s − a2

(a)

cosh(at ) =

(b)

sinh(at ) =

e at − e - at 2 1 1 1  a L [ sinh(at ) ] =  − = 2  2  s − a s + a  s − a2

Chapter 15, Solution 2.

(a)

f ( t ) = cos(ωt ) cos(θ) − sin(ωt ) sin(θ) F(s) = cos(θ) L [ cos(ωt ) ] − sin(θ) L [ sin(ωt ) ] F(s) =

(b)

s cos(θ) − ω sin(θ) s 2 + ω2

f ( t ) = sin(ωt ) cos(θ) + cos(ωt ) sin(θ) F(s) = sin(θ) L [ cos(ωt ) ] + cos(θ) L [ sin(ωt ) ]

F(s) =

s sin(θ) − ω cos(θ) s 2 + ω2

Chapter 15, Solution 3.

(a)

L [ e -2t cos(3t ) u ( t ) ] =

s+2 (s + 2 ) 2 + 9

(b)

L [ e -2t sin(4 t ) u ( t ) ] =

4 (s + 2) 2 + 16

(c)

Since L [ cosh(at ) ] =

s s − a2 2

L [ e -3t cosh(2 t ) u ( t ) ] =

(d)

Since L [ sinh(at ) ] =

L [ e -4t sinh( t ) u ( t ) ] =

(e)

L [ e - t sin( 2t ) ] = If

s+3 (s + 3 ) 2 − 4

a s − a2 2

1 (s + 4) 2 − 1

2 (s + 1) 2 + 4

f (t) ← → F(s)

t f (t) ← →

-d F(s) ds

Thus, L [ t e - t sin(2 t ) ] = =

L [ t e -t sin( 2t ) ] =

[

-d -1 2 ( (s + 1) 2 + 4) ds

2 ⋅ 2 (s + 1) ((s + 1) 2 + 4) 2 4 (s + 1) ((s + 1) 2 + 4) 2

Chapter 15, Solution 4.

(a)

G (s) = 6

(b)

F(s) =

s s2 + 42

2 s2

+5

]

e −s =

e −2s s+3

6se −s s 2 + 16

Chapter 15, Solution 5.

(a)

L [ cos(2t + 30°) ] =

s cos(30°) − 2 sin(30°) s2 + 4

L [ t 2 cos(2t + 30°) ] =

d 2  s cos(30°) − 1  ds 2  s 2 + 4 

=

 -1  d d  3 s − 1 (s 2 + 4)   ds ds  2  

=

 3  -1 -2  d  3 2 s − 1 (s 2 + 4)   (s + 4) − 2s  ds  2   2 

 3   3  3  3   2s   (8s 2 )   2 s 1 s 1 − − (- 2s )  2   2   2   −  +   = 2 − 3 2 2 2 2 2 2 2 s +4 s +4 s +4 s +4

(

)

(

)

(

)

(

 3  (8s 2 )  s − 1 - 3s − 3s + 2 − 3s  2  = + 3 2 2 2 s +4 s +4

(

=

)

(-3 3 s + 2)(s 2 + 4)

(s

2

+ 4)

3

L [ t 2 cos(2t + 30°) ] =

(

+

)

4 3 s3 − 8 s 2

(s

2

+ 4)

3

8 − 12 3 s − 6s 2 + 3s 3

( s 2 + 4) 3

4! 720 5 = (s + 2) (s + 2 ) 5

(b)

L [ 30 t 4 e - t ] = 30 ⋅

(c)

  2 d 2 L  2t u ( t ) − 4 δ( t )  = 2 − 4(s ⋅ 1 − 0) = 2 − 4s   s s dt

)

(d)

2 e -(t-1) u ( t ) = 2 e -t u ( t ) 2e L [ 2 e -(t-1) u ( t ) ] = s+1

(e)

Using the scaling property, 1 1 1 5 L [ 5 u ( t 2) ] = 5 ⋅ ⋅ = 5⋅ 2⋅ = 1 2 s (1 2) 2s s 6 18 = s + 1 3 3s + 1

(f)

L [ 6 e -t 3 u ( t ) ] =

(g)

Let f ( t ) = δ( t ) . Then, F(s) = 1 .  dn  L  n δ( t )  =  dt 

 dn  L  n f ( t )  = s n F(s) − s n −1 f (0) − s n − 2 f ′(0) − "  dt 

 dn  L  n δ( t )  =  dt 

 dn  L  n f ( t )  = s n ⋅ 1 − s n −1 ⋅ 0 − s n − 2 ⋅ 0 − "  dt 

 dn  L  n δ( t )  = s n  dt 

Chapter 15, Solution 6.

(a)

L [ 2 δ( t − 1) ] = 2 e -s

(b)

L [ 10 u ( t − 2) ] =

10 - 2s e s

(c)

L [ ( t + 4) u ( t ) ] =

1 4 + s2 s

(d)

L[ 2e

-t

u ( t − 4) ] = L [ 2 e

-4

e

-(t - 4)

u ( t − 4) ] =

2 e -4s e 4 (s + 1)

Chapter 15, Solution 7.

s , we use the linearity and shift properties to s + 42 10 s e - s obtain L [10 cos(4 ( t − 1)) u ( t − 1) ] = 2 s + 16

(a)

Since L [ cos(4t ) ] =

(b)

Since L [ t 2 ] = L [ t 2 e -2 t ] =

2

2 1 [ ] L = u ( t ) , , s3 s

e -3s 2 [ ] , and L − = u ( t 3 ) s (s + 2) 3

L [ t 2 e -2 t u ( t ) + u ( t − 3) ] =

2 e -3s + (s + 2 ) 3 s

Chapter 15, Solution 8.

(a)

L [ 2 δ(3t ) + 6 u (2t ) + 4 e -2 t − 10 e -3 t

]

1 1 1 4 10 = 2⋅ + 6⋅ ⋅ + − 3 2 s 2 s+2 s+3

= (b)

2 6 4 10 + + − 3 s s+2 s+3

t e -t u ( t − 1) = ( t − 1) e -t u ( t − 1) + e -t u ( t − 1) t e -t u ( t − 1) = ( t − 1) e -(t-1) e -1 u ( t − 1) + e -(t-1) e -1 u ( t − 1) L [ t e - t u ( t − 1) ] =

(c)

e -1 e -s e -1 e -s e -(s+1) e -(s+1) + = + (s + 1) 2 s + 1 (s + 1) 2 s + 1

L [ cos(2 ( t − 1)) u ( t − 1) ] =

s e -s s2 + 4

(d)

Since sin(4 ( t − π)) = sin(4t ) cos(4π ) − sin( 4π) cos(4t ) = sin(4t ) sin(4t ) u ( t − π ) = sin(4 ( t − π )) u ( t − π ) L[ sin( 4 t ) [ u ( t ) − u ( t − π )] ] = L[ sin( 4 t ) u ( t ) ] − L[ sin( 4( t − π )) u ( t − π) ]

=

4 e - πs 4 4 − = 2 ⋅ (1 − e -πs ) 2 2 s + 16 s + 16 s + 16

Chapter 15, Solution 9.

(a)

f ( t ) = ( t − 4) u ( t − 2) = ( t − 2) u ( t − 2) − 2 u ( t − 2) e -2s 2 e -2s F(s) = 2 − 2 s s

(b)

g( t ) = 2 e -4t u ( t − 1) = 2 e -4 e -4(t -1) u ( t − 1) 2 e -s G (s) = 4 e (s + 4)

(c)

h ( t ) = 5 cos(2 t − 1) u ( t ) cos(A − B) = cos(A) cos(B) + sin(A) sin(B) cos(2t − 1) = cos(2t ) cos(1) + sin(2t ) sin(1) h ( t ) = 5 cos(1) cos(2 t ) u ( t ) + 5 sin(1) sin(2t ) u ( t ) H(s) = 5 cos(1) ⋅

H(s) = (d)

s 2 + 5 sin(1) ⋅ 2 s +4 s +4 2

2.702 s 8.415 + s2 + 4 s2 + 4

p( t ) = 6u ( t − 2) − 6u ( t − 4) P(s) =

6 - 2s 6 -4s e − e s s

Chapter 15, Solution 10. (a)

By taking the derivative in the time domain, g ( t ) = (-t e -t + e -t ) cos( t ) − t e -t sin( t ) g ( t ) = e -t cos( t ) − t e -t cos( t ) − t e -t sin( t )

(b)

G (s) =

 s +1 d  s +1  d  1 +  +   2 2 2 (s + 1) + 1 ds  (s + 1) + 1 ds  (s + 1) + 1

G (s) =

s +1 s 2 + 2s 2s + 2 s 2 (s + 2) − − = s 2 + 2s + 2 (s 2 + 2s + 2) 2 (s 2 + 2s + 2) 2 (s 2 + 2s + 2) 2

By applying the time differentiation property, G (s) = sF(s) − f (0) where f ( t ) = t e -t cos( t ) , f (0) = 0 G (s) = (s) ⋅

- d  s +1  (s)(s 2 + 2s) s 2 (s + 2) = = ds  (s + 1) 2 + 1  (s 2 + 2s + 2) 2 (s 2 + 2s + 2) 2

Chapter 15, Solution 11.

(a)

Since L [ cosh(at ) ] =

F(s) =

(b)

s s − a2 2

6 (s + 1) 6 (s + 1) = 2 2 (s + 1) − 4 s + 2s − 3

Since L [ sinh(at ) ] = L [ 3 e -2t sinh(4t ) ] =

a s − a2 2

(3)(4) 12 = 2 2 (s + 2) − 16 s + 4s − 12

F(s) = L [ t ⋅ 3 e -2t sinh(4t ) ] =

-d [ 12 (s 2 + 4s − 12) -1 ] ds

F(s) = (12)(2s + 4)(s 2 + 4s − 12) -2 =

24 (s + 2) (s + 4s − 12) 2 2

(c)

cosh( t ) =

1 ⋅ (e t + e - t ) 2

1 f ( t ) = 8 e -3t ⋅ ⋅ (e t + e - t ) u ( t − 2) 2 = 4 e-2t u ( t − 2) + 4 e-4t u ( t − 2) = 4 e-4 e-2(t - 2) u ( t − 2) + 4 e-8 e-4(t - 2) u ( t − 2) L [ 4 e -4 e -2(t -2) u ( t − 2)] = 4 e -4 e -2s ⋅ L [ e -2 u ( t )] L [ 4 e -4 e -2(t -2) u ( t − 2)] =

4 e -(2s+ 4) s+2

Similarly, L [ 4 e -8 e -4(t -2) u ( t − 2)] =

4 e -(2s+8) s+4

Therefore, 4 e -(2s+ 4) 4 e -(2s+8) e -(2s+ 6) [ (4 e 2 + 4 e -2 ) s + (16 e 2 + 8 e -2 )] + = F(s) = s+2 s+4 s 2 + 6s + 8 Chapter 15, Solution 12. f ( t ) = te −2( t −1) e −2 u ( t − 1) = ( t − 1)e −2 e −2( t −1) u ( t − 1) + e −2 e −2( t −1) u ( t − 1) f (s) = e − s

e−2 (s + 2) 2

+ e−2

e −s e − (s + 2)  1  s + 3 − (s + 2) = e 1 + = s+2 s + 2  s + 2  (s + 2) 2

Chapter 15, Solution 13.

(a) tf (t )

← →

−

d F (s) ds

If f(t) = cost, then F ( s )=

s d ( s 2 + 1)(1) − s (2s + 1) F s and ( ) = ds s2 +1 ( s 2 + 1) 2 L (t cos t ) =

s2 + s −1 ( s 2 + 1) 2

(b) Let f(t) = e-t sin t. F (s) =

1 1 = 2 2 ( s + 1) + 1 s + 2s + 2

dF ( s 2 + 2s + 2)(0) − (1)(2s + 2) = ds ( s 2 + 2s + 2) 2

L (e −t t sin t ) = −

(c )

f (t ) t

dF 2( s + 1) = 2 ds ( s + 2s + 2) 2

∞

← →

∫ F (s)ds s

Let f (t ) = sin βt , then F ( s ) = ∞

β s +β2 2

β 1 s  sin βt  L =∫ 2 ds = β tan −1 2  β β  t  s s +β

∞ s

=

π 2

− tan −1

s

β

Chapter 15, Solution 14.

 5t 0 < t <1 f (t) =  10 − 5t 1 < t < 2 We may write f ( t ) as f ( t ) = 5t [ u ( t ) − u ( t − 1)] + (10 − 5t ) [ u ( t − 1) − u ( t − 2)] = 5t u ( t ) − 10 ( t − 1) u ( t − 1) + 5 ( t − 2) u ( t − 2) F(s) =

5 10 -s 5 -2s − e + 2e s2 s2 s

F(s) =

5 ( 1 − 2 e -s + e - 2 s ) s2

= tan −1

β s

Chapter 15, Solution 15. f ( t ) = 10 [ u ( t ) − u ( t − 1) − u ( t − 1) + u ( t − 2)]

1 2 e -2s  10 F(s) = 10  − e -s + = (1 − e -s ) 2 s  s s s

Chapter 15, Solution 16.

f ( t ) = 5 u ( t ) − 3 u ( t − 1) + 3 u ( t − 3) − 5 u ( t − 4)

F(s) =

1 [ 5 − 3 e -s + 3 e - 3 s − 5 e - 4 s ] s

Chapter 15, Solution 17. 0 t<0  t 2 0 < t < 1 f (t) =  1 1< t < 3  0 t>3 f ( t ) = t 2 [ u ( t ) − u ( t − 1)] + 1[ u ( t − 1) − u ( t − 3)] = t 2 u ( t ) − ( t − 1) 2 u ( t − 1) + (-2t + 1) u ( t − 1) + u ( t − 1) − u ( t − 3) = t 2 u ( t ) − ( t − 1) 2 u ( t − 1) − 2 ( t − 1) u ( t − 1) − u ( t − 3) F(s) =

2 2 -s e -3s -s − − ( 1 e ) e − s3 s2 s

Chapter 15, Solution 18. (a)

g ( t ) = u ( t ) − u ( t − 1) + 2 [ u ( t − 1) − u ( t − 2)] + 3 [ u ( t − 2) − u ( t − 3)] = u ( t ) + u ( t − 1) + u ( t − 2) − 3 u ( t − 3)

1 G (s) = (1 + e -s + e - 2s − 3 e - 3s ) s

(b)

h ( t ) = 2 t [ u ( t ) − u ( t − 1)] + 2 [ u ( t − 1) − u ( t − 3)] + (8 − 2 t ) [ u ( t − 3) − u ( t − 4)]

= 2t u ( t ) − 2 ( t − 1) u ( t − 1) − 2 u ( t − 1) + 2 u ( t − 1) − 2 u ( t − 3) − 2 ( t − 3) u ( t − 3) + 2 u ( t − 3) + 2 ( t − 4) u ( t − 4) = 2t u ( t ) − 2 ( t − 1) u ( t − 1) − 2 ( t − 3) u ( t − 3) + 2 ( t − 4) u ( t − 4) H(s) =

2 2 - 3s 2 - 4 s 2 -s + 2 e = 2 (1 − e -s − e - 3s + e -4s ) 2 (1 − e ) − 2 e s s s s

Chapter 15, Solution 19. Since L[ δ( t )] = 1 and T = 2 , F(s) =

1 1 − e - 2s

Chapter 15, Solution 20.

Let

g 1 ( t ) = sin(πt ), 0 < t < 1 = sin( πt ) [ u ( t ) − u ( t − 1)] = sin(πt ) u ( t ) − sin(πt ) u ( t − 1)

Note that sin(π ( t − 1)) = sin(πt − π) = - sin(πt ) . So, g1 ( t ) = sin( πt) u(t) + sin( π( t - 1)) u(t - 1) G 1 (s) =

π (1 + e -s ) s + π2 2

G 1 (s) π (1 + e -s ) G (s) = = 1 − e -2s (s 2 + π 2 )(1 − e - 2s )

Chapter 15, Solution 21.

T = 2π Let

 t  f 1 ( t ) = 1 −  [ u ( t ) − u ( t − 1)]  2π   t 1 1  f1 ( t ) = u ( t ) − u(t) + ( t − 1) u ( t − 1) − 1 −  u ( t − 1)  2π  2π 2π  1 1 e -s 1  -s 1 [ 2π + (-2π + 1) e -s ] s + [ - 1 + e -s ]  e ⋅ = F1 (s) = − 1 + + + s 2πs 2 2πs 2  2π  s 2πs 2 F(s) =

[ 2π + (-2π + 1) e -s ] s + [ - 1 + e -s ] F1 (s) = 1 − e -Ts 2πs 2 (1 − e - 2 πs )

Chapter 15, Solution 22. (a)

Let

g1 ( t ) = 2t, 0 < t < 1 = 2 t [ u ( t ) − u ( t − 1)] = 2t u ( t ) − 2 ( t − 1) u ( t − 1) + 2 u ( t − 1)

2 2 e -s 2 -s G 1 (s) = 2 − 2 + e s s s G (s) =

G 1 (s) , T =1 1 − e -sT

2 (1 − e -s + s e -s ) G (s) = s 2 (1 − e -s ) (b)

Let h = h 0 + u ( t ) , where h 0 is the periodic triangular wave. Let h 1 be h 0 within its first period, i.e.  2t 0 < t <1 h 1 (t) =   4 − 2t 1 < t < 2 h 1 ( t ) = 2 t u ( t ) − 2 t u ( t − 1) + 4u ( t − 1) − 2 t u ( t − 1) − 2 ( t − 2) u ( t − 2) h 1 ( t ) = 2 t u ( t ) − 4 ( t − 1) u ( t − 1) − 2 ( t − 2) u ( t − 2)

2 4 -s 2 e -2s 2 = 2 (1 − e -s ) 2 2 2 − 2 e − s s s s

H 1 (s) =

2 (1 − e -s ) 2 H 0 (s) = 2 s (1 − e -2s ) 1 2 (1 − e -s ) 2 H(s) = + 2 s s (1 − e - 2s )

Chapter 15, Solution 23.

(a)

Let

1 0 < t <1 f1 ( t ) =  - 1 1 < t < 2

f 1 ( t ) = [ u ( t ) − u ( t − 1)] − [ u ( t − 1) − u ( t − 2)] f 1 ( t ) = u ( t ) − 2 u ( t − 1) + u ( t − 2)

1 1 F1 (s) = (1 − 2 e -s + e -2s ) = (1 − e -s ) 2 s s F(s) =

F1 (s) , T=2 (1 − e -sT )

(1 − e -s ) 2 F(s) = s (1 − e - 2s ) (b)

Let h 1 ( t ) = t 2 [ u ( t ) − u ( t − 2)] = t 2 u ( t ) − t 2 u ( t − 2) h 1 ( t ) = t 2 u ( t ) − ( t − 2) 2 u ( t − 2) − 4 ( t − 2) u ( t − 2) − 4 u ( t − 2) H 1 (s) =

2 4 4 - 2s ) − 2 e -2s − e -2s 3 (1 − e s s s

H(s) =

H 1 (s) , T=2 (1 − e -Ts )

H(s) =

2 (1 − e -2s ) − 4s e -2s (s + s 2 ) s 3 (1 − e - 2s )

Chapter 15, Solution 24.

(a)

10s 4 + s 2 s →∞ s + 6s + 5

f (0) = lim sF(s) = lim s →∞

= lim s →∞

10 +

1 s3

1 6 5 + + s 2 s3 s 4

=

10 =∞ 0

10s 4 + s =0 s → 0 s 2 + 6s + 5

f (∞) = lim sF(s) = lim s →0

(b)

s2 + s =1 2 s →∞ s − 4s + 6

f (0) = lim sF(s) = lim s →∞

The complex poles are not in the left-half plane. f (∞) does not exist (c)

2s 3 + 7s 2 s →∞ (s + 1)(s + 2)(s + 2s + 5)

f (0) = lim sF(s) = lim s →∞

2 7 + 0 s s3 = lim = =0 s →∞  1  2  2 5  1 1 +  1 +  1 + + 2  s  s s   s 

2s 3 + 7s 0 f (∞) = lim sF(s) = lim = =0 2 s →0 s → 0 (s + 1)(s + 2)(s + 2s + 5) 10

Chapter 15, Solution 25.

(a)

(8)(s + 1)(s + 3) s →∞ (s + 2)(s + 4)

f (0) = lim sF(s) = lim s →∞

 1 3  (8) 1 +  1 +  s  s  = lim  =8 s →∞  2  4  1 +  1 +  s  s 

(8)(1)(3) =3 s → 0 ( 2)( 4)

f (∞) = lim sF(s) = lim s →0

(b)

6s (s − 1) 4 s →∞ s − 1

f (0) = lim sF(s) = lim s →∞

1 1 6 2 − 4  s s  0 f (0) = lim = =0 1 s →∞ 1 1− 4 s

All poles are not in the left-half plane. f (∞) does not exist

Chapter 15, Solution 26.

(a)

s 3 + 3s =1 s →∞ s 3 + 4s 2 + 6

f (0) = lim sF(s) = lim s →∞

Two poles are not in the left-half plane. f (∞) does not exist

(b)

s 3 − 2s 2 + s f (0) = lim sF(s) = lim 2 s →∞ s →∞ (s − 2)(s + 2s + 4) 2 1 + 2 s s = lim =1 s →∞  2  2 4  1 −  1 + + 2   s  s s  1−

One pole is not in the left-half plane. f (∞) does not exist

Chapter 15, Solution 27. (a)

f ( t ) = u(t ) + 2 e -t

(b)

G (s) =

3 (s + 4) − 11 11 = 3− s+4 s+4

g( t ) = 3 δ(t ) − 11 e -4t (c)

H(s) =

A = 2, H(s) =

4 A B = + (s + 1)(s + 3) s + 1 s + 3

B = -2 2 2 − s +1 s + 3

h ( t ) = 2 e -t − 2 e -3t

(d)

J (s) =

B=

12 A B C = + 2 2 + (s + 2) (s + 4) s + 2 (s + 2) s+4

12 = 6, 2

C=

12 =3 (-2) 2

12 = A (s + 2) (s + 4) + B (s + 4) + C (s + 2) 2 Equating coefficients : s2 : 0= A+C  → A = -C = -3 s1 : s0 :

0 = 6A + B + 4C = 2A + B  → B = -2A = 6 12 = 8A + 4B + 4C = -24 + 24 + 12 = 12

J (s) =

-3 6 3 + 2 + s + 2 (s + 2) s+4

j( t ) = 3 e -4t − 3 e -2t + 6 t e -2t

Chapter 15, Solution 28.

(a)

2(−2) 2(−4) −2 4 + F(s) = 2 + − 2 = s+5 s+3 s+5 s+3 f ( t ) = (−2e − 3t + 4e − 5t )ut ( t )

(b) H(s) =

3s + 11 (s + 1)(s 2 + 2s + 5)

=

A Bs + C + s + 1 s 2 + 2s + 5

3s + 11 = A(s 2 + 2s + 5) + (Bs + C)(s + 1) = (A + B)s 2 + (2A + B + C)s + 5A + C 5A + C = 11; A = −B; − B + C = 3, B = C − 3 → A = 2; B = −2; C = 1 H(s) =

(

)

2 − 2s + 1 + → h ( t ) = 2e − t − 2e − t cos 2t + 1.5e − t sin 2t u ( t ) s + 1 s 2 + 2s + 5

Chapter 15, Solution 29.

V(s) =

2 As + B ; 2s 2 + 8s + 26 + As 2 + Bs = 2s + 26 → A = −2 and B = −6 + 2 2 s (s + 2) + 3

V(s) =

2 2(s + 2) 2 3 − − s (s + 2) 2 + 3 2 3 (s + 2) 2 + 3 2

2 v(t) = 2u ( t ) − 2e − 2 t cos 3t − e − 2 t sin 3t , t ≥ 0 3

Chapter 15, Solution 30.

(a) H1 (s) =

2(s + 2) + 2 (s + 2) 2 + 32

=

2(s + 2) (s + 2) 2 + 3 2

h1 ( t ) = 2e −2 t cos 3t +

(b) H 2 (s) =

s2 + 4 (s + 1) 2 (s 2 + 2s + 5)

+

2 3 3 (s + 2) 2 + 32

2 −2 t e sin 3t 3 =

A B Cs + D + + (s + 1) (s + 1) 2 (s 2 + 2s + 5)

s 2 + 4 = A(s + 1)(s 2 + 2s + 5) + B(s 2 + 2s + 5) + Cs(s + 1) 2 + D(s + 1) 2 or s 2 + 4 = A(s 3 + 3s 2 + 7s + 5) + B(s 2 + 2s + 5) + C(s 3 + 2s 2 + s) + D(s 2 + 2s + 1) Equating coefficients:

s3 :

0= A+C

s2 :

1 = 3A + B + 2C + D = A + B + D

s:

 →

C = −A

0 = 7 A + 2B + C + 2D = 6A + 2B + 2D = 4A + 2

constant :

4 = 5A + 5B + D = 4A + 4B + 1

 →

 →

A = −1 / 2, C = 1 / 2

B = 5 / 4, D = 1 / 4

H 2 (s) =

1 −2 5 2s + 1  1  − 2 5 2(s + 1) − 1  =  + + + +    4  (s + 1) (s + 1) 2 (s 2 + 2s + 5)  4  (s + 1) (s + 1) 2 (s + 1) 2 + 2 2 ) 

Hence, h 2 (t) =

(c ) H 3 (s) =

h 3 (t) =

(

)

1 − 2e − t + 5te − t + 2e − t cos 2t − 0.5e − t sin 2t u ( t ) 4 (s + 2)e − s

 A B  1 −s  1 1  + = e−s  +  = e  (s + 1)(s + 3)  (s + 1) (s + 3)   (s + 1) (s + 3)  2

(

)

1 −( t −1) e + e −3( t −1) u ( t − 1) 2

Chapter 15, Solution 31.

(a)

F(s) =

10s A B C = + + (s + 1)(s + 2)(s + 3) s + 1 s + 2 s + 3

A = F(s) (s + 1) s= -1 =

- 10 = -5 2

B = F(s) (s + 2) s= -2 =

- 20 = 20 -1

C = F(s) (s + 3) s= -3 =

- 30 = -15 2

F(s) =

-5 20 15 + − s +1 s + 2 s + 3

f ( t ) = - 5 e -t + 20 e -2t − 15 e -3t (b)

F(s) =

2s 2 + 4s + 1 A B C D + + 3 = 2 + (s + 1)(s + 2) s + 1 s + 2 (s + 2) (s + 2) 3

A = F(s) (s + 1) s= -1 = -1 D = F(s) (s + 2) 3

s = -2

= -1

2s 2 + 4s + 1 = A(s + 2)(s 2 + 4s + 4) + B(s + 1)(s 2 + 4s + 4)

+ C(s + 1)(s + 2) + D(s + 1) Equating coefficients : s3 : 0= A+B  → B = -A = 1 s2 : s1 : s0 : F(s) =

2 = 6A + 5B + C = A + C  → C = 2 − A = 3 4 = 12A + 8B + 3C + D = 4A + 3C + D 4 = 6+A+ D  → D = -2 − A = -1 1 = 8A + 4B + 2C + D = 4A + 2C + D = -4 + 6 − 1 = 1 -1 1 3 1 + + 2 − s + 1 s + 2 (s + 2) (s + 2) 3

f(t) = -e - t + e -2t + 3 t e -2t −

t 2 -2t e 2

 t2  f ( t ) = - e -t + 1 + 3 t −  e - 2t 2  (c)

F(s) =

s +1 A Bs + C = + 2 2 (s + 2)(s + 2s + 5) s + 2 s + 2s + 5

A = F(s) (s + 2) s= -2 =

-1 5

s + 1 = A (s 2 + 2s + 5) + B (s 2 + 2s) + C (s + 2) Equating coefficients : 1 5

s2 :

0= A+B  → B = -A =

s1 : s0 :

1 = 2A + 2B + C = 0 + C  → C = 1 1 = 5A + 2C = -1 + 2 = 1

F(s) =

-1 5 1 5⋅ s +1 -1 5 1 5 (s + 1) 45 + + 2 2 = 2 2 + s + 2 (s + 1) + 2 s + 2 (s + 1) + 2 (s + 1) 2 + 2 2

f ( t ) = - 0.2 e -2t + 0.2 e -t cos( 2t ) + 0.4 e -t sin( 2t )

Chapter 15, Solution 32.

(a)

F(s) =

8 (s + 1)(s + 3) A B C = + + s (s + 2)(s + 4) s s + 2 s + 4

A = F(s) s s= 0 =

(8)(3) =3 (2)(4)

B = F(s) (s + 2) s= -2 =

(8)(-1) =2 (-4)

C = F(s) (s + 4) s= -4 =

(8)(-1)(-3) =3 (-4)(-2)

F(s) =

3 2 3 + + s s+2 s+4

f ( t ) = 3 u(t ) + 2 e -2t + 3 e -4t (b)

F(s) =

s 2 − 2s + 4 A B C + + 2 = (s + 1)(s + 2) s + 1 s + 2 (s + 2) 2

s 2 − 2s + 4 = A (s 2 + 4s + 4) + B (s 2 + 3s + 2) + C (s + 1) Equating coefficients : s2 : 1= A+ B  → B = 1 − A 1 s : - 2 = 4A + 3B + C = 3 + A + C 0 s : 4 = 4A + 2B + C = -B − 2  → B = -6 A = 1− B = 7

F(s) =

C = -5 - A = -12

7 6 12 − − s + 1 s + 2 (s + 2) 2

f ( t ) = 7 e -t − 6 (1 + 2 t ) e -2t (c)

F(s) =

s2 +1 A Bs + C = + 2 2 (s + 3)(s + 4s + 5) s + 3 s + 4s + 5

s 2 + 1 = A (s 2 + 4s + 5) + B (s 2 + 3s) + C (s + 3)

Equating coefficients : s2 : 1= A+ B  → B = 1 − A s1 : 0

s :

0 = 4A + 3B + C = 3 + A + C  → A + C = -3 1 = 5A + 3C = -9 + 2A  → A = 5

B = 1 − A = -4 F(s) =

C = -A − 3 = -8

4 (s + 2) 5 4s + 8 5 − = − 2 s + 3 (s + 2) + 1 s + 3 (s + 2) 2 + 1

f ( t ) = 5 e -3t − 4 e -2t cos(t )

Chapter 15, Solution 33.

(a)

F(s) =

6 (s − 1) 6 As + B C = 2 = 2 + 4 s −1 (s + 1)(s + 1) s + 1 s + 1

6 = A (s 2 + s) + B (s + 1) + C (s 2 + 1) Equating coefficients : s2 : 0= A+C  → A = -C s1 :

0= A+B  → B = -A = C

0

6 = B + C = 2B  → B = 3

s :

A = -3 ,

F(s) =

B = 3,

C=3

3 - 3s + 3 3 - 3s 3 = + 2 + 2 + 2 s +1 s +1 s +1 s +1 s +1

f ( t ) = 3 e -t + 3 sin( t ) − 3 cos(t ) (b)

F(s) =

s e - πs s2 +1

f ( t ) = cos(t − π ) u(t − π )

(c)

F(s) =

8 A B C D + + 3 = 2 + s (s + 1) s s + 1 (s + 1) (s + 1) 3

A = 8,

D = -8

8 = A (s 3 + 3s 2 + 3s + 1) + B (s 3 + 2s 2 + s) + C (s 2 + s) + D s Equating coefficients : s3 : 0= A+B  → B = -A

s2 :

0 = 3A + 2B + C = A + C  → C = -A = B

s1 : s0 :

0 = 3A + B + C + D = A + D  → D = -A A = 8, B = −8, C = −8, D = −8

F(s) =

8 8 8 8 − − 2 − s s + 1 (s + 1) (s + 1) 3

f ( t ) = 8 [ 1 − e -t − t e -t − 0.5 t 2 e -t ] u(t )

Chapter 15, Solution 34.

(a)

F(s) = 10 +

s2 + 4 − 3 3 = 11 − 2 2 s +4 s +4

f ( t ) = 11 δ(t ) − 1.5 sin( 2t ) (b)

G (s) =

e -s + 4 e -2s (s + 2)(s + 4)

Let

1 A B = + (s + 2)(s + 4) s + 2 s + 4

A =1 2

G (s) =

B=1 2

 1 1  e -s  1 1    + 2 e -2s   + + s + 2 s + 4 2 s + 2 s + 4

g( t ) = 0.5 [ e -2(t -1) − e -4(t -1) ] u(t − 1) + 2 [ e -2(t - 2) − e -4(t - 2) ] u(t − 2)

(c)

s +1 A B C = + + s (s + 3)(s + 4) s s + 3 s + 4

Let

A = 1 12 ,

B = 2 3,

C = -3 4

1 1 23 3 4  -2s e H(s) =  ⋅ + − 12 s s + 3 s + 4  1 2  3 h ( t ) =  + e - 3(t - 2) − e -4(t - 2)  u(t − 2)  12 3  4

Chapter 15, Solution 35. (a)

G (s) =

Let

B = -1

A = 2, G (s) =

s+3 A B = + (s + 1)(s + 2) s + 1 s + 2

2 1 − s +1 s + 2

 → g( t ) = 2 e - t − e -2t

F(s) = e -6s G (s)  → f ( t ) = g( t − 6) u ( t − 6) -(t - 6) -2(t - 6) ] u( t − 6) f (t) = [ 2 e −e (b)

Let

G (s) =

A = 1 3,

1 A B = + (s + 1)(s + 4) s + 1 s + 4 B = -1 3

G (s) =

1 1 − 3 (s + 1) 3 (s + 4)

g( t ) =

1 -t [ e − e -4t ] 3

F(s) = 4 G (s) − e -2t G (s) f ( t ) = 4 g( t ) u ( t ) − g ( t − 2) u ( t − 2) f (t) =

4 -t [ e − e -4t ] u(t ) − 1 [ e -(t-2) − e -4(t-2) ] u(t − 2) 3 3

(c)

Let

G (s) =

s A Bs + C = + 2 2 (s + 3)(s + 4) s + 3 s + 4

A = - 3 13

s = A (s 2 + 4) + B (s 2 + 3s) + C (s + 3) Equating coefficients : s2 : 0= A+B  → B = -A 1 = 3B + C s1 : 0 0 = 4A + 3C s : A = - 3 13 ,

13 G (s) =

B = 3 13 ,

C = 4 13

- 3 3s + 4 + s + 3 s2 + 4

13 g( t ) = -3 e -3t + 3 cos(2t ) + 2 sin(2t ) F(s) = e -s G (s) f ( t ) = g( t − 1) u ( t − 1)

f (t) =

1 [ - 3 e -3(t-1) + 3 cos(2 (t − 1)) + 2 sin( 2 (t − 1))] u(t − 1) 13

Chapter 15, Solution 36. (a)

X(s) =

1 A B C D = + 2+ + s (s + 2)(s + 3) s s s+2 s+3

B = 1 6,

2

C =1 4,

D = -1 9

1 = A (s 3 + 5s 2 + 6s) + B (s 2 + 5s + 6) + C (s 3 + 3s 2 ) + D (s 3 + 2s 2 ) Equating coefficients : 0 = A+C+D s3 : 2 s : 0 = 5A + B + 3C + 2D = 3A + B + C 1 s : 0 = 6 A + 5B s0 : 1 = 6B  → B = 1 6

A = - 5 6 B = - 5 36

(b)

X(s) =

- 5 36 1 6 1 4 19 + 2 + − s s s+2 s+3

x(t) =

-5 1 1 1 u(t ) + t + e - 2t − e - 3t 36 6 4 9

Y(s) =

1 A B C + + 2 = s (s + 1) s s + 1 (s + 1) 2

A = 1,

C = -1

1 = A (s 2 + 2s + 1) + B (s 2 + s) + C s Equating coefficients : 0= A+B  → B = -A s2 :

s1 : s0 :

0 = 2A + B + C = A + C  → C = -A 1 = A, B = -1, C = -1

1 1 1 Y(s) = − − s s + 1 (s + 1) 2 y( t ) = u(t ) − e -t − t e -t (c)

Z(s) =

A B Cs + D + + 2 s s + 1 s + 6s + 10

A = 1 10 ,

B = -1 5

1 = A (s 3 + 7s 2 + 16s + 10) + B (s 3 + 6s 2 + 10s) + C (s 3 + s 2 ) + D (s 2 + s) Equating coefficients : 0 = A+ B+C s3 : 2 s : 0 = 7 A + 6 B + C + D = 6 A + 5B + D 1 s : 0 = 16A + 10B + D = 10A + 5B  → B = -2A

s0 :

1 = 10A  → A = 1 10

A = 1 10 ,

B = -2A = - 1 5 ,

C = A = 1 10 ,

D = 4A =

4 10

1 2 s+4 10 Z(s) = − + 2 s s + 1 s + 6s + 10 1 2 s+3 1 10 Z(s) = − + + 2 s s + 1 (s + 3) + 1 (s + 3) 2 + 1 z( t ) = 0.1 [ 1 − 2 e -t + e -3t cos(t ) + e -3t sin( t )] u(t )

Chapter 15, Solution 37.

(a)

Let

P(s) =

12 A Bs + C = + 2 2 s (s + 4) s s + 4

A = P(s) s s=0 = 12 4 = 3 12 = A (s 2 + 4) + B s 2 + C s Equating coefficients : s0 : 12 = 4A  → A = 3 1 s : 0=C 2 s : 0= A+B  → B = -A = -3 P(s) =

3 3s − 2 s s +4

p( t ) = 3 u ( t ) − 3 cos(2t ) F(s) = e -2s P(s) f ( t ) = 3 [ 1 − cos( 2(t − 2))] u(t − 2) (b)

Let

G (s) =

2s + 1 As + B Cs + D = 2 + 2 (s + 1)(s + 9) s + 1 s 2 + 9 2

2s + 1 = A (s 3 + 9s) + B (s 2 + 9) + C (s 3 + s) + D (s 2 + 1) Equating coefficients : 0= A+C  → C = -A s3 :

s2 :

0 = B+ D  → D = -B

s1 :

2 = 9A + C = 8A  → A = 2 8, C = - 2 8

0

s :

1 = 9B + D = 8B  → B = 1 8 , D = - 1 8

1  2s + 1  1  2s + 1  −   G (s) =  2 8  s + 1 8  s 2 + 9 

(c)

G (s) =

1 s 1 1 1 s 1 1 ⋅ 2 + ⋅ 2 − ⋅ 2 − ⋅ 2 4 s +1 8 s +1 4 s + 9 8 s + 9

g( t ) =

1 1 1 1 cos(t ) + sin( t ) − cos( 3t ) − sin( 3t ) 4 8 4 24

Let

9s2 36s + 117 H(s) = 2 = 9− 2 s + 4s + 13 s + 4s + 13

H(s) = 9 − 36 ⋅

s+2 3 2 2 − 15 ⋅ (s + 2) + 3 (s + 2) 2 + 3 2

h ( t ) = 9 δ(t ) − 36e -2 t cos( 3t ) − 15e -2 t sin( 3t )

Chapter 15, Solution 38.

(a)

F(s) =

s 2 + 4s s 2 + 10s + 26 − 6s − 26 = s 2 + 10s + 26 s 2 + 10s + 26

F(s) = 1 −

6s + 26 s + 10s + 26

F(s) = 1 −

6 (s + 5) 4 2 2 + (s + 5) + 1 (s + 5) 2 + 12

2

f ( t ) = δ(t ) − 6 e -t cos(5t ) + 4 e -t sin( 5t )

(b)

5s 2 + 7s + 29 A Bs + C = + 2 F(s) = 2 s (s + 4s + 29) s s + 4s + 29

5s 2 + 7s + 29 = A (s 2 + 4s + 29) + B s 2 + C s Equating coefficients : s0 : 29 = 29A  → A = 1

s1 :

7 = 4A + C  → C = 7 − 4A = 3

s2 :

5= A+B  → B = 5 − A = 4 B = 4,

A = 1,

C=3

4 (s + 2) 1 4s + 3 1 5 F(s) = + 2 = + 2 2 − s s + 4s + 29 s (s + 2) + 5 (s + 2) 2 + 5 2 f ( t ) = u(t ) + 4 e -2t cos(5t ) − e -2t sin( 5t )

Chapter 15, Solution 39.

(a)

2s 3 + 4s 2 + 1 As + B Cs + D F(s) = 2 = 2 + 2 2 (s + 2s + 17)(s + 4s + 20) s + 2s + 17 s + 4s + 20 s 3 + 4s 2 + 1 = A(s 3 + 4s 2 + 20s) + B(s 2 + 4s + 20) + C(s3 + 2s 2 + 17s) + D(s 2 + 2s + 17) Equating coefficients : s3 : 2= A+C 2 s : 4 = 4 A + B + 2C + D 1 0 = 20A + 4B + 17C + 2D s : 0 s : 1 = 20B + 17 D Solving these equations (Matlab works well with 4 unknowns), D = 21 A = -1.6 , B = -17.8 , C = 3 .6 , F(s) =

- 1.6s − 17.8 3.6s + 21 + 2 2 s + 2s + 17 s + 4s + 20

F(s) =

(-1.6)(s + 1) (-4.05)(4) (3.6)(s + 2) (3.45)(4) 2 2 + 2 2 + 2 2 + (s + 1) + 4 (s + 1) + 4 (s + 2) + 4 (s + 2) 2 + 4 2

f ( t ) = - 1.6 e -t cos(4t ) − 4.05 e -t sin( 4t ) + 3.6 e -2t cos(4t ) + 3.45 e -2t sin( 4t )

(b)

F(s) =

s2 + 4 As + B Cs + D = 2 + 2 2 2 (s + 9)(s + 6s + 3) s + 9 s + 6s + 3

s 2 + 4 = A (s 3 + 6s 2 + 3s) + B (s 2 + 6s + 3) + C (s 3 + 9s) + D (s 2 + 9) Equating coefficients : s3 : 0= A+C  → C = -A 2 s : 1 = 6A + B + D 1 s : 0 = 3A + 6B + 9C = 6B + 6C  → B = -C = A 4 = 3B + 9D s0 : Solving these equations, A = 1 12 , B = 1 12 , 12 F(s) =

D = 5 12

s +1 -s+5 + 2 2 s + 9 s + 6s + 3

s 2 + 6s + 3 = 0  →

Let

G (s) =

E=

-s+5 s + 5.449

F=

-s+5 s + 0.551

G (s) =

C = - 1 12 ,

- 6 ± 36 - 12 = -0.551, - 5.449 2

-s+5 E F = + s + 6s + 3 s + 0.551 s + 5.449 2

s = -0.551

= 1.133

s = -5.449

= - 2.133

1.133 2.133 − s + 0.551 s + 5.449

12 F(s) =

s 1 3 1.133 2.133 ⋅ 2 − 2 + 2 + s +3 3 s +3 s + 0.551 s + 5.449 2

f ( t ) = 0.08333 cos( 3t ) + 0.02778 sin( 3t ) + 0.0944 e -0.551t − 0.1778 e -5.449t

Chapter 15, Solution 40.  4s 2 + 7s + 13  A Bs + C Let H(s) =  + = 2 2  (s + 2)(s + 2s + 5)  s + 2 s + 2s + 5 4s 2 + 7s + 13 = A(s 2 + 2s + 5) + B(s 2 + 2s) + C(s + 2) Equating coefficients gives: s2 :

4=A+B

s:

7 = 2A + 2B + C

 →

C = −1

13 = 5A + 2C

 →

5A = 15 or A = 3, B = 1

constant : H(s) =

3 s −1 3 (s + 1) − 2 + = + 2 s + 2 s + 2s + 5 s + 2 (s + 1) 2 + 2 2

Hence, h ( t ) = 3e −2 t + e − t cos 2t − e − t sin 2t = 3e −2 t + e − t (A cos α cos 2t − A sin α sin 2t ) where A cos α = 1,

A sin α = 1

Thus, h(t) =

[ 2e

−t

 →

A = 2,

α = 45 o

]

cos(2 t + 45 o ) + 3e −2 t u ( t )

Chapter 15, Solution 41.

Let y(t) = f(t)*h(t). For 0 < t < 1, f ( t − λ) h(λ )

0 t

t

y( t ) = ∫ (1)4λdλ = 2λ2 0 = 2t 2 0

t 1

2

λ

For 1
h (λ )

0 1

t

0

1

1 t

2

t

t

y( t ) = ∫ (1)4λdλ + ∫ (1)(8 − 4λ )dλ = 2λ2 0 + (8λ − 2λ2 ) 1 = 8t − 2t 2 − 4 For 3 < t < 4

h (λ)

0 2

2

1

t-2

y( t ) = ∫ (8 − 4λ )λdλ = 8λ − 2λ2 t − 2 = 32 − 16t + 2t 2 t −2

Thus,  2t 2 , 0 < t < 1   8t - 2t 2 − 4, 1 < t < 3 y( t ) =  2 32 - 16t + 2t , 3 < t < 4  0, otherwise 

2

f ( t − λ)

3

t

4

λ

Chapter 15, Solution 42.

(a)

For 0 < t < 1 , f1 ( t − λ) and f 2 (λ) overlap from 0 to t, as shown in Fig. (a). y( t ) = f1 ( t ) ∗ f 2 ( t ) = ∫0

t

f1(t - λ)

λ2 (1)(λ) dλ = 2

f2(λ)

1

t-1

0 t

1

t2 = 2

t 0

1

0

λ

(a)

t-1 1

t

λ

(b)

For 1 < t < 2 , f1 ( t − λ) and f 2 (λ) overlap as shown in Fig. (b). y( t ) = ∫t −1 (1)(λ) dλ = 1

λ2 2

1 t −1

=t−

t2 2

For t > 2 , there is no overlap. Therefore,

 t 2 2, 0
For 0 < t < 1 , the two functions overlap as shown in Fig. (c). y( t ) = f1 ( t ) ∗ f 2 ( t ) = ∫0 (1)(1) dλ = t t

f1(t - λ)

f2(λ)

1

t-1

0 t

(c)

1

λ

1

0

t-1 1

(d)

t

λ

For 1 < t < 2 , the functions overlap as shown in Fig. (d).

y( t ) = ∫t −1 (1)(1) dλ = λ 1t −1 = 2 − t 1

For t > 2 , there is no overlap. Therefore, 0
(c)

For t < -1 , there is no overlap. For - 1 < t < 0 , f1 ( t − λ) and f 2 (λ) overlap as shown in Fig. (e).   λ2 y( t ) = f 1 ( t ) ∗ f 2 ( t ) = ∫-1 (1)(λ + 1) dλ =  + λ -t1  2 t

y( t ) =

1 2 1 ( t + 2t + 1) = ( t + 1) 2 2 2

f2(t - λ)

f1(λ)

1

-1

t

0

1

λ

1

-1

(e)

0 t

(f)

For 0 < t < 1 , the functions overlap as shown in Fig. (f). y( t ) = ∫-1 (1)(λ + 1) dλ + ∫0 (1)(1 − λ) dλ 0

t

  λ2  λ2  y( t ) =  + λ  0-1 +  λ −  0t 2  2  y( t ) =

1 (1 + 2t − t 2 ) 2

For t > 1 , the two functions overlap.

1

λ

y( t ) = ∫-1 (1)(λ + 1) dλ + ∫0(1)(1 − λ) dλ 0

1

1  1 λ2  1 1  y( t ) = + λ −  0 = + 1 − = 1 2  2 2 2 Therefore,

0, t < -1   0.5(t 2 + 2t + 1), - 1 < t < 0  y( t ) =  2 0.5(-t + 2t + 1), 0 < t < 1  1, t >1

Chapter 15, Solution 43. (a)

For 0 < t < 1 , x ( t − λ) and h (λ) overlap as shown in Fig. (a). y( t ) = x ( t ) ∗ h ( t ) = ∫0 (1)(λ) dλ = t

x(t - λ)

λ2 2

1

t 0

=

t2 2

1 h(λ)

t-1

0 t

1

λ

0

t-1 1

(a)

t

λ

(b)

For 1 < t < 2 , x ( t − λ) and h (λ) overlap as shown in Fig. (b). y( t ) = ∫t −1 (1)(λ) dλ + ∫1 (1)(1) dλ = 1

t

λ2 2

1 t −1

For t > 2 , there is a complete overlap so that y( t ) = ∫t −1 (1)(1) dλ = λ tt −1 = t − ( t − 1) = 1 t

+ λ 1t =

-1 2 t + 2t − 1 2

Therefore,  t 2 2, 02   0, otherwise

(b)

For t > 0 , the two functions overlap as shown in Fig. (c). y( t ) = x ( t ) ∗ h ( t ) = ∫0 (1) 2 e -λ dλ = -2 e -λ t

x(t-λ)

2

t 0

h(λ) = 2e-λ

1

0

t

λ

(c) Therefore, y( t ) = 2 (1 − e -t ), t > 0 (c)

For - 1 < t < 0 , x ( t − λ) and h (λ) overlap as shown in Fig. (d). y( t ) = x ( t ) ∗ h ( t ) = ∫0 (1)(λ) dλ = t +1

x(t - λ)

λ2 2

1

t-1 -1

t 0

t +1 0

=

1 ( t + 1) 2 2

h(λ)

t+1 1

2

λ

(d) For 0 < t < 1 , x ( t − λ) and h (λ) overlap as shown in Fig. (e). y( t ) = ∫0 (1)(λ) dλ + ∫1 (1)(2 − λ) dλ 1

t +1

y( t ) =

λ2 2

1 0

 -1 1 λ2  +  2λ −  1t +1 = t 2 + t + 2 2 2  1

-1 t-1

0 t

1 t+1 2

λ

(e) For 1 < t < 2 , x ( t − λ) and h (λ) overlap as shown in Fig. (f). y( t ) = ∫t −1 (1)(λ) dλ + ∫1 (1)(2 − λ) dλ 1

y( t ) =

λ2 2

2

1 t −1

 -1 1 λ2  +  2λ −  12 = t 2 + t + 2 2 2 

1

0

t-1 1

t

2 t+1

λ

(f)

For 2 < t < 3 , x ( t − λ) and h (λ) overlap as shown in Fig. (g).  λ2   y( t ) = ∫t −1 (1)(2 − λ) dλ = 2λ −  2  2

2 t −1

=

9 1 − 3t + t 2 2 2

1

0

1 t-1 2

(g)

t

t+1

λ

Therefore,  ( t 2 2 ) + t + 1 2, - 1 < t < 0  2 - ( t 2 ) + t + 1 2 , 0 < t < 2 y( t ) =  2  ( t 2 ) − 3t + 9 2, 2 < t < 3  0, otherwise

Chapter 15, Solution 44. (a)

For 0 < t < 1 , x ( t − λ) and h (λ) overlap as shown in Fig. (a). y( t ) = x ( t ) ∗ h ( t ) = ∫0 (1)(1) dλ = t t

x(t - λ)

h(λ)

1

t-1

0 t

1

2

λ

-1

(a) For 1 < t < 2 , x ( t − λ) and h (λ) overlap as shown in Fig. (b). y( t ) = ∫t −1 (1)(1) dλ + ∫1 (-1)(1) dλ = λ 1t −1 − λ 1t = 3 − 2 t 1

t

For 2 < t < 3 , x ( t − λ) and h (λ) overlap as shown in Fig. (c). y( t ) = ∫t −1 (1)(-1) dλ = -λ 2

2 t −1

= t−3

1

1

0

t-1 1

t

2

0

λ

-1

1 t-1

2 t

λ

-1

(b)

(c)

Therefore,

0
For t < 2 , there is no overlap. For 2 < t < 3 , f1 ( t − λ) and f 2 (λ) overlap, as shown in Fig. (d). y( t ) = f 1 ( t ) ∗ f 2 ( t ) =

 λ2  =  λt − 2 

∫

t

2

(1)( t − λ) dλ

 t t2  2 = − 2 t + 2 2 

f1(t - λ)

f2(λ)

1

0

1 t-1 2

t

3

4

5

λ

4

5

λ

(d)

1

0

1

2 t-1 3

(e)

t

For 3 < t < 5 , f1 ( t − λ) and f 2 (λ) overlap as shown in Fig. (e).  1 λ2  t  y( t ) = ∫t −1 (1)( t − λ) dλ = λt −  t −1 = 2 2  t

For 5 < t < 6 , the functions overlap as shown in Fig. (f).  5 -1 λ2  y( t ) = ∫t −1 (1)( t − λ) dλ =  λt −  5t −1 = t 2 + 5t − 12 2 2 

1

0

1

2

3

4 t-1 5

(f) Therefore,  ( t 2 2 ) − 2t + 2, 2
Chapter 15, Solution 45.

(a)

f ( t ) ∗ δ( t ) = ∫0 f (λ) δ( t − λ) dλ = f (λ) λ= t t

f (t ) ∗ δ(t ) = f (t ) (b)

f ( t ) ∗ u ( t ) = ∫0 f (λ) u ( t − λ) dλ t

1 λ< t Since u ( t − λ) =  0 λ> t f ( t ) ∗ u( t ) =

∫

t

o

f ( λ ) dλ

t

λ

Alternatively,

L{ f ( t ) ∗ u ( t )} =

F(s) s

t  F(s)   = f ( t ) ∗ u ( t ) = ∫ f (λ) dλ L−1  o  s 

Chapter 15, Solution 46.

(a)

Let y( t ) = x 1 ( t ) ∗ x 2 ( t ) = ∫0 x 2 ( t ) x 1 ( t − λ) dλ t

For 0 < t < 3 , x 1 ( t − λ) and x 2 (λ) overlap as shown in Fig. (a). y( t ) = ∫0 4 e -2λ e -(t -λ ) dλ = 4 e - t ∫0 e -λ dλ = 4 (e - t − e -2 t ) t

t

4

x2(λ)

x1(t-λ)

0

λ

3

t (a)

For t > 3 , the two functions overlap as shown in Fig. (b). y( t ) = ∫0 4 e -2 λ e -(t -λ ) dλ = 4 e - t ( - e -λ ) 3

3 0

= 4 e - t (1 − e -3 )

4

3

0 (b)

t

λ

Therefore,  4 (e - t − e -2t ), 0 < t < 3 y( t ) =  -t -3 t>3 4e (1 − e ),

(b)

For 1 < t < 2 , x 1 (λ) and x 2 ( t − λ) overlap as shown in Fig. (c). y( t ) = x 1 ( t ) ∗ x 2 ( t ) = ∫1 (1)(1) dλ = λ 1t = t − 1 t

x2(t - λ)

x1(λ)

1

0

t-1 1

t

2

λ

3

(c)

For 2 < t < 3 , the two functions overlap completely. y( t ) = ∫t −1 (1)(1) dλ = λ tt −1 = t − ( t − 1) = 1 t

For 3 < t < 4 , the two functions overlap as shown in Fig. (d). y( t ) = ∫t −1 (1)(1) dλ = λ 3t −1 = 4 − t 3

1

0

1

2 t-1 3

t

λ

(d)

Therefore,  t − 1, 1 < t < 2  1, 2
For - 1 < t < 0 , x 1 ( t − λ) and x 2 (λ) overlap as shown in Fig. (e). y( t ) = x1 ( t ) ∗ x 2 ( t ) = ∫-1 (1) 4 e -(t-λ) dλ t

y( t ) = 4 e -t ∫-1 e λ dλ = 4 [ 1 − e -(t+1) ] t

x1(t - λ)

1 -1

t -1

x2(λ) 1

λ

(e)

For 0 < t < 1 ,

y( t ) = ∫-1 (1) 4 e -(t -λ) dλ + ∫0 (-1) 4 e -(t -λ) dλ 0

y( t ) = 4 e -t e λ

t

− 4 e -t e λ

0 -1

t 0

= 8 e -t − 4 e -(t+1) − 4

For t > 1 , the two functions overlap completely. y( t ) = ∫-1 (1) 4 e -(t-λ ) dλ + ∫0 (-1) 4 e -(t-λ) dλ 0

y( t ) = 4 e -t e λ Therefore,

1

0 -1

− 4 e -t e λ 10 = 8 e -t − 4 e -(t+1) − 4 e -(t−1)

[

]

 4 1 − e -(t +1) , -1 < t < 0  -t -(t +1) − 4, y( t ) =  8 e − 4 e 01 

Chapter 15, Solution 47. f1 ( t ) = f 2 ( t ) = cos( t )

L -1 [ F1 (s) F2 (s)] = ∫0 cos(λ) cos( t − λ) dλ t

cos(A) cos(B) =

1 [ cos(A + B) + cos(A − B)] 2

L -1 [ F1 (s) F2 (s)] =

1 t [cos( t ) + cos( t − 2λ )] dλ 2 ∫0

L -1 [ F1 (s) F2 (s)] =

1 1 sin( t − 2λ) cos(t ) ⋅ λ 0t + ⋅ 2 2 -2

t 0

L-1 [ F1 (s) F2 (s)] = 0.5 t cos(t ) + 0.5 sin( t )

Chapter 15, Solution 48.

(a)

Let G (s) =

2 2 = s + 2s + 5 (s + 1) 2 + 2 2 2

g( t ) = e -t sin(2 t ) F(s) = G (s) G (s) f ( t ) = L -1 [ G (s) G (s)] = ∫0 g (λ) g ( t − λ) dλ t

f ( t ) = ∫0 e -λ sin( 2λ) e -( t −λ ) sin( 2( t − λ)) dλ t

sin(A) sin(B) =

f (t) =

1 [ cos(A − B) − cos(A + B)] 2

1 - t t -λ e ∫ e [ cos(2t ) − cos(2( t − 2λ))] dλ 2 0

t e -t e -t -2 λ f (t) = cos(2 t ) ∫0 e dλ − 2 2

f (t) =

e -t e -2λ cos(2 t ) ⋅ 2 -2

t 0

−

e -t 2

∫

∫

t

0

t

0

e -2λ cos(2 t − 4λ) dλ

e -2λ [ cos(2 t ) cos(4λ) + sin( 2 t ) sin( 4λ)] dλ

t 1 -t e -t -2 t f ( t ) = e cos(2 t ) (-e + 1) − cos(2 t ) ∫0 e -2 λ cos(4λ) dλ 4 2

−

t e -t sin( 2 t ) ∫0 e -2λ sin( 4λ) dλ 2

f (t) =

1 -t e cos(2t ) (1 − e -2 t ) 4  e -2λ  e -t (- 2cos(4λ) − 4 sin(4λ)) 0t − cos(2t )  2  4 + 16  −

f (t) =

(b)

Let

 e -2λ  e -t (- 2sin(4λ) + 4 cos(4λ)) 0t sin(2t )  2  4 + 16 

e -t e -3t e -t e -3t cos( 2t ) − cos( 2t ) − cos( 2t ) + cos( 2t ) cos(4t ) 2 4 20 20 +

e -3t e -t cos( 2t ) sin( 4t ) + sin( 2t ) 10 10

+

e -t e -t sin( 2t ) sin( 4t ) − sin( 2t ) cos(4t ) 20 10

X(s) =

2 , s +1

Y(s) =

x ( t ) = 2 e -t u ( t ) ,

s s+4

y( t ) = cos(2t ) u ( t )

F(s) = X(s) Y(s) f ( t ) = L -1 [ X(s) Y(s)] = ∫0 y(λ) x ( t − λ) dλ ∞

f ( t ) = ∫0 cos(2λ) ⋅ 2 e -(t−λ ) dλ t

f ( t ) = 2 e -t ⋅

eλ (cos(2λ) + 2 sin(2λ)) 0t 1+ 4

f (t) =

2 -t t e [ e ( cos(2t ) + 2 sin(2t ) − 1) ] 5

f (t) =

2 4 2 cos( 2t ) + sin( 2t ) − e -t 5 5 5

Chapter 15, Solution 49.

Let x(t) = u(t) – u(t-1) and y(t) = h(t)*x(t).  4(1 − e − s )   4 1 e −s  y( t ) = L−1 [H(s)X(s)] = L−1  ( − ) = L−1   s   s(s + 2)   s + 2 s But 1 A B 1 1 1  = + =  − s(s + 2) s s + 2 2  s s + 2  1 1 e−s e−s  Y(s) = 2 − − +  s s + 2   s s + 2 y( t ) = 2[1 − e −2 t ]u ( t ) − 4[1 − e −2( t −1) ]u ( t − 1)

Chapter 15, Solution 50.

Take the Laplace transform of each term.

[s

2

V(s) − s v(0) − v ′(0)] + 2 [ s V(s) − v(0)] + 10 V(s) =

s 2 V(s) − s + 2 + 2s V(s) − 2 + 10 V(s) =

3s s +4 2

3s s +4 2

3s s 3 + 7s (s + 2s + 10) V(s) = s + 2 = s + 4 s2 + 4 2

V(s) =

s 3 + 7s As + B Cs + D = 2 + 2 2 2 (s + 4)(s + 2s + 10) s + 4 s + 2s + 10

s 3 + 7s = A (s 3 + 2s 2 + 10s) + B (s 2 + 2s + 10) + C (s 3 + 4s) + D (s 2 + 4) Equating coefficients : 1= A+C  → C = 1 − A s3 : 2 s : 0 = 2A + B + D 1 7 = 10A + 2B + 4C = 6A + 2B + 4 s :

0 = 10B + 4D  → D = -2.5 B

s0 :

Solving these equations yields 9 12 A= , B= , 26 26

C=

17 , 26

D=

- 30 26

V(s) =

1  9s + 12 17s − 30  + 2 2 26  s + 4 s + 2s + 10 

V(s) =

 s +1 47 2 1  9s + 6⋅ 2 + 17 ⋅  2 2 2 − 2 2  (s + 1) + 3 (s + 1) + 3  s +4 26  s + 4

v( t ) =

47 17 6 9 sin( 2t ) + e -t cos( 3t ) − e -t sin( 3t ) cos( 2t ) + 78 26 26 26

Chapter 15, Solution 51.

Taking the Laplace transform of the differential equation yields

[s V(s) − sv(0) − v' (0)]+ 5[sV(s) − v(0)]] + 6V(s) = s10+ 1 2

(

)

or s 2 + 5s + 6 V(s) − 2s − 4 − 10 =

Let V(s) =

A B C , + + s +1 s + 2 s + 3

10 s +1 A = 5,

 →

V(s) =

B = 0,

Hence, v( t ) = (5e − t − 3e −3t )u ( t )

2s 2 + 16s + 24 (s + 1)(s + 2)(s + 3)

C = −3

Chapter 15, Solution 52. Take the Laplace transform of each term.

[s

2

I(s) − s i(0) − i ′(0)] + 3 [ s I(s) − i(0)] + 2 I(s) + 1 = 0

(s 2 + 3s + 2) I(s) − s − 3 − 3 + 1 = 0 I(s) =

s+5 A B = + (s + 1)(s + 2) s + 1 s + 2

A = 4, I(s) =

B = -3

4 3 − s +1 s + 2

i( t ) = (4 e -t − 3 e -2t ) u(t )

Chapter 15, Solution 53.

Take the Laplace transform of each term.

[s

2

Y(s) − s y(0) − y ′(0)] + 5 [ s Y(s) − y(0)] + 6 V(s) =

(s 2 + 5s + 6) Y(s) − s − 4 − 5 =

(s 2 + 5s + 6) Y(s) = s + 9 +

Y(s) =

s s +4 2

s s +4 2

s s + (s + 9)(s 2 + 4) = s2 + 4 s2 + 4

s 3 + 9s 2 + 5s + 36 A B Cs + D = + + 2 2 (s + 2)(s + 3)(s + 4) s + 2 s + 3 s + 4

A = (s + 2) Y(s) s= -2 =

27 , 4

B = (s + 3) Y(s) s= -3 =

- 75 13

When s = 0 , 36 A B D = + + (2)(3)(4) 2 3 4

 → D =

5 26

When s = 1 , 46 + 5 A B C D = + + + (12)(5) 3 4 5 5 Thus, Y(s) =

 → C =

1 52

27 4 75 13 1 52 ⋅ s + 5 26 − + s+2 s+3 s2 + 4

y( t ) =

27 - 2t 75 - 3t 1 5 e − e + cos( 2t ) + sin( 2t ) 4 13 52 52

Chapter 15, Solution 54.

Taking the Laplace transform of the differential equation gives

[ s 2 V(s) − s v(0) − v′(0)] + 3[ s V(s) − v(0)] + 2 V(s) = (s 2 + 3s + 2) V(s) =

5 2−s −1 = s+3 s+3

V(s) =

2−s 2−s = 2 (s + 3)(s + 3s + 2) (s + 1)(s + 2)(s + 3)

V(s) =

A B C + + s +1 s + 2 s + 3

A = 3 2,

V(s) =

B = -4 ,

C=5 2

32 52 4 − + s +1 s + 2 s + 3 v( t ) = (1.5 e -t − 4 e -2t + 2.5 e -3t ) u(t )

5 s+3

Chapter 15, Solution 55.

Take the Laplace transform of each term.

[s

3

Y(s) − s 2 y(0) − s y′(0) − y′′(0)] + 6 [ s 2 Y(s) − s y(0) − y′(0)] + 8 [ s Y(s) − y(0)] =

s +1 (s + 1) 2 + 2 2

Setting the initial conditions to zero gives (s 3 + 6 s 2 + 8s) Y(s) =

Y(s) =

A=

s +1 s + 2s + 5 2

(s + 1) A B C Ds + E = + + + 2 2 s (s + 2)(s + 4)(s + 2s + 5) s s + 2 s + 4 s + 2s + 5

1 , 40

B=

1 , 20

C=

-3 , 104

D=

-3 , 65

E=

-7 65

Y(s) =

3s + 7 1 1 1 1 3 1 1 ⋅ + ⋅ − ⋅ − ⋅ 40 s 20 s + 2 104 s + 4 65 (s + 1) 2 + 2 2

Y(s) =

3 (s + 1) 1 1 1 1 3 1 1 1 4 ⋅ + ⋅ − ⋅ − ⋅ ⋅ 2 2 − 40 s 20 s + 2 104 s + 4 65 (s + 1) + 2 65 (s + 1) 2 + 2 2

y( t ) =

1 1 3 -4t 3 -t 2 u(t ) + e - 2t − e − e cos( 2t ) − e -t sin( 2t ) 40 20 104 65 65

Chapter 15, Solution 56.

Taking the Laplace transform of each term we get: 12 4 [ s V(s) − v(0)] + V(s) = 0 s  12   4 s + s  V(s) = 8 V(s) =

8s 2s = 2 4s + 12 s + 3 2

v( t ) = 2 cos

(

3t

)

Chapter 15, Solution 57. Take the Laplace transform of each term.

[ s Y(s) − y(0)] + 9 Y(s) = s

s s +4 2

s2 + 9  s s2 + s + 4   Y(s) = 1 + 2 = 2 s +4 s +4  s  s 3 + s 2 + 4s As + B Cs + D Y(s) = 2 = + (s + 4)(s 2 + 9) s 2 + 4 s 2 + 9 s 3 + s 2 + 4s = A (s 3 + 9s) + B (s 2 + 9) + C (s 3 + 4s) + D (s 2 + 4) Equating coefficients : s0 : 0 = 9B + 4D 1 s : 4 = 9 A + 4C 2 1= B+ D s : 3 s : 1= A+C

Solving these equations gives A = 0,

Y(s) =

B = - 4 5,

C = 1,

D=9 5

-4 5 s+9 5 -4 5 95 s = 2 + 2 + 2 + 2 2 s +4 s +9 s +4 s +9 s +9 y( t ) = - 0.4 sin( 2t ) + cos( 3t ) + 0.6 sin( 3t )

Chapter 15, Solution 58.

We take the Laplace transform of each term and obtain 10 6V(s) + [sV (s) − v(0)] + V(s) = e − 2s s

V(s) =

(s + 3)e −2s − 3e −2s (s + 3) 2 + 1

 →

V(s) =

se −2s s 2 + 6s + 10

Hence, v(t ) = e −3( t − 2) cos(t − 2) − 3e −3( t − 2) sin(t − 2)  u (t − 2)

Chapter 15, Solution 59.

Take the Laplace transform of each term of the integrodifferential equation.

[ s Y(s) − y(0)] + 4 Y(s) + 3 Y(s) = s

6 s+2

 6  (s 2 + 4s + 3) Y(s) = s  − 1 s + 2  Y(s) =

s ( 4 − s) ( 4 − s) s = 2 (s + 2)(s + 4s + 3) (s + 1)(s + 2)(s + 3)

Y(s) =

A B C + + s +1 s + 2 s + 3

A = 2 .5 ,

Y(s) =

B = 6,

C = -10.5

2.5 6 10.5 + − s +1 s + 2 s + 3

y( t ) = 2.5 e -t + 6 e -2t − 10.5 e -3t

Chapter 15, Solution 60.

Take the Laplace transform of each term of the integrodifferential equation. 3 4 4 2 [ s X(s) − x (0)] + 5 X(s) + X(s) + = 2 s s s + 16 (2s 2 + 5s + 3) X(s) = 2s − 4 +

4s 2s 3 − 4s 2 + 36s − 64 = s 2 + 16 s 2 + 16

2s 3 − 4s 2 + 36s − 64 s 3 − 2s 2 + 18s − 32 X(s) = = (2s 2 + 5s + 3)(s 2 + 16) (s + 1)(s + 1.5)(s 2 + 16)

X(s) =

A B Cs + D + + 2 s + 1 s + 1.5 s + 16

A = (s + 1) X(s) s= -1 = -6.235 B = (s + 1.5) X(s) s = -1.5 = 7.329 When s = 0 , B D - 32 = A+ + 1.5 16 (1.5)(16)

 → D = 0.2579

s3 − 2s 2 + 18s − 32 = A (s3 + 1.5s 2 + 16s + 24) + B (s3 + s 2 + 16s + 16) + C (s3 + 2.5s 2 + 1.5s) + D (s 2 + 2.5s + 1.5) Equating coefficients of the s3 terms,

1= A+ B+C  → C = -0.0935 X(s) =

- 6.235 7.329 - 0.0935s + 0.2579 + + s +1 s + 1.5 s 2 + 16

x ( t ) = - 6.235 e -t + 7.329 e -1.5t − 0.0935 cos(4t ) + 0.0645 sin( 4t )

Chapter 16, Solution 1. Consider the s-domain form of the circuit which is shown below. 1

1/s

I(s)

+ −

1/s

s I(s) =

i( t ) =

1s 1 1 = 2 = 1 + s + 1 s s + s + 1 (s + 1 2) 2 + ( 3 2) 2

 3  e - t 2 sin  t  2 3  

2

i( t ) = 1.155 e -0.5t sin (0.866t ) A

Chapter 16, Solution 2. 8/s

s

4 s

+ −

+ Vx −

2

4

4 s + Vx − 0 + Vx − 0 = 0 8 s 2 4+ s

Vx −

Vx (4s + 8) −

(16s + 32) + (2s 2 + 4s)Vx + s 2 Vx = 0 s

Vx (3s 2 + 8s + 8) =

16s + 32 s

    s+2 0.25 − 0.125  − 0.125  Vx = −16 + = −16 +  s 8 4 8 4 s(3s 2 + 8s + 8) s+ − j s+ + j   3  3 3 3  v x = (−4 + 2e − (1.3333 + j0.9428) t + 2e − (1.3333 − j0.9428) t )u ( t ) V 2 2 vx = 4u ( t ) − e − 4 t / 3 cos  3

 6 − 4t / 3  2 2 t  − e sin  2   3

Chapter 16, Solution 3. s + 1/2

5/s

1/8

Vo

− Current division leads to: 1     5 5 1 5 2 = = Vo = 8s1 1  10 + 16s 16(s + 0.625)  + +s 2 8 

(

)

vo(t) = 0.3125 1 − e −0.625t u ( t ) V

 t  V 

Chapter 16, Solution 4.

The s-domain form of the circuit is shown below. 6

1/(s + 1)

s +

+ −

10/s

Vo(s)

−

Using voltage division,  1   1  10    = 2 s  s + 1  s + 6s + 10  s + 1 

Vo (s) =

10 s s + 6 + 10

Vo (s) =

10 A Bs + C = + 2 (s + 1)(s + 6s + 10) s + 1 s + 6s + 10 2

10 = A (s 2 + 6s + 10) + B (s 2 + s) + C (s + 1) Equating coefficients : s2 : 0= A+B  → B = -A

s1 : 0

s :

0 = 6A + B + C = 5A + C  → C = -5A 10 = 10A + C = 5A  → A = 2, B = -2, C = -10

Vo (s) =

2 (s + 3) 2 2s + 10 2 4 = − − 2 2 2 − s + 1 s + 6s + 10 s + 1 (s + 3) + 1 (s + 3) 2 + 12

v o ( t ) = 2 e -t − 2 e -3t cos(t ) − 4 e -3t sin( t ) V

Chapter 16, Solution 5.

Io

1 s+2

s

2

2 s

     1  1 2s 2s = 1   = V=  s + 2  1 1 s  s + 2  s 2 + s + 2  (s + 2)(s + 0.5 + j1.3229)(s + 0.5 − j1.3229)  + +  s 2 2

Io =

Vs s2 = 2 (s + 2)(s + 0.5 + j1.3229)(s + 0.5 − j1.3229)

(−0.5 − j1.3229) 2 (−0.5 + j1.3229) 2 1 (1.5 − j1.3229)(− j2.646) (1.5 + j1.3229)(+ j2.646) + = + s+2 s + 0.5 + j1.3229 s + 0.5 − j1.3229

(

)

i o ( t ) = e − 2 t + 0.3779e − 90° e − t / 2 e − j1.3229 t + 0.3779e 90° e − t / 2 e j1.3229 t u ( t ) A or

(

)

= e − 2 t − 0.7559 sin 1.3229 t u ( t ) A

Chapter 16, Solution 6. 2 Io

5 s+2

10/s

s

Use current division. Io =

5 5s 5(s + 1) 5 s+2 = = − 2 2 2 10 s + 2 s + 2s + 10 (s + 1) + 3 (s + 1) 2 + 3 2 s+2+ s

5 i o ( t ) = 5e − t cos 3t − e − t sin 3t 3

Chapter 16, Solution 7.

The s-domain version of the circuit is shown below. 1/s 1

+

Ix

2s

2 s +1 –

Z 1 (2s) 1 2s 2s 2 + 2s + 1 Z = 1 + // 2s = 1 + s = 1+ = 1 s 1 + 2s 2 1 + 2s 2 + 2s s

V 2 1 + 2s 2 2s 2 + 1 A Bs + C = = + Ix = = x 2 2 2 Z s + 1 2s + 2s + 1 (s + 1)(s + s + 0.5) (s + 1) (s + s + 0.5) 2s 2 + 1 = A(s 2 + s + 0.5) + B(s 2 + s) + C(s + 1) s2 :

2=A+B

0 = A+B+C = 2+C

s: constant : Ix =

 →

1 = 0.5A + C or 0.5A = 3

C = −2  →

A = 6, B = -4

6 4s + 2 6 4(s + 0.5) − = − 2 s + 1 (s + 0.5) + 0.75 s + 1 (s + 0.5) 2 + 0.866 2

[

]

i x ( t ) = 6 − 4e − 0.5t cos 0.866 t u ( t ) A

Chapter 16, Solution 8.

1 1 (1 + 2s) s 2 + 1.5s + 1 = (a) Z = + 1 //(1 + 2s) = + s s 2 + 2s s(s + 1) 1 1 1 1 3s 2 + 3s + 2 = + + = (b) 1 Z 2 s 2s(s + 1) 1+ s Z=

2s(s + 1) 3s 2 + 3s + 2

Chapter 16, Solution 9.

The s-domain form of the circuit is shown in Fig. (a).

(a)

2 (s + 1 s) 2 (s 2 + 1) Z in = 2 || (s + 1 s) = = 2 + s + 1 s s 2 + 2s + 1 1 s 2

s

1/s

(b)

2

2/s 1

(a) (b) The s-domain equivalent circuit is shown in Fig. (b).

2 || (1 + 2 s) =

2 (1 + 2 s) 2 (s + 2) = 3+ 2 s 3s + 2

1 + 2 || (1 + 2 s) =

5s + 6 3s + 2

 5s + 6   s ⋅  3s + 2   5s + 6  s (5s + 6) = Z in = s ||  = 2  3s + 2   5s + 6  3s + 7s + 6  s +  3s + 2 

Chapter 16, Solution 10.

To find ZTh, consider the circuit below.

1/s

Vx

+ 1V 2 -

Vo

2Vo

Applying KCL gives 1 + 2Vo =

Vx 2 + 1/ s

But Vo =

2 Vx . Hence 2 + 1/ s

1+

4Vx Vx = 2 + 1/ s 2 + 1/ s

ZTh =

 →

Vx = −

(2s + 1) 3s

Vx (2s + 1) =− 1 3s

To find VTh, consider the circuit below.

1/s

Vy

+ 2 s +1

2

Vo -

Applying KCL gives 2 V + 2Vo = o s +1 2

 →

Vo = −

4 3(s + 1)

2Vo

1 But − Vy + 2Vo • + Vo = 0 s

2 4  s + 2  − 4(s + 2) VTh = Vy = Vo (1 + ) = −  = s 3(s + 1)  s  3s(s + 1)

Chapter 16, Solution 11.

The s-domain form of the circuit is shown below. 4/s

1/s

+ −

s

I1

2

+ −

I2

4/(s + 2)

Write the mesh equations. 1  4 =  2 +  I1 − 2 I 2 s  s

(1)

-4 = -2 I1 + (s + 2) I 2 s+2

(2)

Put equations (1) and (2) into matrix form.  1 s   2 + 4 s - 2   I1   - 4 (s + 2)  =  - 2 s + 2  I 2     s 2 − 4s + 4 , s (s + 2)

∆=

2 2 (s + 2s + 4) , s

I1 =

∆1 1 2 ⋅ (s 2 − 4s + 4) A Bs + C = = + 2 2 ∆ (s + 2)(s + 2s + 4) s + 2 s + 2s + 4

∆1 =

∆2 =

-6 s

1 2 ⋅ (s 2 − 4s + 4) = A (s 2 + 2s + 4) + B (s 2 + 2s) + C (s + 2) Equating coefficients : s2 : 1 2= A+B 1 s : - 2 = 2A + 2B + C

2 = 4 A + 2C

s0 :

Solving these equations leads to

A = 2,

B = -3 2,

I1 =

- 3 2s − 3 2 + s + 2 (s + 1) 2 + ( 3 ) 2

I1 =

2 -3 (s + 1) -3 3 + ⋅ + ⋅ 2 2 2 s + 2 2 (s + 1) + ( 3 ) 2 3 (s + 1) + ( 3 ) 2

C = -3

i1 ( t ) = [ 2 e -2t − 1.5 e -t cos(1.732t ) − 0.866 sin(1.732t )] u(t ) A

I2 =

s -3 ∆2 - 6 = ⋅ = 2 2 s 2 (s + 2s + 4) (s + 1) + ( 3 ) 2 ∆

i 2 (t) =

-3 3

e - t sin( 3t ) = - 1.732 e -t sin(1.732t ) u(t ) A

Chapter 16, Solution 12.

We apply nodal analysis to the s-domain form of the circuit below. s

10/(s + 1)

+ −

Vo

1/(2s)

4

10 − Vo 3 V s +1 o + = + 2sVo s s 4 (1 + 0.25s + s 2 ) Vo = Vo =

10 10 + 15s + 15 + 15 = s +1 s +1

15s + 25 A Bs + C = + 2 2 (s + 1)(s + 0.25s + 1) s + 1 s + 0.25s + 1

3/s

A = (s + 1) Vo

s = -1

=

40 7

15s + 25 = A (s 2 + 0.25s + 1) + B (s 2 + s) + C (s + 1) Equating coefficients : s2 : 0= A+B  → B = -A 1 s : 15 = 0.25A + B + C = -0.75A + C 0 25 = A + C s : A = 40 7 ,

B = - 40 7 ,

C = 135 7

- 40 135 40 3 1 s+ + s  155 2  40 1 40 7 7 7 2 2  + = − + ⋅ Vo = 2 2 2 7 + 7 s 1 7 s +1  1   3  1  1 3 3 3 s +  + s +  + s +  +  2  2  2 4 4 4

v o (t) =

 3  (155)(2)  3  40 - t 40 - t 2 e − e cos  t  + e - t 2 sin  t  7 7  2  (7)( 3 )  2 

v o ( t ) = 5.714 e -t − 5.714 e -t 2 cos(0.866t ) + 25.57 e -t 2 sin( 0.866t ) V

Chapter 16, Solution 13.

Consider the following circuit. 1/s

2s

Vo

Io 2

1/(s + 2)

Applying KCL at node o, Vo Vo 1 s +1 = + = V s + 2 2s + 1 2 + 1 s 2s + 1 o

1

Vo =

2s + 1 (s + 1)(s + 2)

Io =

Vo 1 A B = = + 2s + 1 (s + 1)(s + 2) s + 1 s + 2

A = 1, Io =

B = -1

1 1 − s +1 s + 2

i o ( t ) = ( e -t − e -2t ) u(t ) A

Chapter 16, Solution 14.

We first find the initial conditions from the circuit in Fig. (a). 1Ω

4Ω +

5V

+ −

vc(0)

io

− (a)

i o (0 − ) = 5 A , v c (0 − ) = 0 V We now incorporate these conditions in the s-domain circuit as shown in Fig.(b). 1

4

Vo Io

15/s

+ −

2s

(b)

At node o, Vo − 15 s Vo 5 Vo − 0 + + + =0 1 2s s 4 + 4 s

5/s

4/s

1 s  15 5  V − = 1 + + s s  2s 4 (s + 1)  o 5s 2 + 6s + 2 10 4s 2 + 4s + 2s + 2 + s 2 Vo Vo = = 4s (s + 1) s 4s (s + 1)

Vo =

40 (s + 1) 5s 2 + 6s + 2

Vo 5 4 (s + 1) 5 + + = 2 2s s s (s + 1.2s + 0.4) s 5 A Bs + C Io = + + 2 s s s + 1.2s + 0.4

Io =

4 (s + 1) = A (s 2 + 1.2s + 0.4) + B s s + C s Equating coefficients : s0 : 4 = 0.4A  → A = 10

s1 : 2

s :

4 = 1.2A + C  → C = -1.2A + 4 = -8 0= A+B  → B = -A = -10

Io =

5 10 10s + 8 + − 2 s s s + 1.2s + 0.4

Io =

10 (s + 0.6) 10 (0.2) 15 − 2 2 − s (s + 0.6) + 0.2 (s + 0.6) 2 + 0.2 2

i o ( t ) = [ 15 − 10 e -0.6t ( cos(0.2 t ) − sin( 0.2 t )) ] u(t ) A

Chapter 16, Solution 15.

First we need to transform the circuit into the s-domain. s/4

10

Vo +

3Vx

+ −

5/s

Vx

− + −

5 s+2

5 Vo − Vo − 3Vx Vo − 0 s+2 =0 + + s/4 5/s 10 5s 5s = 0 = (2s 2 + s + 40)Vo − 120Vx − 40Vo − 120Vx + 2s 2 Vo + sVo − s+2 s+2 5 5 → Vo = Vx + s+2 s+2

But, Vx = Vo −

We can now solve for Vx. 5  5s  (2s 2 + s + 40) Vx + =0  − 120Vx − s + 2 s+2  2(s 2 + 0.5s − 40)Vx = −10

Vx = − 5

(s 2 + 20) s+2

(s 2 + 20) (s + 2)(s 2 + 0.5s − 40)

Chapter 16, Solution 16.

We first need to find the initial conditions. For t < 0 , the circuit is shown in Fig. (a). To dc, the capacitor acts like an open circuit and the inductor acts like a short circuit. 2Ω

+

−

1Ω

1F Vo/2

Vo

+ −

(a)

+ −

1H

io

3V

Hence, i L (0) = i o =

-3 = -1 A , 3

v o = -1 V

 - 1 v c (0) = -(2)(-1) −   = 2.5 V 2 We now incorporate the initial conditions for t > 0 as shown in Fig. (b). 2

+

Vo

−

1

1/s

s 5/(s + 2)

+ −

I1

2.5/s

+ −

Vo/2

+ −

I2 − +

-1 V

Io (b)

For mesh 1, - 5  1 1 2.5 Vo +  2 +  I1 − I 2 + + =0 s+2  s s s 2 But,

Vo = I o = I 2  1  1 1 5 2.5  2 +  I1 +  −  I 2 = −  2 s s s+2 s

(1)

For mesh 2, V  1 1 2.5 1 + s +  I 2 − I1 + 1 − o − =0  s s 2 s 1 1 1 2.5 - I1 +  + s +  I 2 = −1 2 s s s

(2)

Put (1) and (2) in matrix form.  1  5 1 1  2.5  −   I1   − 2 + s 2 s   s+2 s     =   1 1 1 2 . 5     + s +  I 2   −1   s  s  2 s 3 ∆ = 2s + 2 + , s Io = I2 =

∆ 2 = -2 +

4 5 + s s (s + 2)

∆2 - 2s 2 + 13 A Bs + C = = + 2 2 ∆ (s + 2)(2s + 2s + 3) s + 2 2s + 2s + 3

- 2s 2 + 13 = A (2s 2 + 2s + 3) + B (s 2 + 2s) + C (s + 2) Equating coefficients : s2 : - 2 = 2A + B 1 0 = 2A + 2 B + C s : 0 s : 13 = 3A + 2C Solving these equations leads to A = 0.7143 , B = -3.429 , C = 5.429 0.7143 3.429 s − 5.429 0.7143 1.7145 s − 2.714 − = − s+2 2s 2 + 2s + 3 s+2 s 2 + s + 1.5 0.7143 1.7145 (s + 0.5) (3.194)( 1.25 ) Io = − + s+2 (s + 0.5) 2 + 1.25 (s + 0.5) 2 + 1.25

Io =

[

]

i o ( t ) = 0.7143 e -2t − 1.7145 e -0.5t cos(1.25t ) + 3.194 e -0.5t sin(1.25t ) u(t ) A

Chapter 16, Solution 17. We apply mesh analysis to the s-domain form of the circuit as shown below. 2/(s+1) + −

I3 1/s

1

s

I1

I2

1

4 For mesh 3,  1 2 1 +  s +  I 3 − I1 − s I 2 = 0 s +1  s s

For the supermesh,  1 1  1 +  I1 + (1 + s) I 2 −  + s  I 3 = 0  s s  But

I1 = I 2 − 4

Substituting (3) into (1) and (2) leads to   1  1 1  2 + s +  I 2 − s +  I 3 = 4 1 +    s  s s  1  1 -4 2 - s +  I 2 + s +  I 3 = −  s  s s s +1 Adding (4) and (5) gives 2 2 I2 = 4 − s +1

I2 = 2 −

1 s +1

(1)

(2) (3)

(4)

(5)

i o ( t ) = i 2 ( t ) = ( 2 − e -t ) u(t ) A

Chapter 16, Solution 18. 3 e −s 3 = (1 − e − s ) vs(t) = 3u(t) – 3u(t–1) or Vs = − s s s

1Ω

Vs

+ + −

1/s

2Ω

Vo −

V Vo − Vs + sVo + o = 0 → (s + 1.5)Vo = Vs 2 1 Vo =

3 2  2 −s (1 − e − s ) =  − (1 − e ) s(s + 1.5) s s + 1 . 5  

v o ( t ) = [(2 − 2e −1.5t )u ( t ) − (2 − 2e −1.5( t −1) )u ( t − 1)] V

Chapter 16, Solution 19.

We incorporate the initial conditions in the s-domain circuit as shown below. 2

V1

2I

Vo

− +

I 4/(s + 2)

+ −

1/s 1/s

s

2

At the supernode, V1 1 4 (s + 2) − V1 +2= + + sVo s s 2  1 1 1 2 + 2 =  +  V1 + + s Vo 2 s s s+2

But

Vo = V1 + 2 I and Vo = V1 +

2 (V1 + 1) s

I=

(1)

V1 + 1 s

 → V1 =

Vo − 2 s s Vo − 2 = (s + 2) s s+2

Substituting (2) into (1) 2 1  2s + 1  s  2    Vo − + 2− = + s Vo s+2 s  s  s + 2  s + 2  2 1 2 (2s + 1)  2s + 1   +s V +2− + =  s+2 s s (s + 2)  s + 2   o 2s 2 + 9s 2s + 9 s 2 + 4s + 1 = = Vo s (s + 2) s+2 s+2

Vo =

2s + 9 A B = + s + 4s + 1 s + 0.2679 s + 3.732 2

A = 2.443 ,

Vo =

B = -0.4434

2.443 0.4434 − s + 0.2679 s + 3.732

Therefore, v o ( t ) = ( 2.443 e -0.2679t − 0.4434 e -3.732t ) u(t ) V

(2)

Chapter 16, Solution 20.

We incorporate the initial conditions and transform the current source to a voltage source as shown. 1

2/s

1/s

Vo

+ −

1/(s + 1)

+ −

1

At the main non-reference node, KCL gives 1 (s + 1) − 2 s − Vo Vo Vo 1 = + + 1+1 s 1 s s s s +1 − 2 − s Vo = (s + 1)(s + 1 s) Vo + s +1 s s s +1 − − 2 = (2s + 2 + 1 s) Vo s +1 s Vo =

- 2s 2 − 4s − 1 (s + 1)(2s 2 + 2s + 1)

Vo =

- s − 2s − 0.5 A Bs + C = + 2 2 (s + 1)(s + s + 0.5) s + 1 s + s + 0.5

A = (s + 1) Vo

s = -1

=1

- s 2 − 2s − 0.5 = A (s 2 + s + 0.5) + B (s 2 + s) + C (s + 1) Equating coefficients : s2 : -1 = A + B  → B = -2

s1 : s0 : Vo =

-2 = A+ B+C  → C = -1 - 0.5 = 0.5A + C = 0.5 − 1 = -0.5 2 (s + 0.5) 1 2s + 1 1 = − − 2 s + 1 s + s + 0.5 s + 1 (s + 0.5) 2 + (0.5) 2

v o ( t ) = [ e -t − 2 e -t 2 cos(t 2)] u(t ) V

s

1/s

Chapter 16, Solution 21. The s-domain version of the circuit is shown below. 1

s V1

+

Vo

2/s

2

1/s

10/s At node 1, 10 − V1 V − Vo s s = 1 + Vo 1 s 2

At node 2, V1 − Vo Vo = + sVo s 2

 →

 →

10 = ( s + 1)V1 + (

s2 − 1)Vo 2

s V1 = Vo ( + s 2 + 1) 2

(1)

(2)

Substituting (2) into (1) gives s2 10 = ( s + 1)( s + s / 2 + 1)Vo + ( − 1)Vo = s ( s 2 + 2s + 1.5)Vo 2 2

Vo =

A Bs + C 10 = + 2 s ( s + 2s + 1.5) s s + 2s + 1.5 2

10 = A( s 2 + 2 s + 1.5) + Bs 2 + Cs s2 : 0 = A+ B s: 0 = 2A + C constant : 10 = 1.5 A  → Vo =

A = 20 / 3, B = -20/3, C = -40/3

 0.7071 20  1 s+2 s +1  20  1 − 1.414 = − 2  − 2 2 2 2    3  s s + 2 s + 1.5  3  s ( s + 1) + 0.7071 ( s + 1) + 0.7071 

Taking the inverse Laplace tranform finally yields v o (t) =

[

]

20 1 − e − t cos 0.7071t − 1.414e − t sin 0.7071t u ( t ) V 3

Chapter 16, Solution 22.

The s-domain version of the circuit is shown below. 4s V1

12 s +1

1

At node 1, V V − V2 12 = 1+ 1 s +1 1 4s

At node 2, V1 − V2 V2 s = + V2 4s 2 3

V2

2

→

3/s

12 1 V  = V1 1 +  − 2 s +1  4s  4s

(1)

4  V1 = V2  s 2 + 2s + 1 3 

(2)

→

Substituting (2) into (1),  4 12 1  1  4 7 3  = V2  s 2 + 2s + 11 +  −  =  s 2 + s + V2 s +1 3 2  4s  4s   3  3 V2 =

9 7 9 (s + 1)(s 2 + s + ) 4 8

9 = A(s 2 +

=

A Bs + C + (s + 1) (s 2 + 7 s + 9 ) 4 8

7 9 s + ) + B(s 2 + s) + C(s + 1) 4 8

Equating coefficients: s2 :

0=A+B

s:

0=

7 3 A+B+C = A+C 4 4

constant :

9=

9 3 A + C= A  → 8 8

 →

3 C=− A 4

A = 24, B = -24, C = -18

V2 =

3 24s + 18 24 24(s + 7 / 8) + = − 7 23 7 9 7 23 (s + 1) (s + ) 2 + (s 2 + s + ) (s + ) 2 + 8 64 4 8 8 64

24 − (s + 1)

Taking the inverse of this produces:

[

]

v 2 ( t ) = 24e − t − 24e −0.875t cos(0.5995t ) + 5.004e −0.875t sin(0.5995t ) u ( t ) Similarly, 4  9 s 2 + 2s + 1 Es + F 3  = D + V1 = 7 9 7 9 (s + 1) (s + 1)(s 2 + s + ) (s 2 + s + ) 4 8 4 8

7 9 4  9 s 2 + 2s + 1 = D(s 2 + s + ) + E(s 2 + s) + F(s + 1) 4 8 3  Equating coefficients: s2 : s:

constant :

V1 =

Thus,

12 = D + E 18 = 9=

7 3 D + E + F or 6 = D + F 4 4

9 3 D + F or 3 = D  → 8 8

8 + (s + 1)

 →

3 F = 6− D 4

D = 8, E = 4, F = 0

4s 8 4(s + 7 / 8) 7/2 = + − 7 9 7 23 7 23 (s + 1) (s 2 + s + ) (s + ) 2 + (s + ) 2 + 4 8 8 64 8 64

[

]

v1 ( t ) = 8e − t + 4e −0.875t cos(0.5995t ) − 5.838e −0.875t sin(0.5995t ) u ( t )

Chapter 16, Solution 23.

The s-domain form of the circuit with the initial conditions is shown below. V I 4/s

R

sL

-2/s

1/sC

5C

At the non-reference node, 4 2 V V + + 5C = + + sCV s s R sL s 1  6 + 5 sC CV  2 s +  = + RC LC  s s  V=

But

5s + 6 C s + s RC + 1 LC 2

1 1 = = 8, RC 10 80 V=

1 1 = = 20 LC 4 80

5s + 480 5 (s + 4) (230)(2) = 2 2 + s + 8s + 20 (s + 4) + 2 (s + 4) 2 + 22 2

v( t ) = 5 e -4t cos( 2t ) + 230 e -4t sin( 2t ) V I=

V 5s + 480 = sL 4s (s 2 + 8s + 20)

I=

1.25s + 120 A Bs + C = + 2 2 s (s + 8s + 20) s s + 8s + 20

A = 6,

I=

B = -6 ,

C = -46.75

6 6s + 46.75 6 6 (s + 4) (11.375)(2) − 2 = − 2 2 − s s + 8s + 20 s (s + 4) + 2 (s + 4) 2 + 22

i( t ) = 6 u(t ) − 6 e -4t cos( 2t ) − 11.375 e -4t sin( 2t ), t > 0

Chapter 16, Solution 24.

At t = 0-, the circuit is equivalent to that shown below. + 4Ω

9A

5Ω

vo -

v o (0) = 5x

4 (9) = 20 4+5

For t > 0, we have the Laplace transform of the circuit as shown below after transforming the current source to a voltage source. 4Ω

16 Ω Vo

+ 36V

10A

2/s

5Ω

-

Applying KCL gives 36 − Vo sV V + 10 = o + o 20 2 5 Thus,

 →

Vo =

[

3.6 + 20s A B = + , s(s + 0.5) s s + 0.5

]

v o ( t )= 7.2 − 12.8e −0.5t u ( t )

A = 7.2, B = −12.8

Chapter 16, Solution 25.

For t > 0 , the circuit in the s-domain is shown below. 6

s

I + 9/s

(2s)/(s2 + 16)

+ −

V

−

+ −

2/s

Applying KVL, − 2s  9 2 + 6 + s +  I + = 0 s s s + 16  2

I=

4s 2 + 32 (s 2 + 6s + 9)(s 2 + 16)

V=

9 2 2 36s 2 + 288 I+ = + s s s s (s + 3) 2 (s 2 + 16) =

2 A B C Ds + E + + + + 2 2 s s s + 3 (s + 3) s + 16

36s 2 + 288 = A (s 4 + 6s 3 + 25s 2 + 96s + 144) + B (s 4 + 3s 3 + 16s 2 + 48s) + C (s 3 + 16s) + D (s 4 + 6s 3 + 9s 2 ) + E (s 3 + 6s 2 + 9s) Equating coefficients : 288 = 144A s0 : 1 s : 0 = 96A + 48B + 16C + 9E 2 36 = 25A + 16B + 9D + 6E s : 3 0 = 6A + 3B + C + 6D + E s : 4 s : 0 = A+ B+ D Solving equations (1), (2), (3), (4) and (5) gives A = 2 , B = -1.7984 , C = -8.16 , D = -0.2016 ,

(1) (2) (3) (4) (5) E = 2.765

V(s) =

4 1.7984 8.16 0.2016 s (0.6912)(4) − − + 2 − s s + 3 (s + 3) s 2 + 16 s 2 + 16

v( t ) = 4 u(t ) − 1.7984 e -3t − 8.16 t e -3t − 0.2016 cos(4t ) + 0.6912 sin( 4t ) V

Chapter 16, Solution 26.

Consider the op-amp circuit below. R2

1/sC R1

0

+ −

Vs

− +

At node 0, Vs − 0 0 − Vo = + (0 − Vo ) sC R1 R2  1  + sC  ( - Vo ) Vs = R 1  R2 

Vo -1 = Vs sR 1C + R 1 R 2 But

R 1 20 = = 2, R 2 10

So,

Vo -1 = Vs s + 2 Vs = 3 e -5t

R 1C = (20 × 103 )(50 × 10-6 ) = 1

 → Vs = 3 (s + 5)

+ Vo −

Vo =

-3 (s + 2)(s + 5)

- Vo =

3 A B = + (s + 2)(s + 5) s + 2 s + 5

A = 1, Vo =

B = -1

1 1 − s+5 s+2 v o ( t ) = ( e -5t − e -2t ) u(t )

Chapter 16, Solution 27.

Consider the following circuit. 2s

10/(s + 3)

+ −

I1

s

2s

1

I2

1

For mesh 1, 10 = (1 + 2s) I1 − I 2 − s I 2 s+3 10 = (1 + 2s) I1 − (1 + s) I 2 s+3 For mesh 2, 0 = (2 + 2s) I 2 − I1 − s I1 0 = -(1 + s) I1 + 2 (s + 1) I 2 (1) and (2) in matrix form,

10 (s + 3)   2s + 1 - (s + 1)  I1   =  - (s + 1) 2 (s + 1)  I   0  2     ∆ = 3s 2 + 4s + 1

(1)

(2)

∆1 =

20 (s + 1) s+3

∆2 =

10 (s + 1) s+3

Thus I1 =

20 (s + 1) ∆1 = ∆ (s + 3)( 3s 2 + 4s + 1)

I2 =

10 (s + 1) ∆2 I = = 1 2 ∆ (s + 3)( 3s + 4s + 1) 2

Chapter 16, Solution 28.

Consider the circuit shown below. s 1 + + −

6/s

I1

2s

s

I2

For mesh 1, 6 = (1 + 2s) I1 + s I 2 s

Vo

−

2

(1)

For mesh 2, 0 = s I1 + (2 + s) I 2  2 I1 = - 1 +  I 2  s Substituting (2) into (1) gives  2 6 - (s 2 + 5s + 2) I2 = -(1 + 2s)1 +  I 2 + s I 2 =  s s s or

I2 =

-6 s + 5s + 2 2

(2)

Vo = 2 I 2 =

- 12 - 12 = s + 5s + 2 (s + 0.438)(s + 4.561) 2

Since the roots of s 2 + 5s + 2 = 0 are -0.438 and -4.561, Vo =

A B + s + 0.438 s + 4.561

A=

- 12 = -2.91 , 4.123

B=

- 12 = 2.91 - 4.123

- 2.91 2.91 + s + 0.438 s + 4.561

Vo (s) =

v o ( t ) = 2.91 [ e -4.561t − e 0.438t ] u(t ) V

Chapter 16, Solution 29.

Consider the following circuit. 1

10/(s + 1)

Let

Io

1:2

+ −

4/s

Z L = 8 ||

4 (8)(4 s) 8 = = s 8 + 4 s 2s + 1

When this is reflected to the primary side, Zin = 1 +

ZL , n=2 n2

Zin = 1 +

2 2s + 3 = 2s + 1 2s + 1

Io =

10 1 10 2s + 1 ⋅ = ⋅ s + 1 Zin s + 1 2s + 3

8

Io =

10s + 5 A B = + (s + 1)(s + 1.5) s + 1 s + 1.5

A = -10 ,

I o (s) =

B = 20

- 10 20 + s + 1 s + 1.5

[

]

i o ( t ) = 10 2 e -1.5t − e − t u(t ) A

Chapter 16, Solution 30.

Y(s) = H(s) X(s) ,

X(s) =

4 12 = s + 1 3 3s + 1

Y(s) =

12 s 2 4 8s + 4 3 − 2 = (3s + 1) 3 (3s + 1) 2

Y(s) =

4 8 s 4 1 − ⋅ ⋅ 2 − 3 9 (s + 1 3) 27 (s + 1 3) 2

Let G (s) =

-8 s ⋅ 9 (s + 1 3) 2

Using the time differentiation property,  -8 d - 8  -1 g( t ) = ⋅ ( t e -t 3 ) =  t e -t 3 + e -t 3   9 dt 93 g( t ) =

8 -t 3 8 -t 3 te − e 27 9

y( t ) =

4 8 -t 3 8 -t 3 4 -t 3 u(t) + te − e − te 3 27 9 27

y( t ) =

4 -t 3 8 4 te u( t ) − e - t 3 + 27 9 3

Hence,

Chapter 16, Solution 31.

x(t) = u(t)  → X(s) =

1 s

y( t ) = 10 cos(2t )  → Y(s) =

H(s) =

10s s2 + 4

Y(s) 10s 2 = X(s) s 2 + 4

Chapter 16, Solution 32.

(a)

Y(s) = H(s) X(s) =

s+3 1 ⋅ s + 4s + 5 s

=

s+3 A Bs + C = + 2 s (s + 4s + 5) s s + 4s + 5

2

2

s + 3 = A (s 2 + 4s + 5) + Bs 2 + Cs Equating coefficients : 3 = 5A  → A = 3 5 s0 :

s1 :

1 = 4A + C  → C = 1 − 4A = - 7 5

s2 :

0= A+B  → B = -A = - 3 5

Y(s) =

35 1 3s + 7 − ⋅ 2 s 5 s + 4s + 5

Y(s) =

0.6 1 3 (s + 2) + 1 − ⋅ s 5 (s + 2) 2 + 1

y( t ) = [ 0.6 − 0.6 e -2t cos(t ) − 0.2 e -2t sin( t )] u(t )

(b)

x ( t ) = 6 t e -2t

 → X(s) =

Y(s) = H(s) X(s) =

Y(s) =

6 (s + 2) 2

s+3 6 ⋅ s + 4s + 5 (s + 2) 2 2

6 (s + 3) A B Cs + D = + 2 2 2 + 2 (s + 2) (s + 4s + 5) s + 2 (s + 2) s + 4s + 5

Equating coefficients : s3 : 0= A+C  → C = -A 2 0 = 6 A + B + 4C + D = 2 A + B + D s : 1 s : 6 = 13A + 4B + 4C + 4D = 9A + 4B + 4D 0 18 = 10A + 5B + 4D = 2A + B s : Solving (1), (2), (3), and (4) gives A=6, B = 6, C = -6 ,

(1) (2) (3) (4)

D = -18

Y(s) =

6 6 6s + 18 + 2 − s + 2 (s + 2) (s + 2) 2 + 1

Y(s) =

6 6 6 (s + 2) 6 − + 2 − 2 s + 2 (s + 2) (s + 2) + 1 (s + 2) 2 + 1

y( t ) = [ 6 e -2t + 6 t e -2t − 6 e -2t cos(t ) − 6 e -2t sin( t )] u(t )

Chapter 16, Solution 33.

1 s

H(s) =

Y(s) , X(s)

Y(s) =

4 1 2s (3)(4) + − − 2 s 2 (s + 3) (s + 2) + 16 (s + 2) 2 + 16

H(s) = s Y(s) = 4 +

X(s) =

s 2s2 12 s − 2 − 2 2 (s + 3) s + 4s + 20 s + 4s + 20

Chapter 16, Solution 34. Consider the following circuit.

2

s

Vo +

+ −

Vs

4

10/s

Vo(s) −

Using nodal analysis, Vs − Vo Vo Vo = + s+2 4 10 s

 1   1 1 s 1 Vs = (s + 2)  + +  Vo = 1 + (s + 2) + (s 2 + 2s)  Vo  4   s + 2 4 10  10 Vs =

1 ( 2s 2 + 9s + 30) Vo 20

20 Vo = 2 Vs 2s + 9s + 30

Chapter 16, Solution 35. Consider the following circuit. I

Vs

2/s

s

V1

+

+ −

2I

Vo −

At node 1, 2I + I =

V1 , s+3

where I =

Vs − V1 2s

3

3⋅

Vs − V1 V = 1 2s s+3

3s V1 3s = Vs − V1 2 s+3 2  1 3s 3s   +  V1 = Vs s + 3 2  2 V1 =

3s (s + 3) V 3s 2 + 9s + 2 s

Vo =

9s 3 V V1 = 2 3s + 9s + 2 s s+3

H(s) =

9s Vo = 2 Vs 3s + 9s + 2

Chapter 16, Solution 36. From the previous problem, 3I = I=

But

V1 3s V = 2 s + 3 3s + 9s + 2 s

s V 3s + 9s + 2 s

Vs =

2

3s 2 + 9s + 2 Vo 9s

V s 3s 2 + 9 s + 2 I= 2 ⋅ Vo = o 3s + 9 s + 2 9s 9

H(s) =

Vo =9 I

Chapter 16, Solution 37. (a)

Consider the circuit shown below. 3

Vs

+ −

2s +

I1

Vx

2/s

I2

+ −

4Vx

−

For loop 1,  2 2 Vs = 3 +  I1 − I 2  s s

(1)

For loop 2,  2 2 4Vx +  2s +  I 2 − I1 = 0  s s But,

2 Vx = (I1 − I 2 )   s

So,

 2 2 8 (I1 − I 2 ) +  2s +  I 2 − I1 = 0  s s s 0=

 6 -6 I1 +  − 2s  I 2  s s

(2)

In matrix form, (1) and (2) become  Vs  3 + 2 s - 2 s  I1   0  =  - 6 s 6 s − 2s  I   2        6  2  2  6 ∆ = 3 +  − 2s  −       s  s  s  s ∆=

18 − 6s − 4 s

6  ∆ 1 =  − 2s  Vs , s 

∆2 =

6 V s s

I1 =

∆1 (6 s − 2s) = V ∆ 18 s − 4 − 6s s

I1 3 s−s s2 − 3 = = 2 Vs 9 s − 2 − 3 3s + 2s − 9

(b)

I2 =

∆2 ∆

Vx =

2 2  ∆1 − ∆ 2   ( I1 − I 2 ) =  s s ∆ 

Vx =

2 s Vs (6 s − 2s − 6 s) - 4Vs = ∆ ∆

6 s Vs - 3 I2 = = Vx - 4Vs 2s

Chapter 16, Solution 38. (a)

Consider the following circuit.

Is

1

V1

s

Vo

Io +

Vs

+ −

1/s

1/s

1

Vo −

At node 1, Vs − V1 V1 − Vo = s V1 + 1 s  1 1 Vs = 1 + s +  V1 − Vo  s s

(1)

At node o, V1 − Vo = s Vo + Vo = (s + 1) Vo s V1 = (s 2 + s + 1) Vo

(2)

Substituting (2) into (1) Vs = (s + 1 + 1 s)(s 2 + s + 1)Vo − 1 s Vo Vs = (s 3 + 2s 2 + 3s + 2)Vo H 1 (s) =

(b)

Vo 1 = 3 2 Vs s + 2s + 3s + 2

I s = Vs − V1 = (s 3 + 2s 2 + 3s + 2)Vo − (s 2 + s + 1)Vo I s = (s 3 + s 2 + 2s + 1)Vo H 2 (s) =

(c)

(d).

Io =

Vo 1 = 3 2 Is s + s + 2s + 1

Vo 1

H 3 (s) =

I o Vo 1 = = H 2 (s) = 3 2 Is Is s + s + 2s + 1

H 4 (s) =

I o Vo 1 = = H 1 (s) = 3 2 Vs Vs s + 2s + 3s + 2

Chapter 16, Solution 39. Consider the circuit below.

Va Vb Vs

+ −

1/sC

− +

+

R Io

Vo −

Since no current enters the op amp, I o flows through both R and C.  1 Vo = -I o  R +   sC  Va = Vb = Vs = H(s) =

- Io sC

Vo R + 1 sC = = sRC + 1 Vs 1 sC

Chapter 16, Solution 40.

(a)

(b)

H(s) =

Vo R R L = = Vs R + sL s + R L

h(t) =

R - Rt L e u( t ) L

v s (t) = u(t)  → Vs (s) = 1 s

Vo =

R L R L A B Vs = = + s+R L s (s + R L) s s + R L

A = 1,

B = -1

1 1 Vo = − s s+R L v o ( t ) = u ( t ) − e -Rt L u ( t ) = (1 − e -Rt L ) u(t )

Chapter 16, Solution 41.

Y(s) = H(s) X(s) h ( t ) = 2 e -t u ( t )  →

H(s) =

2 s +1

v i (t) = 5 u(t)  → Vi (s) = X(s) = 5 s Y(s) =

10 A B = + s (s + 1) s s + 1

A = 10 ,

Y(s) =

B = -10

10 10 − s s +1

y( t ) = 10 (1 − e -t ) u(t )

Chapter 16, Solution 42.

2s Y(s) + Y(s) = X(s) (2s + 1) Y(s) = X(s) H(s) =

Y(s) 1 1 = = X(s) 2s + 1 2 (s + 1 2)

h ( t ) = 0.5 e -t 2 u(t )

Chapter 16, Solution 43. 1Ω

u(t)

+ −

i(t)

1F

1H First select the inductor current iL and the capacitor voltage vC to be the state variables. Applying KVL we get:

− u ( t ) + i + v C + i' = 0; i = v 'C

Thus, v 'C = i i ' = −v C − i + u(t)

Finally we get,  v ′   0 1   v C  0  v  +   u ( t ) ; i( t ) = [0 1]  C  + [0]u ( t )  C =     i   i ′  − 1 − 1  i  1

Chapter 16, Solution 44. 1/8 F

1H

4u ( t )

+ −

+ vx

2Ω

4Ω

−

First select the inductor current iL and the capacitor voltage vC to be the state variables. Applying KCL we get: v − iL + x + 2

v 'C = 0; or v 'C = 8i L − 4v x 8

i 'L = 4u ( t ) − v x v 'C v 'C v x = vC + 4 = vC + = v C + 4i L − 2v x ; or v x = 0.3333v C + 1.3333i L 8 2 v 'C = 8i L − 1.3333v C − 5.333i L = −1.3333v C + 2.666i L i 'L = 4u ( t ) − 0.3333v C − 1.3333i L

Now we can write the state equations.  v 'C   − 1.3333 2.666   v C  0 0.3333  v C  +   u ( t ); v x =   ' =      1.3333  i L   i L  − 0.3333 − 1.3333  i L  4

Chapter 16, Solution 45.

First select the inductor current iL (current flowing left to right) and the capacitor voltage vC (voltage positive on the left and negative on the right) to be the state variables. Applying KCL we get:

v 'C v o − + + i L = 0 or v 'C = 4i L + 2 v o 4 2 i 'L = v o − v 2 v o = − v C + v1 v 'C = 4i L − 2 v C + 2 v1 i 'L = − v C + v1 − v 2

 i ′  0 − 1  i L  1 − 1  v1 ( t )  i   v (t)  + ; v o ( t ) = [0 − 1]  L  + [1 0]  1   L ′=        v 2 ( t ) v C   v C  4 − 2  v C  2 0   v 2 ( t ) Chapter 16, Solution 46.

First select the inductor current iL (left to right) and the capacitor voltage vC to be the state variables. Letting vo = vC and applying KCL we get: v − i L + v 'C + C − i s = 0 or v 'C = −0.25v C + i L + i s 4 i 'L = − v C + v s Thus,  v '  − 0.25 1  v '  0 1  v s  1  v C  0 0  v s  C + = ; v ( t )   'C  =   o 0   i  + 0 0   i  0  i 'L  1 0  i s    L    s   i L   − 1

Chapter 16, Solution 47.

First select the inductor current iL (left to right) and the capacitor voltage vC (+ on the left) to be the state variables. Letting i1 =

v 'C and i2 = iL and applying KVL we get: 4

Loop 1:  v'  − v1 + v C + 2 C − i L  = 0 or v 'C = 4i L − 2 v C + 2 v1  4   

Loop 2:

 v 'C  '  2 iL − + i + v 2 = 0 or  L  4   4i − 2v C + 2v1 − v 2 = − v C + v1 − v 2 i 'L = −2i L + L 2

i1 =

i ′   L ′=  v C 

4i L − 2 v C + 2 v1 = i L − 0.5v C + 0.5v1 4

0 − 1  i L  1 − 1  v1 ( t )   4 − 2  v  +  2 0   v ( t )  ;   C   2 

 i1 ( t )  1 − 0.5  i L  0.5 0  v1 ( t )  + i ( t ) = 1 0   v C   0 0  v 2 ( t ) 2  

Chapter 16, Solution 48.

Let x1 = y(t). Thus, x1' = y ' = x 2 and x '2 = y′′ = −3x1 − 4 x 2 + z( t ) This gives our state equations.  x1'   0 1   x 1  0  x  +   z( t ); y( t ) = [1 0] 1  + [0]z( t )  ' =    x 2   x 2  − 3 − 4  x 2  1 Chapter 16, Solution 49.

Let x1 = y( t ) and x 2 = x1' − z = y ' − z or y ' = x 2 + z Thus, x '2 = y ′′ − z ' = −6x1 − 5( x 2 + z) + z ' + 2z − z ' = −6x1 − 5x 2 − 3z This now leads to our state equations,  x1'   0 1   x1   1  x  +   z( t ); y( t ) = [1 0]  1  + [0]z( t )  ' =     x 2  − 6 − 5  x 2  − 3 x 2  Chapter 16, Solution 50.

Let x1 = y(t), x2 = x1' , and x 3 = x '2 . Thus, x "3 = −6x1 − 11x 2 − 6x 3 + z( t ) We can now write our state equations.  x1'   0 1 0   x 1  0  x1   '        0 1   x 2  + 0 z( t ); y( t ) = [1 0 0] x 2  + [0]z( t ) x 2  =  0  x '  − 6 − 11 − 6  x  1  x 3   3     3  

Chapter 16, Solution 51. We transform the state equations into the s-domain and solve using Laplace transforms.

1 sX(s) − x (0) = AX(s) + B  s Assume the initial conditions are zero.

1 (sI − A)X(s) = B  s s + 4 − 4 X(s) =  s   2

−1

4  0  0  1  s 1  2  s  = 2  2 s + 4  ( 2 / s )     s + 4s + 8   

1 −s−4 = + s(s 2 + 4s + 8) s s 2 + 4s + 8 1 1 −2 − (s + 2) −s−4 + = + = + 2 2 2 2 s (s + 2) + 2 s (s + 2) + 2 (s + 2) 2 + 2 2 8

Y(s) = X1 (s) =

(

)

y(t) = 1 − e − 2 t (cos 2t + sin 2t ) u ( t )

Chapter 16, Solution 52.

Assume that the initial conditions are zero. Using Laplace transforms we get, 1  s + 2 X(s) =    − 2 s + 4 X1 =

=

3s + 8 2

2

s((s + 3) + 1 )

−1

=

s + 4 − 1  3 / s  1 1  1 / s  1   4 0  2 / s  = 2 s + 2 4 / s     s + 6s + 10  2 0.8 − 0.8s − 1.8 + s (s + 3) 2 + 12

0.8 s+3 1 − 0.8 + .6 s (s + 3) 2 + 12 (s + 3) 2 + 12

x1 ( t ) = (0.8 − 0.8e −3t cos t + 0.6e −3t sin t )u ( t )

X2 =

=

4s + 14 s((s + 3) 2 + 12

=

1.4 − 1.4s − 4.4 + s (s + 3) 2 + 12

1.4 s+3 1 − 1.4 − 0.2 2 2 s (s + 3) + 1 (s + 3) 2 + 12

x 2 ( t ) = (1.4 − 1.4e −3t cos t − 0.2e −3t sin t )u ( t ) y1 ( t ) = −2x1 ( t ) − 2x 2 ( t ) + 2u ( t ) = (−2.4 + 4.4e − 3t cos t − 0.8e − 3t sin t )u ( t ) y 2 ( t ) = x1 ( t ) − 2u ( t ) = (−1.2 − 0.8e −3t cos t + 0.6e −3t sin t )u ( t )

Chapter 16, Solution 53.

If Vo is the voltage across R, applying KCL at the non-reference node gives Is =

Vo V 1 1 + sC Vo + o =  + sC +  Vo R sL  R sL  Is

Vo =

Io =

1 1 + sC + R sL

=

sRL Is sL + R + s 2 RLC

Vo sL Is = 2 R s RLC + sL + R

H(s) =

Io sL s RC = 2 = 2 Is s RLC + sL + R s + s RC + 1 LC

The roots s1, 2 =

-1 1 1 ± 2 − 2RC (2RC) LC

both lie in the left half plane since R, L, and C are positive quantities. Thus, the circuit is stable.

Chapter 16, Solution 54.

(a)

H1 (s) =

3 , s +1

H(s) = H1 (s) H 2 (s) =

H 2 (s) =

1 s+4

3 (s + 1)(s + 4)

 A B  + h ( t ) = L-1 [ H(s)] = L-1   s + 1 s + 4  A = 1, B = -1 -t -4t h ( t ) = ( e − e ) u( t ) (b)

Since the poles of H(s) all lie in the left half s-plane, the system is stable.

Chapter 16, Solution 55.

Let

Vo1 be the voltage at the output of the first op amp. Vo1 − 1 sC − 1 = = , Vs R sRC H(s) =

Vo 1 = 2 2 2 Vs s R C

h(t) =

t R C2

Vo −1 = Vo1 sRC

2

lim h ( t ) = ∞ , i.e. the output is unbounded. t →∞

Hence, the circuit is unstable.

Chapter 16, Solution 56.

1 sL ⋅ 1 sC = sL sL || = 1 1 + s 2 LC sC sL + sC sL 2 V2 sL = 1 + s LC = 2 sL V1 s RLC + sL + R R+ 2 1 + s LC V2 = V1

1 RC 1 1 s2 + s ⋅ + RC LC s⋅

Comparing this with the given transfer function, 1 1 2= , 6= RC LC If R = 1 kΩ ,

C=

1 = 500 µF 2R

L=

1 = 333.3 H 6C

Chapter 16, Solution 57.

The circuit in the s-domain is shown below. R1

Vi

L

V1

+

+ −

C

R2

Vx

− Z

Z=

(1 sC) ⋅ (R 2 + sL) R 2 + sL 1 || (R 2 + sL) = = sC R 2 + sL + 1 sC 1 + s 2 LC + sR 2 C

V1 =

Z V R1 + Z i

Vo =

R2 R2 Z ⋅ V1 = V R 2 + sL R 2 + sL R 1 + Z i

R 2 + sL Vo R2 R2 1 + s 2 LC + sR 2 C Z = ⋅ = ⋅ R 2 + sL Vi R 2 + sL R 1 + Z R 2 + sL R1 + 1 + s 2 LC + sR 2 C Vo R2 = 2 Vi s R 1 LC + sR 1 R 2 C + R 1 + R 2 + sL

R2 Vo R 1 LC = R2 Vi 1  R1 + R 2 + s 2 + s +  L R 1C  R 1 LC

Comparing this with the given transfer function, R2 R2 R1 + R 2 1 5= 6= 25 = + R 1 LC L R 1C R 1 LC Since R 1 = 4 Ω and R 2 = 1 Ω , 1 1 5=  → LC = 4 LC 20 6=

1 1 + L 4C

25 =

5 4 LC

(2)

 → LC =

1 20

Substituting (1) into (2), 1 6 = 20 C +  → 80 C 2 − 24 C + 1 = 0 4C Thus, C =

1 , 4

1 20

(1)

When C =

1 , 4

L=

1 1 = . 20 C 5

When C =

1 , 20

L=

1 = 1. 20 C

Therefore, there are two possible solutions. C = 0.25 F L = 0.2 H or C = 0.05 F

L = 1H

Chapter 16, Solution 58.

We apply KCL at the noninverting terminal at the op amp. (Vs − 0) Y3 = (0 − Vo )(Y1 − Y2 ) Y3 Vs = - (Y1 + Y2 )Vo Vo - Y3 = Vs Y1 + Y2

Let

Y1 = sC1 ,

Y2 = 1 R 1 ,

Y3 = sC 2

Vo - sC 2 - sC 2 C1 = = Vs sC1 + 1 R 1 s + 1 R 1C1

Comparing this with the given transfer function, C2 1 = 1, = 10 R 1 C1 C1 If R 1 = 1 kΩ , C1 = C 2 =

1 = 100 µF 10 4

Chapter 16, Solution 59.

Consider the circuit shown below. We notice that V3 = Vo and V2 = V3 = Vo . Y4 Y1

Vin

Y2

V2

V1

+ −

− +

Vo

Y3

At node 1, (Vin − V1 ) Y1 = (V1 − Vo ) Y2 + (V1 − Vo ) Y4 Vin Y1 = V1 (Y1 + Y2 + Y4 ) − Vo (Y2 + Y4 ) At node 2, (V1 − Vo ) Y2 = (Vo − 0) Y3 V1 Y2 = (Y2 + Y3 ) Vo V1 =

(1)

Y2 + Y3 Vo Y2

(2)

Substituting (2) into (1), Y2 + Y3 Vin Y1 = ⋅ (Y1 + Y2 + Y4 ) Vo − Vo (Y2 + Y4 ) Y2 Vin Y1 Y2 = Vo (Y1 Y2 + Y22 + Y2 Y4 + Y1 Y3 + Y2 Y3 + Y3 Y4 − Y22 − Y2 Y4 ) Vo Y1 Y2 = Vin Y1 Y2 + Y1 Y3 + Y2 Y3 + Y3 Y4 Y1 and Y2 must be resistive, while Y3 and Y4 must be capacitive.

Let

Y1 =

1 , R1

Y2 =

1 , R2

Y3 = sC1 ,

Y4 = sC 2

1 Vo R 1R 2 = sC1 sC1 1 Vin + + + s 2 C1 C 2 R 1R 2 R 1 R 2 1 Vo R 1 R 2 C1C 2 =  R1 + R 2  Vin 1 + s2 + s ⋅  R 1 R 2 C 2  R 1 R 2 C1 C 2

Choose R 1 = 1 kΩ , then 1 = 10 6 R 1 R 2 C1 C 2

and

R1 + R 2 = 100 R 1R 2 C 2

We have three equations and four unknowns. Thus, there is a family of solutions. One such solution is R 2 = 1 kΩ , C1 = 50 nF , C 2 = 20 µF

Chapter 16, Solution 60. With the following MATLAB codes, the Bode plots are generated as shown below. num=[1 1]; den= [1 5 6]; bode(num,den);

Chapter 16, Solution 61. We use the following codes to obtain the Bode plots below. num=[1 4]; den= [1 6 11 6]; bode(num,den);

Chapter 16, Solution 62. The following codes are used to obtain the Bode plots below. num=[1 1]; den= [1 0.5 1]; bode(num,den);

Chapter 16, Solution 63. We use the following commands to obtain the unit step as shown below. num=[1 2]; den= [1 4 3]; step(num,den);

Chapter 16, Solution 64. With the following commands, we obtain the response as shown below. t=0:0.01:5; x=10*exp(-t); num=4; den= [1 5 6]; y=lsim(num,den,x,t); plot(t,y)

Chapter 16, Solution 65. We obtain the response below using the following commands. t=0:0.01:5; x=1 + 3*exp(-2*t); num=[1 0]; den= [1 6 11 6]; y=lsim(num,den,x,t); plot(t,y)

Chapter 16, Solution 66. We obtain the response below using the following MATLAB commands. t=0:0.01:5; x=5*exp(-3*t); num=1; den= [1 1 4]; y=lsim(num,den,x,t); plot(t,y)

Chapter 16, Solution 67. Using the result of Practice Problem 16.14, Vo - Y1 Y2 = Vi Y2 Y3 + Y4 (Y1 + Y2 + Y3 )

When Y1 = sC1 ,

C1 = 0.5 µF

1 , R1

R 1 = 10 kΩ

Y2 =

Y3 = Y2 ,

Y4 = sC 2 ,

C 2 = 1 µF

Vo - sC1 R 1 - sC1 R 1 = = 2 Vi 1 R 1 + sC 2 (sC1 + 2 R 1 ) 1 + sC 2 R 1 (2 + sC1 R 1 ) Vo - sC1 R 1 = 2 Vi s C1C 2 R 12 + s ⋅ 2C 2 R 1 + 1 Vo - s (0.5 × 10 -6 )(10 × 10 3 ) = Vi s 2 (0.5 × 10 -6 )(1 × 10 -6 )(10 × 10 3 ) 2 + s (2)(1 × 10 -6 )(10 × 10 3 ) + 1 Vo - 100 s = 2 Vi s + 400 s + 2 × 10 4

Therefore, a = - 100 ,

b = 400 ,

c = 2 × 10 4

Chapter 16, Solution 68.

(a)

Let

Y(s) =

K (s + 1) s+3

K (s + 1) K (1 + 1 s) = lim =K s →∞ s →∞ 1 + 3 s s+3

Y(∞) = lim i.e.

0.25 = K .

Hence, Y(s) =

(b)

s+1 4 (s + 3)

Consider the circuit shown below. t=0

Vs = 8 V

+ −

I

YS

Vs = 8 u ( t )  → Vs = 8 s I=

Vs 8 s + 1 2 (s + 1) = Y(s) Vs (s) = ⋅ = Z 4s s + 3 s (s + 3)

I=

A B + s s+3

A = 2 3,

i( t ) =

B= -4 3

1 [ 2 − 4 e -3t ] u(t ) A 3

Chapter 16, Solution 69.

The gyrator is equivalent to two cascaded inverting amplifiers. Let V1 be the voltage at the output of the first op amp. V1 =

-R V = -Vi R i

Vo =

- 1 sC 1 V1 = V R sCR i

Io =

Vo Vo = R sR 2 C

Vo = sR 2 C Io

Vo = sL, when L = R 2 C Io

Chapter 17, Solution 1. (a)

This is periodic with ω = π which leads to T = 2π/ω = 2.

(b) y(t) is not periodic although sin t and 4 cos 2πt are independently periodic. (c)

Since sin A cos B = 0.5[sin(A + B) + sin(A – B)], g(t) = sin 3t cos 4t = 0.5[sin 7t + sin(–t)] = –0.5 sin t + 0.5 sin7t which is harmonic or periodic with the fundamental frequency ω = 1 or T = 2π/ω = 2π.

(d) h(t) = cos 2 t = 0.5(1 + cos 2t). Since the sum of a periodic function and a constant is also periodic, h(t) is periodic. ω = 2 or T = 2π/ω = π. (e)

The frequency ratio 0.6|0.4 = 1.5 makes z(t) periodic. ω = 0.2π or T = 2π/ω = 10.

(f)

p(t) = 10 is not periodic.

(g)

g(t) is not periodic.

Chapter 17, Solution 2. (a)

The frequency ratio is 6/5 = 1.2. The highest common factor is 1. ω = 1 = 2π/T or T = 2π.

(b)

ω = 2 or T = 2π/ω = π.

(c)

f3(t) = 4 sin2 600π t = (4/2)(1 – cos 1200π t) ω = 1200π or T = 2π/ω = 2π/(1200π) = 1/600.

(d)

f4(t) = ej10t = cos 10t + jsin 10t. ω = 10 or T = 2π/ω = 0.2π.

Chapter 17, Solution 3. T = 4, ωo = 2π/T = π/2 g(t) = 5, 10, 0,

0
T

1

0

0

2

ao = (1/T) ∫ g( t )dt = 0.25[ ∫ 5dt + ∫ 10dt ] = 3.75 an = (2/T)

1

T

∫ g( t ) cos(nωo t )dt = (2/4)[ 0

1

∫ 5 cos( 0

2 nπ nπ t )dt + ∫ 10 cos( t )dt ] 1 2 2

2

1

nπ 2 2 nπ t + 10 sin t ] = (–1/(nπ))5 sin(nπ/2) = 0.5[ 5 sin 2 0 nπ nπ 2 1

an =

bn = (2/T)

(5/(nπ))(–1)(n+1)/2, 0,

T

∫ g( t ) sin(nω t )dt = o

0

(2/4)[

0

1

= 0.5[

nπ

1

∫ 5 sin( 2

n = odd n = even 2

t )dt + ∫ 10 sin( 1

nπ t )dt ] 2

2

nπ nπ − 2x5 2 x10 t – t ] = (5/(nπ))[3 – 2 cos nπ + cos(nπ/2)] cos cos 2 0 2 1 nπ nπ

Chapter 17, Solution 4.

f(t) = 10 – 5t, 0 < t < 2, T = 2, ωo = 2π/T = π ao = (1/T) an = (2/T) =

T

2

2

0

0

0

2 ∫ f ( t )dt = (1/2) ∫ (10 − 5t )dt = 0.5[10t − (5t / 2)] = 5 T

∫ f ( t ) cos(nωo t )dt = (2/2) 0

2

∫ (10) cos(nπt )dt – 0

2

2

∫ (10 − 5t ) cos(nπt )dt 0

2

∫ (5t ) cos(nπt )dt 0

2

−5 5t sin nπt = [–5/(n2π2)](cos 2nπ – 1) = 0 = 2 2 cos nπt + n π nπ 0 0

bn = (2/2)

2

∫ (10 − 5t ) sin(nπt )dt 0

2

2

0

0

∫ (10) sin(nπt )dt – ∫ (5t ) sin(nπt )dt

=

2

2

−5 5t cos nπt = 0 + [10/(nπ)](cos 2nπ) = 10/(nπ) = 2 2 sin nπt + n π nπ 0 0 f(t) = 5 +

Hence

10 ∞ 1 ∑ sin(nπt ) π n =1 n

Chapter 17, Solution 5. T = 2π,

ω = 2π / T = 1 T

1 1 a o = ∫ z( t )dt = [1xπ − 2 xπ] = −0.5 T 2π 0

an =

T

π

2π

0

0

π

T

π

2π

0

0

2 1 1 z( t ) cos nωo dt = ∫ 1 cos ntdt − ∫ T π π

2 1 1 b n = ∫ z( t ) cos nωo dt = ∫ 1sin ntdt − π π T Thus, z( t ) = − 0.5 +

∞

6 sin nt n =1 nπ

∑

n =odd

1

∫ 2 cos ntdt = nπ sin ..nt

2 π 2π − sin nt π = 0 0 nπ

 6 1 2 2π  π , n = odd ∫ 2 sin ntdt = − nπ cos nt 0 + nπ cos nt π =  nπ 0, n = even π

Chapter 17, Solution 6.

T = 2, ωo =

ao =

2π =π 2

1 2 1 6 y( t )dt = (4 x1 + 2 x1) = = 3 ∫ 2 0 2 2

Since this is an odd function, a n = 0. bn =

1 2 2 2 y( t ) sin( nωo t )dt = ∫ 4 sin( nπt )dt + ∫ 2 sin( nπt )dt ∫ 0 1 2 0

=

−4 −4 2 2 1 2 (cos(2nπ) − cos(nπ)) (cos(nπ) − 1) − cos(nπt ) 1 = cos(nπt ) 0 − nπ nπ nπ nπ

=

2 2 4 0, n = even (1 − cos(nπ)) = 4 (1 − cos(nπ)) = (1 − cos(nπ)) − nπ nπ nπ , n = odd nπ

4 ∞ 1 y( t ) = 3 + ∑ sin(nπt ) π n =1 n n = odd

Chapter 17, Solution 7.

T = 12,

ω = 2π / T =

π , 6

a0 = 0

T

4

10

0

−2

4

1 2 a n = ∫ f ( t ) cos nωo dt = [ ∫ 10 cos nπt / 6dt + ∫ (−10) cos nπt / 6dt ] 6 T

=

10 10 4 10 10 [2 sin 2nπ / 3 + sin nπ / 3 − sin 5nπ / 3] sin nπt / 6 − 2 − sin nπt / 6 4 = nπ nπ nπ T

4

10

0

−2

4

1 2 b n = ∫ f ( t ) sin nωo dt = [ ∫ 10 sin nπt / 6dt + ∫ (−10) sin nπt / 6dt ] 6 T

=−

10 10 4 10 10 [cos 5nπ / 3 + cos nπ / 3 − 2 sin 2nπ / 3] cos nπt / 6 − 2 + cos nπnt / 6 4 = nπ nπ nπ

f (t) =

∞

∑ (a n cos nπt / 6 + b n sin nπt / 6)

n =1

where an and bn are defined above.

Chapter 17, Solution 8. f ( t ) = 2(1 + t ), - 1 < t < 1, T

T = 2,

1

ωo = 2π / T = π

1 1 a o = ∫ f ( t )dt = ∫ 2( t + 1)dt = t 2 + t T 2 −1

0

T

1

=2 −1 1

1

an =

 1  2 2 t 1 f ( t ) cos nωo dt = ∫ 2( t + 1) cos nπtdt = 2 cos nπt + sin nπt + sin nπt  = 0 ∫ T 2 nπ nπ  n 2π2  −1 −1 0

bn =

  t 1 4 2 2 1 sin nπt − cos nπt − cos nπt  = − cos nπ f ( t ) sin nωo dt = ∫ 2( t + 1) sin nπtdt = 2 − ∫ 2 2 nπ nπ nπ T 2  n π  −1

T

1

0

−1

f (t) = 2 −

1

4 ∞ (−1) n cos nπt ∑ π n =1 n

Chapter 17, Solution 9.

f(t) is an even function, bn=0. T = 8,

ao =

ω = 2π / T = π / 4

2 T  10 4 1 2 ( ) f t dt =  ∫ 10 cos πt / 4dt + 0 = ( ) sin πt / 4 ∫ 8 0 T 0  4 π

2 0

=

10

π

= 3.183

an =

4 T

T /2

∫

f (t ) cos nω o dt =

0

2

2

40 [ 10 cos πt / 4 cos nπt / 4dt +0] = 5∫ [cos πt (n + 1) / 4 + cos πt (n − 1) / 4]dt 8 ∫0 0

For n = 1, 2

2

2  a1 = 5∫ [cos πt / 2 + 1]dt = 5 sin πt / 2dt + t  = 10 π 0 0 For n>1, 2

20 20 20 20 π (n + 1)t π (n − 1) π (n + 1) π (n − 1) an = sin sin sin sin + = + π (n + 1) π (n − 1) π (n + 1) π (n − 1) 4 4 2 2 0

a2 =

10

π

sin π +

20

π

sin π / 2 = 6.3662,

a3 =

20 10 sin 2π + sin π = 0 4π π

Thus, a 0 = 3.183,

a1 = 10,

a 2 = 6.362,

a3 = 0,

b1 = 0 = b2 = b3

Chapter 17, Solution 10. T = 2,

ωo = 2π / T = π

cn =

T  − jnπt 1 2e − jnπt 2  2 1 1  1 − jnπt − jnωo t − jnπt  1 4e h ( t ) e dt 4 e dt ( 2 ) e dt = = + − ∫1  2  − jnπ 0 − − jnπ 1  T∫ 2  ∫0   0

cn =

 6j j j − , n = odd 4e − jπn − 4 − 2e − j2nπ + 2e − jnπ = [6 cos nπ − 6] =  nπ , 2nπ 2nπ  0, n = even

[

]

Thus,

f (t ) =

∞

 − j6  jnπt  e n =−∞  nπ 

∑

n =odd

Chapter 17, Solution 11. T = 4,

ω o = 2π / T = π / 2 T

1 1 1 0 c n = ∫ y( t )e − jnωo t dt =  ∫ ( t + 1)e − jnπt / 2 dt + ∫ (1)e − jnπt / 2 dt  0 T 4  −1  0

cn =

=

1  e − jnπt / 2 2 − jnπt / 2 0 2 − jnπt / 2 1  − e e  2 2 (− jnπt / 2 − 1) − −1 jnπ 0 4  − n π / 4 jnπ 

1 4 2 4 2 jnπ / 2 2 − jnπ / 2 2  e jnπ / 2 ( jnπ / 2 − 1) + e e − + − +  4  n 2 π 2 jnπ n 2 π 2 jnπ jnπ jnπ 

But e jnπ / 2 = cos nπ / 2 + j sin nπ / 2 = j sin nπ / 2,

cn =

1 n 2π2

e − jnπ / 2 = cos nπ / 2 − j sin nπ / 2 = − j sin nπ / 2

[1 + j( jnπ / 2 − 1) sin nπ / 2 + nπ sin nπ / 2]

y( t ) =

∞

∑

n = −∞

1 2 2

n π

[1 + j( jnπ / 2 − 1) sin nπ / 2 + nπ sin nπ / 2]e jnπt / 2

Chapter 17, Solution 12.

A voltage source has a periodic waveform defined over its period as v(t) = t(2π - t) V, for all 0 < t < 2π Find the Fourier series for this voltage. v(t) = 2π t – t2, 0 < t < 2π, T = 2π, ωo = 2π/T = 1

ao = T

(1/T) ∫ f ( t )dt = 0

1 2π 1 (πt 2 − t 3 / 3) (2πt − t 2 )dt = ∫ 0 2π 2π

2π 0

=

4π 3 2π 2 (1 − 2 / 3) = 2π 3

2π

2 T 1  2π 2πt  an = ∫ (2πt − t 2 ) cos(nt )dt =  2 cos(nt ) + sin(nt ) 0 T n π n 0

bn = =

[

−

1 2nt cos(nt ) − 2 sin(nt ) + n 2 t 2 sin( nt ) 3 πn

=

−4 2 1 (1 − 1) − 3 4nπ cos(2πn ) = 2 2 πn n n

]

2π 0

2 T 1 (2nt − t 2 ) sin( nt )dt = ∫ (2nt − t 2 ) sin(nt )dt ∫ T 0 π 2π 2n 1 1 π (sin(nt ) − nt cos(nt )) 0 − 3 (2nt sin(nt ) + 2 cos(nt ) − n 2 t 2 cos(nt )) 2 0 π n πn

=

Hence,

f(t) =

− 4 π 4π + =0 n n

2π 2 ∞ 4 − ∑ 2 cos(nt ) 3 n =1 n

Chapter 17, Solution 13.

T = 2π, ωo = 1 T

ao = (1/T) ∫ h( t )dt = 0

= an = (2/T)

2π 1 π [ ∫ 10 sin t dt + ∫ 20 sin( t − π) dt ] π 2π 0

[

]

1 30 π 2π − 10 cos t 0 − 20 cos( t − π) π = 2π π T

∫ h( t ) cos(nω t )dt 0

o

π = [2/(2π)]  ∫ 10 sin t cos( nt )dt +  0

∫

2π

π

20 sin( t − π) cos( nt )dt  

Since sin A cos B = 0.5[sin(A + B) + sin(A – B)] sin t cos nt = 0.5[sin((n + 1)t) + sin((1 – n))t] sin(t – π) = sin t cos π – cost sin π = –sin t sin(t – π)cos(nt) = –sin(t)cos(nt)

an =

2π 1  π 10∫ [sin([1 + n ]t ) + sin([1 − n ]t )]dt − 20∫ [sin([1 + n ]t ) + sin([1 − n ]t )]dt   π 2π  0

5 = π

 cos([1 + n ]t ) cos([1 − n ]t )  π  2 cos([1 + n ]t ) 2 cos([1 − n ]t )  2 π  − +  −  +   1+ n 1− n 1+ n 1− n 0   π  

But,

3 3 cos([1 + n ]π) 3 cos([1 − n ]π)   3 − 1 + n + 1 − n −  1+ n 1− n

5 π

an =

[1/(1+n)] + [1/(1-n)] = 1/(1–n2)

cos([n–1]π) = cos([n+1]π) = cos π cos nπ – sin π sin nπ = –cos nπ an = (5/π)[(6/(1–n2)) + (6 cos(nπ)/(1–n2))] = [30/(π(1–n2))](1 + cos nπ) = [–60/(π(n–1))], n = even = 0, n = odd T

bn = (2/T) ∫ h ( t ) sin nωo t dt 0

π

2π

0

π

= [2/(2π)][ ∫ 10 sin t sin nt dt + ∫ 20( − sin t ) sin nt dt But,

sin A sin B = 0.5[cos(A–B) – cos(A+B)] sin t sin nt = 0.5[cos([1–n]t) – cos([1+n]t)] π

bn = (5/π){[(sin([1–n]t)/(1–n)) – (sin([1+n]t)/ (1 + n )] 0 2π

+ [(2sin([1-n]t)/(1-n)) – (2sin([1+n]t)/ (1 + n )] π } =

Thus,

5 π

 sin([1 − n ]π) sin([1 + n ]π)  +  −  = 0 1− n 1+ n h(t) =

30 60 ∞ cos( 2kt ) − ∑ π π k = 1 ( 4k 2 − 1)

Chapter 17, Solution 14.

Since cos(A + B) = cos A cos B – sin A sin B. ∞ 10   10 cos(nπ / 4) cos( 2nt ) − 3 sin(nπ / 4) sin( 2nt )  f(t) = 2 + ∑  3 n +1  n =1  n + 1

Chapter 17, Solution 15.

(a)

Dcos ωt + Esin ωt = A cos(ωt - θ) where

f(t) = 10 +

A =

D 2 + E 2 , θ = tan-1(E/D)

A =

16 1 + 6 , θ = tan-1((n2+1)/(4n3)) 2 ( n + 1) n

∞

∑

n =1

(b)

2

2  16 1 −1 n + 1   cos 10nt − tan + ( n 2 + 1) 2 n 6 4n 3  

Dcos ωt + Esin ωt = A sin(ωt + θ) where

D 2 + E 2 , θ = tan-1(D/E)

A =

f(t) = 10 +

∞

∑

n =1

 16 1 4n 3  −1  sin 10nt + tan + ( n 2 + 1) 2 n 6 n 2 + 1  

Chapter 17, Solution 16.

If v2(t) is shifted by 1 along the vertical axis, we obtain v2*(t) shown below, i.e. v2*(t) = v2(t) + 1. v2*(t) 2 1 -2 -1

0

1

2

3

4

5

t

Comparing v2*(t) with v1(t) shows that v2*(t) = 2v1((t + to)/2) where (t + to)/2 = 0 at t = -1 or to = 1 Hence

v2*(t) = 2v1((t + 1)/2)

But

v2*(t) = v2(t) + 1 v2(t) + 1 = 2v1((t+1)/2) v2(t) = -1 + 2v1((t+1)/2)

= -1 + 1 −

v2(t) = −

8 π2

8 π2

   t + 1  t + 1 1  t + 1 1 cos π 2  + 9 cos 3π 2  + 25 cos 5π 2  + "        

  πt π  1   5πt 5π   3πt 3π  1 cos 2 + 2  + 9 cos 2 + 2  + 25 cos 2 + 2  + "        

v2(t) = −

8 π2

  π t  1  3 πt   1  5 πt  sin 2  + 9 sin 2  + 25 sin 2  + "        

Chapter 17, Solution 17.

We replace t by –t in each case and see if the function remains unchanged. (a)

1 – t,

neither odd nor even.

(b)

t2 – 1,

even

(c)

cos nπ(-t) sin nπ(-t) = - cos nπt sin nπt,

odd

(d)

sin2 n(-t) = (-sin πt)2 = sin2 πt,

even

(e)

e t,

neither odd nor even.

Chapter 17, Solution 18.

(a)

T = 2 leads to ωo = 2π/T = π f1(-t) = -f1(t), showing that f1(t) is odd and half-wave symmetric.

(b)

T = 3 leads to ωo = 2π/3 f2(t) = f2(-t), showing that f2(t) is even.

(c)

T = 4 leads to ωo = π/2 f3(t) is even and half-wave symmetric.

Chapter 17, Solution 19.

This is a half-wave even symmetric function. ao = 0 = bn, ωo = 2π/T π/2 an =

4 T

∫

T/2

0

4t   1 − T  cos(nωo t )dt

= [4/(nπ)2](1 − cos nπ)

f (t) =

8 π2

∞

∑

n = odd

= 8/(n2π2), = 0,

1  nπt  cos  2 n  2 

Chapter 17, Solution 20. This is an even function. bn = 0, T = 6, ω = 2π/6 = π/3 ao =

2 T

∫

T/2

0

f ( t )dt =

3 2 2 − ( 4 t 4 ) dt 4 dt  ∫ ∫   1 2 6

n = odd n = even

2 1 2 ( 2 t − 4 t ) + 4(3 − 2) = 2 1  3 

=

4 T

an =

∫

T/4

0

f ( t ) cos( nπt / 3)dt 2

= (4/6)[ ∫ ( 4 t − 4) cos( nπt / 3)dt + 1

∫

3

2

4 cos( nπt / 3)dt ] 2

3

16  3 3 16  9  nπt   nπt   nπt   nπt  3t = sin sin sin cos   + − + 2 2   6  nπ  3  2 6 n π  3  nπ  3  nπ  3 1 = [24/(n2π2)][cos(2nπ/3) − cos(nπ/3)] f(t) = 2 +

Thus

24 ∞ 1 ∑ π 2 n =1 n2

  2πn   nπt   πn   cos 3  − cos 3   cos 3        

At t = 2, f(2) = 2 + (24/π2)[(cos(2π/3) − cos(π/3))cos(2π/3) + (1/4)(cos(4π/3) − cos(2π/3))cos(4π/3) + (1/9)(cos(2π) − cos(π))cos(2π) + -----] = 2 + 2.432(0.5 + 0 + 0.2222 + -----) f(2) = 3.756

Chapter 17, Solution 21. This is an even function. bn = 0, T = 4, ωo = 2π/T = π/2. f(t) = 2 − 2t, = 0,

0
 t2  2 1 ao = 2(1 − t )dt =  t −  = 0.5 2 0 4 ∫0  an =

4 T

∫

T/2

0

f ( t ) cos( nωo t )dt =

4 1  nπt  2(1 − t ) cos dt ∫ 4 0  2 

= [8/(π2n2)][1 − cos(nπ/2)] ∞

1 + 2

f(t) =

8

∑n π n=1

2

2

  nπt   nπ   1 − cos 2   cos 2      

Chapter 17, Solution 22. Calculate the Fourier coefficients for the function in Fig. 16.54. f(t) 4

-5 -4 -3 -2 -1

0

1

Figure 17.61

2

3

4

5

t

For Prob. 17.22

This is an even function, therefore bn = 0. In addition, T=4 and ωo = π/2. ao =

an =

2 T

∫

4 T

T2

0

∫

f ( t )dt =

T2

0

1 2 1 4 tdt = t 2 = 1 ∫ 0 0 4

f ( t ) cos(ωo nt )dt =

4 1 4 t cos( nπt / 2)dt 4 ∫0 1

2t  4  sin( nπt / 2) = 4  2 2 cos( nπt / 2) + nπ n π 0 an =

16 8 sin( nπ / 2) (cos( nπ / 2) − 1) + 2 2 nπ n π

Chapter 17, Solution 23.

f(t) is an odd function. f(t) = t, −1< t < 1 ao = 0 = an, T = 2, ωo = 2π/T = π

bn =

=

4 T

∫

T/2

0

f ( t ) sin( nωo t )dt =

4 1 t sin( nπt )dt 2 ∫0

2 [sin(nπt ) − nπt cos(nπt )] 10 2 n π 2

= −[2/(nπ)]cos(nπ) = 2(−1)n+1/(nπ) f(t) =

2 π

( −1) n + 1 sin( nπt ) n n =1 ∞

∑

Chapter 17, Solution 24.

(a)

This is an odd function. ao = 0 = an, T = 2π, ωo = 2π/T = 1 bn =

4 T

∫

T/2

0

f ( t ) sin(ωo nt )dt f(t) = 1 + t/π,

bn =

4 2π

∫

π

0

0
(1 + t / π) sin( nt )dt π

=

2 1 1 t  − cos( nt ) + 2 sin( nt ) − cos( nt )  π n n π nπ 0

= [2/(nπ)][1 − 2cos(nπ)] = [2/(nπ)][1 + 2(−1)n+1] a2 = 0, b2 = [2/(2π)][1 + 2(−1)] = −1/π = −0.3183 (b)

ωn = nωo = 10 or n = 10 a10 = 0, b10 = [2/(10π)][1 − cos(10π)] = −1/(5π) Thus the magnitude is A10 = and the phase is

2 a 210 + b10 = 1/(5π) = 0.06366

φ10 = tan−1(bn/an) = −90°

∞

(c)

f(t) =

2

∑ nπ [1 − 2 cos(nπ)] sin(nt ) π n =1

f(π/2) =

∞

2

∑ nπ [1 − 2 cos(nπ)] sin(nπ / 2) π n =1

For n = 1,

f1 = (2/π)(1 + 2) = 6/π

For n = 2,

f2 = 0

For n = 3,

f3 = [2/(3π)][1 − 2cos(3π)]sin(3π/2) = −6/(3π)

For n = 4,

f4 = 0

For n = 5,

f5 = 6/(5π), ----

Thus, f(π/2) = 6/π − 6/(3π) + 6/(5π) − 6/(7π) --------= (6/π)[1 − 1/3 + 1/5 − 1/7 + --------] f(π/2) ≅ 1.3824 which is within 8% of the exact value of 1.5. (d)

From part (c) f(π/2) = 1.5 = (6/π)[1 − 1/3 + 1/5 − 1/7 + - - -] (3/2)(π/6) = [1 − 1/3 + 1/5 − 1/7 + - - -] or π/4 = 1 − 1/3 + 1/5 − 1/7 + - - -

Chapter 17, Solution 25.

This is an odd function since f(−t) = −f(t). ao = 0 = an, T = 3, ωo = 2π/3. bn =

4 T

∫

T/2

0

f ( t ) sin( nωo t )dt =

4 1 t sin(2πnt / 3)dt 3 ∫0

1

3t 4 9  2πnt   2πnt  = cos sin −  2 2  3  4π n  3  2nπ  3  0 =

f(t) =

3 4 9  2 πn    2 πn  cos sin  − 2 2  3  4π n  3   3  2nπ ∞

 3

∑ π n n =1

2



2

2  2 πn   2π n    2π t  cos sin −    sin  3  nπ  3   3 

Chapter 17, Solution 26.

T = 4, ωo = 2π/T = π/2 ao =

1 1 1 T 1 dt + = f ( t ) dt 4  ∫0 T ∫0

an =

2 T f ( t ) cos( nωo t )dt T ∫0

an =

2 2 1 cos( nπt / 2)dt + 4  ∫1

∫

3

1

∫

3

2

2 dt + ∫ 1 dt  = 1  3 4

2 cos( nπt / 2)dt + ∫ 1 cos( nπt / 2)dt   3 4

2 3 4 2 nπt  nπt 2 nπt 4 sin + = 2  sin sin +  2 3  2 2 nπ 2 1 nπ  nπ

=

4 nπ

nπ   3nπ sin 2 − sin 2 

bn =

2 T f ( t ) sin( nωo t )dt T ∫0

=

nπt 2 2 dt + 1 sin ∫  1 2 4

∫

3

2

2 sin

nπt dt + 2

∫

4

3

1 sin

nπt  dt  2 

2 3 4  2 nπt  nπt 2 nπt 4 cos − cos − cos = 2−  2 3  2 2 nπ 2 1 nπ  nπ

=

4 [cos(nπ) − 1] nπ

Hence f(t) = 1+

∞

4

∑ nπ [(sin( 3nπ / 2) − sin(nπ / 2)) cos( nπt / 2) + (cos( nπ) − 1) sin(nπt / 2)] n =1

Chapter 17, Solution 27.

(a) (b)

odd symmetry.

ao = 0 = an, T = 4, ωo = 2π/T = π/2 f(t) = t, 0 < t < 1 = 0,

1
nπt  nπt 2 t nπt 4 1  4 bn = cos dt =  2 2 sin t sin − ∫ 2  0 2 nπ 2 4 0 n π =

nπ nπ 2 4 − −0 cos sin 2 2 2 nπ n π 2

= 4(−1)(n−1)/2/(n2π2),

n = odd

−2(−1)n/2/(nπ),

n = even

a3 = 0, b3 = 4(−1)/(9π2) = 0.045 (c)

b1 = 4/π2, b2 = 1/π, b3 = −4/(9π2), b4 = −1/(2π), b5 = (25π2) Frms =

a 2o +

1 ∑ (a 2n + b 2n ) 2

Frms2 = 0.5Σbn2 = [1/(2π2)][(16/π2) + 1 + (16/(8π2)) + (1/4) + (16/(625π2))] = (1/19.729)(2.6211 + 0.27 + 0.00259) Frms =

0.14659 = 0.3829

Compare this with the exact value of Frms =

2 T

1

∫ t dt 0

2

= 1 / 6 = 0.4082

Chapter 17, Solution 28. This is half-wave symmetric since f(t − T/2) = −f(t). ao = 0, T = 2, ωo = 2π/2 = π an =

4 T

∫

T/2

0

f ( t ) cos( nωo t )dt =

4 1 ( 2 − 2 t ) cos( nπt )dt 2 ∫0 1

t 1 1  = 4  sin( nπt ) − 2 2 cos( nπt ) − sin( nπt ) nπ n π  nπ 0 = [4/(n2π2)][1 − cos(nπ)] =

8/(n2π2), 0,

n = odd n = even

1

bn = 4 ∫ (1 − t ) sin( nπt )dt 0

1

t 1  1  = 4 − cos( nπt ) cos( nπt ) − 2 2 sin( nπt ) + nπ n π  nπ 0 = 4/(nπ), n = odd f(t) =

∞



∑  n k =1

8 4  cos( nπt ) + sin(nπt )  , n = 2k − 1 2 nπ π 

2

Chapter 17, Solution 29.

This function is half-wave symmetric. T = 2π, ωo = 2π/T = 1, f(t) = −t, 0 < t < π For odd n,

an =

2 T

bn =

2 π

∫

π

∫

π

0

0

( − t ) cos( nt )dt = − ( − t ) sin( nt )dt = −

2 [cos(nt ) + nt sin(nt )] 0π = 4/(n2π) 2 n π

2 [sin(nt ) − nt cos(nt )] 0π = −2/n 2 n π

Thus, ∞ 1  2  f(t) = 2∑  2 cos( nt ) − sin(nt ) , n  k =1  n π

n = 2k − 1

Chapter 17, Solution 30.

1 cn = T

(a)

T/2

∫

f ( t )e − jnωo t dt =

−T / 2

T/2 1  T/2 f ( t ) cos nω o tdt − j∫ f ( t ) sin nω o tdt  ∫  −T / 2 T  −T / 2 

The second term on the right hand side vanishes if f(t) is even. Hence cn =

(b)

(1)

2 T

T/2

∫ f (t ) cos nωo tdt

0

The first term on the right hand side of (1) vanishes if f(t) is odd. Hence, j2 cn = − T

T/2

∫ f (t ) sin nωo tdt

0

Chapter 17, Solution 31.

If h ( t ) = f (αt ),

T' = T / α

an '=

Let αt = λ, ,

2π 2π = αωo = T' T / α

T'

T'

0

0

T

αT ' = T

2α f (λ) cos nωo λdλ / α = a n T ∫ 0

Similarly,

ωo ' =

2 2 h ( t ) cos nωo ' tdt = ∫ f (αt ) cos nωo ' tdt ∫ T' T'

d t = dλ / α , an '=

 →

bn ' = bn

Chapter 17, Solution 32. When is = 1 (DC component) i = 1/(1 + 2) = 1/3 ωn = 3n, Is = 1/n2∠0°

For n ≥ 1,

I = [1/(1 + 2 + jωn2)]Is = Is/(3 + j6n) 1 ∠0° 2 1 n = ∠ − tan(2n ) = 3 1 + 4n 2 ∠ tan −1 (6n / 3) 3n 2 1 + 4n 2 Thus, i(t) =

1 + 3

∞

∑ n =1

1 3n

1 + 4n

2

2

cos( 3n − tan −1 ( 2n ))

Chapter 17, Solution 33. For the DC case, the inductor acts like a short, Vo = 0. For the AC case, we obtain the following: Vo − Vs V jnπVo =0 + o + 10 j2nπ 4  5   1 + j 2.5nπ −  Vo = Vs nπ     Vo =

Vs 5   1 + j 2.5nπ −  nπ  

A n ∠Θ n =

An =

4 nπ

1 5   1 + j 2.5nπ −  nπ  

=

4 nπ + j(2.5n 2 π 2 − 5)

 2.5n 2 π 2 − 5   ; Θ n = − tan −1    2 2 2 2 2 n π n π + (2.5n π − 5)   4

v o (t) =

∞

∑ A n sin(nπt + Θ n ) V

n =1

Chapter 17, Solution 34. For any n, V = [10/n2]∠(nπ/4), ω = n. 1 H becomes jωnL = jn and 0.5 F becomes 1/(jωnC) = −j2/n 2Ω

jn +

+ −

V

−j2/n

Vo −

Vo = {−j(2/n)/[2 + jn − j(2/n)]}V = {−j2/[2n + j(n2 − 2)]}[(10/n2)∠(nπ/4)] = =

20∠((nπ / 4) − π / 2) n

2

4n + (n 2 − 2) 2 ∠ tan −1 ((n 2 − 2) / 2n ) 2

20 n

2

n +4 2

vo(t) =

∠[(nπ / 4) − (π / 2) − tan −1 ((n 2 − 2) / 2n )] ∞

∑ n =1

 nπ π n2 − cos nt + − − tan −1 4 2 2n n2 + 4  20

n2

2  

Chapter 17, Solution 35. If vs in the circuit of Fig. 17.72 is the same as function f2(t) in Fig. 17.57(b), determine the dc component and the first three nonzero harmonics of vo(t). 1Ω

1H +

vS

+ −

1F

1Ω

vo −

Figure 17.72

For Prob. 17.35

f2(t) 2 1 -2 -1

0

1

2

Figure 17.57(b)

3

4

t

5

For Prob. 17.35

The signal is even, hence, bn = 0. In addition, T = 3, ωo = 2π/3. vs(t)

ao =

an =

=

= 1 for all 0 < t < 1 = 2 for all 1 < t < 1.5

2 1 1dt + 3  ∫0

4 2dt  =  3

1.5

∫

1

4 1 cos(2nπt / 3)dt + 3  ∫0

1.5

∫

1

2 cos(2nπt / 3)dt  

4 3 6 2 1 1.5  sin( 2 n t / 3 ) π + π sin( 2 n t / 3 ) = − sin(2nπ / 3) 0 1  3  2nπ 2nπ nπ 4 2 ∞ 1 vs(t) = − ∑ sin(2nπ / 3) cos(2nπt / 3) 3 π n =1 n

Now consider this circuit, 1Ω

j2nπ/3

+ vS

+ −

-j3/(2nπ)

1Ω

vo

−

Let Z = [-j3/(2nπ)](1)/(1 – j3/(2nπ)) = -j3/(2nπ - j3) Therefore, vo = Zvs/(Z + 1 + j2nπ/3). Simplifying, we get vo =

− j9 v s 12nπ + j( 4n 2 π 2 − 18)

For the dc case, n = 0 and vs = ¾ V and vo = vs/2 = 3/8 V. We can now solve for vo(t) 3 ∞  2nπt  vo(t) =  + ∑ A n cos + Θ n   volts  3   8 n =1

where A n =

6 sin( 2nπ / 3) nπ

3   nπ − and Θ n = 90 o − tan −1   2 3 2 n π 2 2     4n π − 6  16n 2 π 2 +    3

where we can further simplify An to this, A n =

9 sin( 2nπ / 3) nπ 4n 4 π 4 + 81

Chapter 17, Solution 36. ∞

vs(t) =

∑A

n =1 n = odd

n

cos( nt − θ n )

where θn = tan−1[(3/(nπ))/(−1/(nπ))] = tan−1(−3) = 100.5° An =

πn nπ 1 9 1 + 2 2 sin 2 = 9 + sin 2 2 2 2 nπ n π n π 2

ωn = n and 2 H becomes jωnL = j2n Let

Z = 1||j2n = j2n/(1 + j2n)

If Vo is the voltage at the non-reference node or across the 2-H inductor. Vo = ZVs/(1 + Z) = [j2n/(1 + j2n)]Vs/{1 + [j2n/(1 + j2n)]} = j2nVs/(1 + j4n) But

Vs = An∠−θn Vo = j2n An∠−θn/(1 + j4n)

Io = Vo/j = [2n An∠−θn]/ 1 + 16n 2 ∠tan−14n 1  nπ   9 + sin 2 2n nπ  2  ∠−100.5° − tan−14n = 2 1 + 16n

Since sin(nπ/2) = (−1)(n−1)/2 for n = odd, sin2(nπ/2) = 1 2 10 ∠ − 100.5° − tan −1 4n π Io = 1 + 16n 2 io(t) =

2 10 ∞ ∑ π n =1

n = odd

1 1 + 16n

2

cos(nt − 100.5° − tan −1 4n )

Chapter 17, Solution 37.

From Example 15.1, vs(t) = 5 +

20 ∞ 1 ∑ sin(nπt ), π k =1 n

n = 2k − 1

For the DC component, the capacitor acts like an open circuit. Vo = 5

For the nth harmonic,

Vs = [20/(nπ)]∠0° 10 mF becomes 1/(jωnC) = −j/(nπx10x10−3) = −j100/(nπ) 100 5 Vs 100∠ − 90° + tan −1 − j 100 20 nπ nπ = = vo = 2 2 100 nπ 25 + n π − j + 20 20nπ − j100 nπ nπ − j

vo(t) =

100 π

∑

1 n 25 + n 2 π 2

sin(nπt − 90° + tan −1

5 ) nπ

Chapter 17, Solution 38. 1 2 ∞ 1 v s ( t ) = + ∑ sin nπt , 2 π k =1n Vo =

jω n Vs , 1 + jω n

For dc, ω n = 0,

Vo =

ω n = nπ

Vs = 0.5,

For nth harmonic, Vs =

n = 2k + 1

Vo = 0

2 ∠ − 90 o nπ

2 2∠ − tan −1 nπ ∠90 o = 1 + n 2 π 2 ∠ tan −1 nπ nπ 1 + n 2π2 nπ∠90 o

v o (t) =

∞

∑

k =1

2 2 2

1+ n π

•

cos(nπt − tan −1 nπ),

n = 2k − 1

Chapter 17, Solution 39. Comparing vs(t) with f(t) in Figure 15.1, vs is shifted by 2.5 and the magnitude is 5 times that of f(t). Hence 10 ∞ 1 vs(t) = 5 + n = 2k − 1 ∑ sin(nπt ), π k =1 n T = 2, ωo = 2π//T = π, ωn = nωo = nπ For the DC component, For the kth harmonic,

io = 5/(20 + 40) = 1/12 Vs = (10/(nπ))∠0°

100 mH becomes jωnL = jnπx0.1 = j0.1nπ 50 mF becomes 1/(jωnC) = −j20/(nπ)

I 20 Ω VS

40 Ω

Io

−j20/(nπ)

+ −

j0.1nπ

Z

j20 ( 40 + j0.1nπ) n π Let Z = −j20/(nπ)||(40 + j0.1nπ) = j20 − + 40 + j0.1nπ nπ −

− j20( 40 + j0.1nπ 2nπ − j800 = 2 2 − j20 + 40nπ + j0.1n π 40nπ + j(0.1n 2 π 2 − 20)

=

Zin = 20 + Z =

I =

802nπ + j( 2n 2 π 2 − 1200) 40nπ + j(0.1n 2 π 2 − 20)

Vs 400nπ + j( n 2 π 2 − 200) = Z in nπ[802nπ + j( 2n 2 π 2 − 1200)] j20 I nπ

− Io =

=

=

j20 − + ( 40 + j0.1nπ) nπ

− j20I 40nπ + j(0.1n 2 π 2 − 20)

=

− j200 nπ[802nπ + j( 2n 2 π 2 − 1200)] 200∠ − 90° − tan −1{(2n 2 π 2 − 1200) /(802nπ)} nπ (802) 2 + ( 2n 2 π 2 − 1200) 2

Thus io(t) =

where

1 200 + π 20

∞

∑I k =1

θ n = 90° + tan −1 In =

n

sin(nπt − θ n ) ,

2n 2 π 2 − 1200 802nπ

1 n (804nπ) + (2n 2 π 2 − 1200) 2

n = 2k − 1

Chapter 17, Solution 40. T = 2, ωo = 2π/T = π 1 ao = T an =

2 T

1

 1 1 t2  v ( t ) dt ( 2 2 t ) dt t = 1/ 2 = − = −  ∫0 2 ∫0 2  0  T

T

∫ v( t ) cos(nπt )dt = 0

1

∫ 2(1 − t ) cos(nπt )dt 0

1

1 t 1  sin( nπt ) = 2  sin( nπt ) − 2 2 cos( nπt ) − n π nπ  nπ 0 2 = 2 2 (1 − cos nπ) = n π

bn =

n = even

0,

4 4 , n = odd = 2 2 n π π ( 2n − 1) 2 2

1 2 T v ( t ) sin( nπt )dt = 2 ∫ (1 − t ) sin( nπt )dt ∫ 0 T 0 1

1 t 2  1  cos( nπt ) − 2 2 sin( nπt ) + cos( nπt ) = = 2− n π nπ  nπ  0 nπ vs(t) =

1 + 2

∑A

where φn = tan −1

n

cos( nπt − ϕ n )

π( 2n − 1) 2 , An = 2n

4 16 + 4 2 n π π ( 2n − 1) 4 2

For the DC component, vs = 1/2. As shown in Figure (a), the capacitor acts like an open circuit. 1Ω

0.5V

+ −

Vx i

− +

2Vx

+ Vx

− (a)

Vo +

3Ω

Vo

−

1Ω

Vx

− +

2Vx

Vo +

VS

+ −

Vo

3Ω

(1/4)F

− (b)

Applying KVL to the circuit in Figure (a) gives

But Adding (1) and (2),

–0.5 – 2Vx + 4i = 0

(1)

–0.5 + i + Vx = 0 or –1 + 2Vx + 2i = 0

(2)

–1.5 + 6i = 0 or i = 0.25 Vo = 3i = 0.75

For the nth harmonic, we consider the circuit in Figure (b). ωn = nπ, Vs = An∠–φ, 1/(jωnC) = –j4/(nπ) At the supernode, (Vs – Vx)/1 = –[nπ/(j4)]Vx + Vo/3 Vs = [1 + jnπ/4]Vx + Vo/3 But

(3)

–Vx – 2Vx + Vo = 0 or Vo = 3Vx

Substituting this into (3), Vs = [1 + jnπ/4]Vx + Vx = [2 + jnπ/4]Vx = (1/3)[2 + jnπ/4]Vo = (1/12)[8 + jnπ]Vo Vo = 12Vs/(8 + jnπ) =

Vo =

12 64 + n π 2

2

12A n ∠ − φ 64 + n 2 π 2 ∠ tan −1 (nπ / 8)

4 16 + 4 ∠[tan −1 (nπ / 8) − tan −1 (π(2n − 1) /(2n ))] 2 4 n π π (2n − 1) 2

Thus vo(t) =

where

Vn =

3 + 4

∞

∑V n =1

n

cos( nπt + θ n )

12 64 + n 2 π 2

4 16 + 4 2 n π π ( 2n − 1) 4 2

θn = tan–1(nπ/8) – tan–1(π(2n – 1)/(2n)) Chapter 17, Solution 41.

For the full wave rectifier, T = π, ωo = 2π/T = 2, ωn = nωo = 2n Hence vin(t) =

2 4 ∞ 1 − ∑ 2 cos (2nt ) π π n =1 4n − 1

For the DC component, Vin = 2/π The inductor acts like a short-circuit, while the capacitor acts like an open circuit. Vo = Vin = 2/π For the nth harmonic, Vin = [–4/(π(4n2 – 1))]∠0° 2 H becomes jωnL = j4n 0.1 F becomes 1/(jωnC) = –j5/n Z = 10||(–j5/n) = –j10/(2n – j) Vo = [Z/(Z + j4n)]Vin = –j10Vin/(4 + j(8n – 10)) = −

 j10 4∠0°   −  4 + j(8n − 10)  π(4n 2 − 1) 

=

40∠{90° − tan −1 (2n − 2.5)} π(4n 2 − 1) 16 + (8n − 10) 2

Hence

vo(t) =

2 + π

∞

∑A n =1

n

cos( 2nt + θ n )

where An =

20 π( 4n 2 − 1) 16n 2 − 40n + 29

θn = 90° – tan–1(2n – 2.5) Chapter 17, Solution 42.

20 ∞ 1 vs = 5 + ∑ sin nπt, n = 2k - 1 π k =1n Vs − 0 = jω n C(0 − Vo ) R

→

Vo =

j Vs , ω n = nω o = nπ ω n RC

For n = 0 (dc component), Vo=0. For the nth harmonic, 1∠90 o 20 20 10 5 o Vo = ∠ − 90 = = nπRC nπ n 2 π 2 x10 4 x 40 x10 −9 2n 2 π 2

Hence, v o (t) =

10 5 ∞ 1

∑

2π 2 k =1 n 2

cos nπt , n = 2k - 1

Alternatively, we notice that this is an integrator so that v o (t) = −

1 10 5 ∞ 1 v dt = ∑ cos nπt, n = 2k - 1 s RC ∫ 2π 2 k =1n 2

Chapter 17, Solution 43.

a 02 +

(a)

Vrms =

(b)

Irms =

(c)

P = VdcIdc +

1 ∞ 2 1 (a n + b 2n ) = 30 2 + (20 2 + 10 2 ) = 33.91 V ∑ 2 n =1 2

1 6 2 + (4 2 + 2 2 ) = 6.782 A 2

1 ∑ Vn I n cos(Θ n − Φ n ) 2

= 30x6 + 0.5[20x4cos(45o-10o) – 10x2cos(-45o+60o)] = 180 + 32.76 – 9.659 = 203.1 W Chapter 17, Solution 44.

[

]

1 60 cos 25 o + 10 cos 45 o + 0 = 27.19 + 3.535 + 0 = 30.73 W 2

(a)

p = vi =

(b)

The power spectrum is shown below. p

27.19

3.535 0

1

2

3

ω

Chapter 17, Solution 45.

ωn = 1000n

jωnL = j1000nx2x10–3 = j2n 1/(jωnC) = –j/(1000nx40x10–6) = –j25/n

Z = R + jωnL + 1/(jωnC) = 10 + j2n – j25/n I = V/Z For n = 1, V1 = 100, Z = 10 + j2 – j25 = 10 – j23 I1 = 100/(10 – j23) = 3.987∠73.89° For n = 2, V2 = 50, Z = 10 + j4 – j12.5 = 10 – j8.5 I2 = 50/(10 – j8.5) = 3.81∠40.36° For n = 3, V3 = 25, Z = 10 + j6 – j25/3 = 10 – j2.333 I3 = 25/(10 – j2.333) = 2.435∠13.13° Irms = 0.5 3.987 2 + 3.812 + 2.435 2 = 3.014 A 1 3 p = VDCIDC + ∑ Vn I n cos(θ n − φ n ) 2 n =1 = 0 + 0.5[100x3.987cos(73.89°) + 50x3.81cos(40.36°) + 25x2.435cos(13.13°)] = 0.5[110.632 + 145.16 + 59.28] = 157.54 watts Chapter 17, Solution 46. (a)

This is an even function Irms =

f(t) =

1 T 2 f ( t )dt = T ∫0

2 − 2t, 0,

2 T/2 2 f ( t )dt T ∫0

0 < t <1 1< t < 2

T = 4, ωo = 2π/T = π/2 Irms2 =

1 2 1 4(1 − t ) 2 dt = 2( t − t 2 + t 3 / 3) ∫ 0 4 0

= 2(1 – 1 + 1/3) = 2/3 or Irms (b)

= 0.8165 A

From Problem 16.14, an = [8/(n2π2)][1 – cos(nπ/2)], ao = 0.5 a1 = 8/π2, a2 = 4/π2, a3 = 8/(9π2), a4 = 0, a5 = 9/(25π2), a6 = 4/(9π2) Irms =

ao +

1 ∞ 2 ∑ An ≅ 2 n =1

64 64 16  1 1  + 4  64 + 16 + + +  = 0.66623 81 625 81  4 2π 

Irms = 0.8162 A Chapter 17, Solution 47. Let I = IDC + I1 + I2 For the DC component IDC = [5/(5 + 10)](3) = 1 A I

j8

5Ω

10 Ω

Is

For AC, ω = 100

For Is = 0.5∠–60°

jωL = j100x80x10–3 = j8 In = 5Is/(5 + 10 + j8)

I1 = 10∠–60°/(15 + j8) or |I1| = 10/ 15 2 + 8 2 For Is = 0.5∠–120° I2 = 2.5∠–120°/(15 + j8) or |I2| = 2.5/ 15 2 + 8 2 p10 = (IDC2 + |I1|2/2 + |I2|2/2)10 = (1 + [100/(2x289)] + [6.25/(2x289)])x10 p10 = 11.838 watts

Chapter 17, Solution 48. (a) For the DC component, i(t) = 20 mA. The capacitor acts like an open circuit so that v = Ri(t) = 2x103x20x10–3 = 40 For the AC component, ωn = 10n, n = 1,2 1/(jωnC) = –j/(10nx100x10–6) = (–j/n) kΩ Z = 2||(–j/n) = 2(–j/n)/(2 – j/n) = –j2/(2n – j) V = ZI = [–j2/(2n – j)]I For n = 1,

V1 = [–j2/(2 – j)]16∠45° = 14.311∠–18.43° mV

For n = 2,

V2 = [–j2/(4 – j)]12∠–60° = 5.821∠–135.96° mV

v(t) = 40 + 0.014311cos(10t – 18.43°) + 0.005821cos(20t – 135.96°) V (b)

p = VDCIDC +

1 ∞ ∑ Vn I n cos(θ n − φ n ) 2 n =1

= 20x40 + 0.5x10x0.014311cos(45° + 18.43°) +0.5x12x0.005821cos(–60° + 135.96°) = 800.1 mW Chapter 17, Solution 49. (a)

T 2π  π 1 2 1  1  ∫ 1dt + ∫ 4dt  = Z rms = ∫ z ( t )dt = (5π) = 2.5 T 2π  2π  0 π  0 2

Z rms = 1.581

(b) Z 2 rms = a 2 o +

1 ∞ 2 1 1 ∞ 36 1 1 2 ( a b ) + = + ∑ = + n n ∑ 2 n =1 4 2 n =1n 2 π 2 4 18π 2 Z rms = 1.7086

(c )

 1.7086  %error =  − 1 x100 = 8.071   1.581

1   1 1 1 + ... = 2.9193 1 + + + +   4 9 16 25

Chapter 17, Solution 50.

cn =

=

1 T

∫

T

0

f ( t )e − jωo nt dt,

ωo =

2n =π 1

1 1 − jnπt te dt 2 ∫−

Using integration by parts, u = t and du = dt dv = e–jnπtdt which leads to v = –[1/(2jnπ)]e–jnπt t e − jnπt cn = − 2 jnπ

1

+ −1

[

1 1 − jnπt e dt 2 jnπ ∫−1

]

j − jnπ 1 e e − jnπt = + e jnπt + 2 2 2 nπ 2n π ( − j)

1

−1

= [j/(nπ)]cos(nπ) + [1/(2n2π2)](e–jnπ – ejnπ) j( −1) n 2j j( −1) n cn = sin( nπ) = + nπ 2n 2 π 2 nπ Thus ∞

f(t) =

∑ c n e jnωot =

n = −∞

∞

∑ ( −1)

n = −∞

n

j jnπt e nπ

Chapter 17, Solution 51. T = 2,

ωo = 2π / T = π T

cn =

cn =

(

2

)

1 1 2 − jnπt 1 e − jnπt 2 − jnωo t f ( t ) e dt = t e dt = − n 2 π 2 t 2 + 2 jnπt + 2 0 T∫ 2∫ 2 (− jnπ) 3 0 0

1 j2n 3 π 3

f (t) =

(−4n 2 π 2 + j4nπ) =

∞

∑

n = −∞ n

2 2 2

π

2 n 2π2

(1 + jnπ)e jnπt

(1 + jnπ)

Chapter 17, Solution 52. cn =

=

1 T

∫

T

0

f ( t )e − jωo nt dt,

ωo =

2n =π 1

1 1 − jnπt te dt 2 ∫−

Using integration by parts, u = t and du = dt dv = e–jnπtdt which leads to v = –[1/(2jnπ)]e–jnπt t cn = − e − jnπt 2 jnπ

1

1 1 − jnπt e dt 2 jnπ ∫−1

+ −1

[

]

j − jnπ 1 = + e jnπt + e e − jnπt 2 2 2 nπ 2n π ( − j)

1

−1

= [j/(nπ)]cos(nπ) + [1/(2n2π2)](e–jnπ – ejnπ) cn =

j( −1) n 2j j( −1) n + sin( n π ) = nπ 2n 2 π 2 nπ

Thus f(t) =

∞

∑c e n

jnωo t

=

n = −∞

∞

∑ ( −1)

n = −∞

n

j jnπt e nπ

Chapter 17, Solution 53. ωo = 2π/T = 2π cn =

∫

T

0

1

e − t e − jnωo t dt = ∫ e −(1+ jnωo ) t dt 0

−1 e − (1 + j2 nπ ) t = 1 + j2πn

1

= 0

[

= [1/(j2nπ)][1 – e–1(cos(2πn) – jsin(2nπ))] = (1 – e–1)/(1 + j2nπ) = 0.6321/(1 + j2nπ 0.6321e j2 nπt f(t) = ∑ n = −∞ 1 + j2nπ ∞

]

−1 e − (1 + j 2 n π ) − 1 1 + j2nπ

Chapter 17, Solution 54. T = 4, ωo = 2π/T = π/2 cn =

1 T

∫

T

0

f ( t )e − jωo nt dt

=

1  1 − jnπt / 2 2e dt + 4  ∫0

=

j 2e − jnπ / 2 − 2 + e − jnπ − e − jnπ / 2 − e − j2 nπ + e − jnπ 2nπ

=

j 3e − jnπ / 2 − 3 + 2e − jnπ 2nπ

f(t) =

∫

2

1

1e − jnπt / 2 dt −

∫

4

2

1e − jnπt / 2 dt  

[ [

∞

∑c e n

]

]

jnωo t

n = −∞

Chapter 17, Solution 55. T = 2π, ωo = 2π/T = 1 1 T

cn =

But

i(t) =

cn =

1 2π

∫

π

0

∫

T

0

i( t )e − jnωo t dt

sin( t ), 0,

0
sin( t )e − jnπt dt =

1 2π

∫

π

0

1 jt (e − e − jt )e − jnt dt 2j

π

1  e jt (1 − n ) e − jt (1 + n )  = + 4πj  j(1 − n ) j(1 + n )  0

=−

1  e jπ (1− n ) − 1 e − jπ ( n + 1) − 1 + 4  1 − n 1 + n 

=

[

1 e jπ (1 − n ) − 1 + ne jπ (1 − n ) − n + e − jπ (1 + n ) − 1 − ne − jπ (1+ n ) + n 2 4π( n − 1)

]

But ejπ = cos(π) + jsin(π) = –1 = e–jπ 1 + e − jnπ 1 − jnπ − jnπ − jnπ − jnπ cn = −e −e − ne + ne −2 = 2 π(1 − n 2 ) 4 π( n 2 − 1)

[

]

Thus i(t) =

∞

1 + e − jnπ

∑ 2π(1 − n

n = −∞

2

)

e jnπt

Chapter 17, Solution 56. co = ao = 10, ωo = π co = (an – jbn)/2 = (1 – jn)/[2(n2 + 1)] ∞

f(t) = 10 +

(1 − jn) jnπt e 2 + 1)

∑ 2(n

n = −∞ n≠0

Chapter 17, Solution 57. ao = (6/–2) = –3 = co cn = 0.5(an –jbn) = an/2 = 3/(n3 – 2) ∞

f(t) = − 3 +

∑n

n = −∞ n≠0

3

3 e j50nt −2

Chapter 17, Solution 58. cn = (an – jbn)/2, (–1) = cos(nπ), ωo = 2π/T = 1 cn = [(cos(nπ) – 1)/(2πn2)] – j cos(nπ)/(2n) Thus f(t) =

π cos(nπ )  jnt  cos(nπ ) − 1 −j + ∑ e 2 4 2n   2πn

Chapter 17, Solution 59. For f(t), T = 2π, ωo = 2π/T = 1. ao = DC component = (1xπ + 0)/2π = 0.5 For

h(t), T = 2, ωo = 2π/T = π. ao = (3x1 – 2x1)/2 = 0.5

Thus by replacing ωo = 1 with ωo = π and multiplying the magnitude by five, we obtain ∞ 1 j5e j( 2n +1) πt h(t) = − ∑ 2 n = −∞ ( 2n + 1)π n≠0

Chapter 17, Solution 60. From Problem 16.17, ao = 0 = an, bn = [2/(nπ)][1 – 2 cos(nπ)], co = 0 cn = (an – jbn)/2 = [j/(nπ)][2 cos(nπ) – 1], n ≠ 0. Chapter 17, Solution 61.

ωo = 1.

(a) f(t) = ao +

∑A

n

cos(nω o t − φ n )

= 6 + 4cos(t + 50°) + 2cos(2t + 35°) + cos(3t + 25°) + 0.5cos(4t + 20°) = 6 + 4cos(t)cos(50°) – 4sin(t)sin(50°) + 2cos(2t)cos(35°) – 2sin(2t)sin(35°) + cos(3t)cos(25°) – sin(3t)sin(25°) + 0.5cos(4t)cos(20°) – 0.5sin(4t)sin(20°) = 6 + 2.571cos(t) – 3.73sin(t) + 1.635cos(2t) – 1.147sin(2t) + 0.906cos(3t) – 0.423sin(3t) + 0.47cos(4t) – 0.171sin(4t)

(b)

frms =

1 ∞ 2 ∑ An 2 n =1

a o2 +

frms2 = 62 + 0.5[42 + 22 + 12 + (0.5)2] = 46.625 frms = 6.828

Chapter 17, Solution 62.

(a) ωo = 20 = 2π / T

(b) f ( t ) = a o +

 →

T=

2π = 0.3141s 20

∞

∑ A n cos(nωo t + φ n ) = 3 + 4 cos(20t + 90 o ) +5.1cos(40t + 90 o ) + ...

n =1

f ( t ) = 3 − 4 sin 20 t − 5.1sin 40t − 2.7 sin 60 t − 1.8 sin 80t − ....

Chapter 17, Solution 63.

This is an even function. T = 3, ωo = 2π/3, bn = 0. f(t) =

ao =

an =

1, 0 < t < 1 2, 1 < t < 1.5 1.5 2 T/2 2 1 f ( t )dt =  ∫ 1dt + ∫ 2 dt  = (2/3)[1 + 1] = 4/3 ∫  1 T 0 3  0

1.5 4 T/2 4 1 f ( t ) cos(nωo t )dt =  ∫ 1cos(2nπt / 3)dt + ∫ 2 cos(2nπt / 3)dt  ∫  1 T 0 3  0

4 3 6  2nπt   2nπt  =  sin  sin   +  3  2nπ  3  0 2nπ  3  1 1

= [–2/(nπ)]sin(2nπ/3)

1.5

  

4 2 ∞ 1  3nπ   2nπt  f2(t) = − ∑ sin   cos  3 π n =1 n  3   3  ao = 4/3 = 1.3333, ωo = 2π/3, an = –[2/(nπ)]sin(2nπt/3) An =

a 2n + b 2n =

2  2nπ  sin   nπ  3 

A1 = 0.5513, A2 = 0.2757, A3 = 0, A4 = 0.1375, A5 = 0.1103 The amplitude spectra are shown below. 1.333

An

0.551

0.275 0.1378 0.1103 0

0

1

2

3

4

5

n

Chapter 17, Solution 64. The amplitude and phase spectra are shown below.

An 3.183 2.122 1.591 0.4244 0

2π

4π

2π

4π

6π

ω

φn

0

6π

ω

-180o

Chapter 17, Solution 65. an = 20/(n2π2), bn = –3/(nπ), ωn = 2n An = a 2n + b 2n =

=

400 9 + 2 2 4 4 n π n π

3 44.44 1 + 2 2 , n = 1, 3, 5, 7, 9, etc. nπ n π

n 1 3 5 7 9

An 2.24 0.39 0.208 0.143 0.109

φn = tan–1(bn/an) = tan–1{[–3/(nπ)][n2π2/20]} = tan–1(–nx0.4712) n 1 3 5 7 9 ∞

φn –25.23° –54.73° –67° –73.14° –76.74° –90°

2.24

0

2

6

10

14

18 ωn

–30°

–25.23°

An 0.39

0.208

0

2

6

10

–60°

–54.73°

0.0.143 0.109

14

18

ωn

–90°

φn

–67° –73.14°

–76.74°

Chapter 17, Solution 66. The schematic is shown below. The waveform is inputted using the attributes of VPULSE. In the Transient dialog box, we enter Print Step = 0.05, Final Time = 12, Center Frequency = 0.5, Output Vars = V(1) and click enable Fourier. After simulation, the output plot is shown below. The output file includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 5.099510E+00 HARMONIC NO 1 2 3 4 5 6 7 8 9

FREQUENCY FOURIER NORMALIZED (HZ) COMPONENT COMPONENT 5.000E-01 1.000E+00 1.500E+00 2.000E+00 2.500E+00 3.000E+00 3.500E+00 4.000E+00 4.500E+00

3.184E+00 1.593E+00 1.063E+00 7.978E-01 6.392E-01 5.336E-01 4.583E-01 4.020E-01 3.583E-01

1.000E+00 5.002E-01 3.338E-01 2.506E-01 2.008E-01 1.676E-01 1.440E-01 1.263E-01 1.126E-01

PHASE (DEG) 1.782E+00 3.564E+00 5.347E+00 7.129E+00 8.911E+00 1.069E+01 1.248E+01 1.426E+01 1.604E+01

NORMALIZED PHASE (DEG) 0.000E+00 1.782E+00 3.564E+00 5.347E+00 7.129E+00 8.911E+00 1.069E+01 1.248E+01 1.426E+01

TOTAL HARMONIC DISTORTION = 7.363360E+01 PERCENT

Chapter 17, Solution 67. The Schematic is shown below. In the Transient dialog box, we type “Print step = 0.01s, Final time = 36s, Center frequency = 0.1667, Output vars = v(1),” and click Enable Fourier. After simulation, the output file includes the following Fourier components,

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 2.000396E+00 HARMONIC FREQUENCY FOURIER NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) 1 2 3 4 5 6 7 8 9

1.667E-01 3.334E-01 5.001E-01 6.668E-01 8.335E-01 1.000E+00 1.167E+00 1.334E+00 1.500E+00

2.432E+00 6.576E-04 5.403E-01 3.343E-04 9.716E-02 7.481E-06 4.968E-02 1.613E-04 6.002E-02

1.000E+00 2.705E-04 2.222E-01 1.375E-04 3.996E-02 3.076E-06 2.043E-02 6.634E-05 2.468E-02

PHASE NORMALIZED PHASE (DEG)

-8.996E+01 -8.932E+01 9.011E+01 9.134E+01 -8.982E+01 -9.000E+01 -8.975E+01 -8.722E+01 9.032E+01

0.000E+00 6.467E-01 1.801E+02 1.813E+02 1.433E-01 -3.581E-02 2.173E-01 2.748E+00 1.803E+02

TOTAL HARMONIC DISTORTION = 2.280065E+01 PERCENT

Chapter 17, Solution 68. The schematic is shown below. We set the final time = 6T=12s and the center frequency = 1/T = 0.5. When the schematic is saved and run, we obtain the Fourier series from the output file as shown below.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = HARMONIC NORMALIZED NO (DEG) 1 2 3 4 5 6 7 8 9

1.990000E+00

FREQUENCY

FOURIER

NORMALIZED

PHASE

(HZ)

COMPONENT

COMPONENT

(DEG)

1.273E+00 6.367E-01 4.246E-01 3.185E-01 2.549E-01 2.125E-01 1.823E-01 1.596E-01 1.419E-01

1.000E+00 5.001E-01 3.334E-01 2.502E-01 2.002E-01 1.669E-01 1.431E-01 1.253E-01 1.115E-01

5.000E-01 1.000E+00 1.500E+00 2.000E+00 2.500E+00 3.000E+00 3.500E+00 4.000E+00 4.500E+00

Chapter 17, Solution 69.

9.000E-01 -1.782E+02 2.700E+00 -1.764E+02 4.500E+00 -1.746E+0 6.300E+00 -1.728E+02 8.100E+00

PHASE 0.000E+00 1.791E+02 1.800E+00 -1.773E+02 3.600E+00 -1.755E+02 5.400E+00 -1.737E+02 7.200E+00

The schematic is shown below. In the Transient dialog box, set Print Step = 0.05 s, Final Time = 120, Center Frequency = 0.5, Output Vars = V(1) and click enable Fourier. After simulation, we obtain V(1) as shown below. We also obtain an output file which includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 5.048510E-01 HARMONIC FREQUENCY FOURIER NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) 1 2 3 4 5 6 7 8 9

PHASE NORMALIZED PHASE (DEG)

5.000E-01 4.056E-01 1.000E+00 -9.090E+01 0.000E+00 1.000E+00 2.977E-04 7.341E-04 -8.707E+01 3.833E+00 1.500E+00 4.531E-02 1.117E-01 -9.266E+01 -1.761E+00 2.000E+00 2.969E-04 7.320E-04 -8.414E+01 6.757E+00 2.500E+00 1.648E-02 4.064E-02 -9.432E+01 -3.417E+00 3.000E+00 2.955E-04 7.285E-04 -8.124E+01 9.659E+00 3.500E+00 8.535E-03 2.104E-02 -9.581E+01 -4.911E+00 4.000E+00 2.935E-04 7.238E-04 -7.836E+01 1.254E+01 4.500E+00 5.258E-03 1.296E-02 -9.710E+01 -6.197E+00

TOTAL HARMONIC DISTORTION = 1.214285E+01 PERCENT

Chapter 17, Solution 70. The schematic is shown below. In the Transient dialog box, we set Print Step = 0.02 s, Final Step = 12 s, Center Frequency = 0.5, Output Vars = V(1) and V(2), and click enable Fourier. After simulation, we compare the output and output waveforms as shown. The output includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 7.658051E-01 HARMONIC FREQUENCY FOURIER NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) 1 2 3 4 5 6 7 8 9

5.000E-01 1.070E+00 1.000E+00 3.758E-01 1.500E+00 2.111E-01 2.000E+00 1.247E-01 2.500E+00 8.538E-02 3.000E+00 6.139E-02 3.500E+00 4.743E-02 4.000E+00 3.711E-02 4.500E+00 2.997E-02

1.000E+00 3.512E-01 1.973E-01 1.166E-01 7.980E-02 5.738E-02 4.433E-02 3.469E-02 2.802E-02

1.004E+01 -3.924E+01 -3.985E+01 -5.870E+01 -5.680E+01 -6.563E+01 -6.520E+01 -7.222E+01 -7.088E+01

PHASE NORMALIZED PHASE (DEG)

0.000E+00 -4.928E+01 -4.990E+01 -6.874E+01 -6.685E+01 -7.567E+01 -7.524E+01 -8.226E+01 -8.092E+01

TOTAL HARMONIC DISTORTION = 4.352895E+01 PERCENT

Chapter 17, Solution 71. The schematic is shown below. We set Print Step = 0.05, Final Time = 12 s, Center Frequency = 0.5, Output Vars = I(1), and click enable Fourier in the Transient dialog box. After simulation, the output waveform is as shown. The output file includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE I(L_L1) DC COMPONENT = 8.374999E-02 HARMONIC FREQUENCY FOURIER NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) 1 2 3 4 5 6 7 8 9

PHASE NORMALIZED PHASE (DEG)

5.000E-01 2.287E-02 1.000E+00 -6.749E+01 0.000E+00 1.000E+00 1.891E-04 8.268E-03 8.174E+00 7.566E+01 1.500E+00 2.748E-03 1.201E-01 -8.770E+01 -2.021E+01 2.000E+00 9.583E-05 4.190E-03 -1.844E+00 6.565E+01 2.500E+00 1.017E-03 4.446E-02 -9.455E+01 -2.706E+01 3.000E+00 6.366E-05 2.783E-03 -7.308E+00 6.018E+01 3.500E+00 5.937E-04 2.596E-02 -9.572E+01 -2.823E+01 4.000E+00 6.059E-05 2.649E-03 -2.808E+01 3.941E+01 4.500E+00 2.113E-04 9.240E-03 -1.214E+02 -5.387E+01

TOTAL HARMONIC DISTORTION = 1.314238E+01 PERCENT

Chapter 17, Solution 72. T = 5, ωo = 2π/T = 2π/5 f(t) is an odd function. ao = 0 = an

4 T/2 4 10 f ( t ) sin(nωo t )dt = ∫ 10 sin(0.4nπt )dt ∫ T 0 5 0

bn =

1

20 8x 5 [1 − cos(0.4nπ)] = − cos(0.4πnt ) = 2 nπ nπ 0 f(t) =

20 ∞ 1 ∑ [1 − cos(0.4nπ)]sin(0.4nπt ) π n =1 n

Chapter 17, Solution 73.

p =

2 VDC 1 V2 + ∑ n R 2 R

= 0 + 0.5[(22 + 12 + 12)/10] = 300 mW

Chapter 17, Solution 74.

(a)

An =

a 2n + b 2n ,

φ = tan–1(bn/an)

A1 =

6 2 + 8 2 = 10,

φ1 = tan–1(6/8) = 36.87°

A2 =

3 2 + 4 2 = 5,

φ2 = tan–1(3/4) = 36.87°

i(t) = {4 + 10cos(100πt – 36.87°) – 5cos(200πt – 36.87°)} A (b)

p = I 2DC R + 0.5∑ I 2n R = 2[42 +0.5(102 + 52)] = 157 W

Chapter 17, Solution 75.

The lowpass filter is shown below. R +

+ C

vs -

vs =

vo -

Aτ 2A ∞ 1 nπτ + sin cos nωo t ∑ T T n =1n T

1 jω n C 1 Vs = Vs , Vo = 1 1 + jω n RC R+ jω n C For n=0, (dc component), Vo = Vs =

Aτ T

ω n = nωo = 2nπ / T

(1)

For the nth harmonic, 2A nπτ sin ∠ − 90 o 2 2 2 −1 nT T 1 + ω n R C ∠ tan ω n RC 1

Vo =

When n=1, | Vo |=

2A nπτ sin • T T

•

1 1+

(2)

2

4π R 2C 2 T

From (1) and (2), Aτ 2A π sin = 50 x T T 10

1+

1 1+

4π 2 2 2 R C T

4π 2 2 2 R C = 1010 T

→

C=

 →

1+

4π 2 2 2 30.9 R C = = 3.09 x10 4 T τ

T 10 −2 x 3.09 x10 4 10 5 = = 24.59 mF 2πR 4πx10 3

Chapter 17, Solution 76.

vs(t) is the same as f(t) in Figure 16.1 except that the magnitude is multiplied by 10. Hence 20 ∞ 1 vo(t) = 5 + ∑ sin( nπt ) , n = 2k – 1 π k =1 n T = 2, ωo = 2π/T = 2π, ωn = nωo = 2nπ jωnL = j2nπ; Z = R||10 = 10R/(10 + R) Vo = ZVs/(Z + j2nπ) = [10R/(10R + j2nπ(10 + R))]Vs Vo =

10R∠ − tan −1{(nπ / 5R )(10 + R )} 100R 2 + 4n 2 π 2 (10 + R ) 2

Vs

Vs = [20/(nπ)]∠0° The source current Is is 20 (10 + R ) Vs Vs nπ Is = = = 10R Z + j2nπ 10R + j2nπ(10 + R ) + j2nπ 10 + R

20 ∠ − tan −1{( nπ / 3)(10 + R )} nπ 100R 2 + 4n 2 π 2 (10 + R ) 2

(10 + R )

=

1 ∑ Vsn I sn cos(θ n − φ n ) 2

ps = VDCIDC +

For the DC case, L acts like a short-circuit. Is =

5(10 + R ) 5 , Vs = 5 = Vo = 10R 10R 10 + R

  −1  π  tan  (10 + R )    2 (10 + R ) cos  25(10 + R ) 1  20  5   +   ps = 10R 2  π  100R 2 + 4π 2 (10 + R ) 2      2π  (10 + R ) 2 cos tan −1  (10 + R )     5   10   +  + " 2 2 2  π   100R + 16π (10 + R )   2

ps =

=

VDC 1 ∞ Von + ∑ R 2 n =1 R  25 1  100R 100R +  + + " 2 2 2 2 2 2 R 2 100R + 4π (10 + R ) 100R + 10π (10 + R ) 

We want po = (70/100)ps = 0.7ps. Due to the complexity of the terms, we consider only the DC component as an approximation. In fact the DC component has the latgest share of the power for both input and output signals. 25 7 25(10 + R ) = x R 10 10R 100 = 70 + 7R which leads to R = 30/7 = 4.286 Ω

Chapter 17, Solution 77.

(a) For the first two AC terms, the frequency ratio is 6/4 = 1.5 so that the highest common factor is 2. Hence ωo = 2. T = 2π/ωo = 2π/2 = π (b) The average value is the DC component = –2 Vrms =

(c)

ao +

1 ∞ 2 (a n + b 2n ) ∑ 2 n =1

1 2 Vrms = (−2) 2 + (10 2 + 8 2 + 6 2 + 3 2 + 12 ) = 121.5 2 Vrms = 11.02 V Chapter 17, Solution 78. 2

(a)

2 2 Vn ,rms VDC Vn2 VDC 1 p = + ∑ = +∑ R 2 R R R

= 0 + (402/5) + (202/5) + (102/5) = 420 W (b)

5% increase = (5/100)420 = 21 pDC = 21 W =

2 VDC 2 which leads to VDC = 21R = 105 R

VDC = 10.25 V Chapter 17, Solution 79.

From Table 17.3, it is evident that an = 0, bn = 4A/[π(2n – 1)], A = 10. A Fortran program to calculate bn is shown below. The result is also shown.

C

FOR PROBLEM 17.79 DIMENSION B(20)

10

A = 10 PIE = 3.142 C = 4.*A/PIE DO 10 N = 1, 10 B(N) = C/(2.*FLOAT(N) – 1.) PRINT *, N, B(N) CONTINUE STOP END n 1 2 3 4 5 6 7 8 9 10

bn 12.731 4.243 2.546 1.8187 1.414 1.1573 0.9793 0.8487 0.7498 0.67

Chapter 17, Solution 80.

From Problem 17.55, cn = [1 + e–jnπ]/[2π(1 – n2)] This is calculated using the Fortran program shown below. The results are also shown. C FOR PROBLEM 17.80 COMPLEX X, C(0:20)

10

PIE = 3.1415927 A = 2.0*PIE DO 10 N = 0, 10 IF(N.EQ.1) GO TO 10 X = CMPLX(0, PIE*FLOAT(N)) C(N) = (1.0 + CEXP(–X))/(A*(1 – FLOAT(N*N))) PRINT *, N, C(N) CONTINUE STOP END

n 0 1 2 3 4 5 6 7 8 9 10

cn 0.3188 + j0 0 –0.1061 + j0 0 –0.2121x10–1 + j0 0 –0.9095x10–2 + j0 0 –0.5052x10–2 + j0 0 –0.3215x10–2 + j0

Chapter 17, Solution 81.

(a) A

0

f(t) =

2A 4A ∞ 1 − cos(nωo t ) ∑ 2 π π n =1 4n − 1

The total average power is

pavg = Frms2R = Frms2 since R = 1 ohm.

Pavg = Frms2 = (b)

2T

T

1 T 2 f ( t )dt = 0.5A2 ∫ 0 T

From the Fourier series above |co| = 2A/2, |cn| = 4A/[π(4n2 – 1)]

3T

n 0 1 2 3 4

ωo 0 2ωo 4ωo 6ωo 8ωo

|cn| 2A/π 2A/(3π) 2A/(15π) 2A/(35π) 2A/(63π)

(c)

81.1%

(d)

0.72%

2|cn|2 4A2/(π2) 8A2/(9π2) 2A2/(225π2) 8A2/(1225π2) 8A2/(3969π2)

Chapter 17, Solution 82. 2 VDC 1 ∞ Vn2 P = + ∑ R 2 n =1 R

Assuming V is an amplitude-phase form of Fourier series. But |An| = 2|Cn|, co = ao |An|2 = 4|Cn|2

Hence,

∞ c o2 c 2n + 2∑ P = R n =1 R

Alternatively, 2 Vrms P = R

where 2 Vrms = a o2 +

∞ ∞ 1 ∞ 2 2 2 A = c + 2 c = c 2n ∑ ∑ ∑ n o n 2 n =1 n =1 n = −∞

= 102 + 2(8.52 + 4.22 + 2.12 + 0.52 + 0.22) = 100 + 2x94.57 = 289.14 P = 289.14/4 = 72.3 W

% power 81.1% 18.01% 0.72% 0.13% 0.04%

Chapter 18, Solution 1.

f ' ( t ) = δ( t + 2) − δ( t + 1) − δ( t − 1) + δ( t − 2) jωF(ω) = e j2 ω − e jω − e − jω + e − jω2 = 2 cos 2ω − 2 cos ω F(ω) =

2[cos 2ω − cos ω] jω

Chapter 18, Solution 2. t, f (t) =  0,

0 < t <1 otherwise f ”(t)

f ‘(t) 1

δ(t)

0

t

t

1

–δ’(t-1) -δ(t-1)

-δ(t-1)

f"(t) = δ(t) - δ(t - 1) - δ'(t - 1) Taking the Fourier transform gives -ω2F(ω) = 1 - e-jω - jωe-jω F(ω) =

(1 + jω)e jω − 1 ω2

1

or F(ω) = ∫ t e − jωt dt 0

eax But ∫ x e dx = 2 (ax − 1) + c a ax

F(ω) =

e − jω

(− jω)

2

(− jωt − 1) 10 =

[

]

1 (1 + jω)e − jω − 1 2 ω

Chapter 18, Solution 3.

f (t) =

1 t , − 2 < t < 2, 2

1 f ' (t) = , − 2 < t < 2 2

1 jωt e − jωt t e dt = (− jωt − 1) 2− 2 2 −2 2 2(− jω)

F(ω) = ∫

F(ω) =

2

[

]

[

]

=−

1 e − jω2 (− jω2 − 1) − e jω2 ( jω2 − 1) 2 2ω

=−

1 − jω2(e jω2 + e jω2 ) + e jω2 − e − jω2 2 2ω

=−

1 (− jω4 cos 2ω + j2 sin 2ω) 2ω 2

j (sin 2ω − 2ω cos 2ω) ω2

Chapter 18, Solution 4. 2δ(t+1) g’ 2 –1 0

1

t

–2 –2δ(t–1)

4δ(t) 2δ’(t+1) g” –1 0 –2δ(t+1)

1

t

–2 –2δ(t–1) –2δ’(t–1)

g ′′ = −2δ( t + 1) + 2δ′( t + 1) + 4δ( t ) − 2δ( t − 1) − 2δ′( t − 1) ( jω) 2 G (ω) = −2e jω + 2 jωe jω + 4 − 2e − jω − 2 jωe − jω = −4 cos ω − 4ω sin ω + 4

G (ω) =

4 ω2

(cos ω + ω sin ω − 1)

Chapter 18, Solution 5. h’(t) 1

0

–1

t

1

–2δ(t) h”(t) 1

δ(t+1)

1 –1 –2δ’(t)

t

0

–δ(t–1)

h ′′( t ) = δ( t + 1) − δ( t − 1) − 2δ′( t ) ( jω) 2 H(ω) = e jω − e − jω − 2 jω = 2 j sin ω − 2 jω

H(ω) =

2j 2j − sin ω ω ω2

Chapter 18, Solution 6. 0

F(ω) = ∫ (−1)e

− jωt

−1

1

dt + ∫ te − jωt dt 0

0

1

−1

0

Re F(ω) = − ∫ cos ωtdt + ∫ t cos ωtdt =−

 1  1 1 t 1 0 sin ωt −1 +  cos ωt + sin ωt  0 = (cos ω − 1) 2 ω ω ω2  ω

Chapter 18, Solution 7. (a)

f1 is similar to the function f(t) in Fig. 17.6.

f 1 ( t ) = f ( t − 1)

Since F(ω) =

2(cos ω − 1) jω

2e − jω (cos ω − 1) F1 (ω) = e F(ω) = jω jω

Alternatively, f 1' ( t ) = δ( t ) − 2δ( t − 1) + δ( t − 2) jωF1 (ω) = 1 − 2e − jω + e − j2 ω = e − jω (e jω − 2 + e jω ) = e − jω (2 cos ω − 2) F1(ω) = (b)

2e − jω (cos ω − 1) jω

f2 is similar to f(t) in Fig. 17.14. f2(t) = 2f(t) F2(ω) =

4(1 − cos ω) ω2

Chapter 18, Solution 8. 1

(a)

F(ω) = ∫ 2e

dt + ∫ (4 − 2 t )e − jωt dt

0

=

1

2 − jωt 1 4 − jωt 2 2 − jωt 2 e + e − e (− jωt − 1) 1 1 0 2 − jω − jω −ω

F(ω) = (b)

2

− jωt

2 ω

2

+

2 − jω 2 4 − j2ω 2 e + − e − (1 + j2ω)e − j2ω 2 jω jω jω ω

g(t) = 2[ u(t+2) – u(t-2) ] - [ u(t+1) – u(t-1) ] G (ω) =

4 sin 2ω 2 sin ω − ω ω

Chapter 18, Solution 9. (a)

y(t) = u(t+2) – u(t-2) + 2[ u(t+1) – u(t-1) ] Y(ω) =

2 4 sin 2ω + sin ω ω ω

1

(b) Z(ω) = ∫ (−2 t )e − jωt dt = 0

− 2e − jωt − ω2

2 2e − j ω 1 (− jωt − 1) 0 = − (1 + jω) 2 2 ω

Chapter 18, Solution 10. (a)

x(t) = e2tu(t) X(ω) = 1/(2 + jω)

(b)

e − t , t > 0 e −( t ) =  t e , t < 0 1

0

1

−1

−1

0

Y(ω) = ∫ y( t )e jωt dt = ∫ e t e jωt dt + ∫ e − t e − jωt dt

ω

=

e (1− jω) t 1 − jω

=

 cos ω + jsin ω cos ω − jsin ω  2 + − e −1   2 1 − jω 1 + jω 1+ ω  

Y(ω) =

0 −1

+

e − (1+ jω) t − (1 + jω)

1 0

[

2 1 − e −1 (cos ω − ω sin ω) 2 1+ ω

Chapter 18, Solution 11. f(t) = sin π t [u(t) - u(t - 2)] 2

F(ω) = ∫ sin πt e − jωt dt = 0

(

)

1 2 j πt e − e − j πt e − jωt dt ∫ 0 2j

=

1  2 + j( − ω + π ) t + e − j( ω + π ) t )dt  (e  2 j  ∫0

=

1  1 e − j( ω+ π ) t 2  e − j( ω− π ) t 02 +  0 − j(ω + π)  2 j  − j(ω − π)

=

1  1 − e − j2 ω 1 − e − j2 ω    + 2  π − ω π + ω 

=

1 2π + 2πe − j2 ω 2 2(π − ω )

(

2

F(ω) =

(

)

)

π e − jω 2 − 1 2 ω −π 2

Chapter 18, Solution 12. (a)

∞

2

0

0

F(ω) = ∫ e t e − jωt dt = ∫ e (1− jω) t dt

=

1 e (1− jω) t 1 − jω

2 0

=

e 2− jω 2 − 1 1 − jω

]

(b)

0

1

−1

0

H(ω) = ∫ e − jωt dt + ∫ (−1)e − jωt dt

=−

=

(

)

(

)

1 1 − jω 1 1 − e jω + e −1 = (−2 + 2 cos ω) jω jω jω

− 4 sin 2 ω / 2  sin ω / 2  = jω  jω  ω/ 2 

2

Chapter 18, Solution 13.

(a) We know that F[cos at ] = π[δ(ω − a ) + δ(ω + a )] . Using the time shifting property, F[cos a ( t − π / 3a )] = πe − jωπ / 3a [δ(ω − a ) + δ(ω + a )] = πe − jπ / 3δ(ω − a ) + πe jπ / 3δ(ω + a )

(b) sin π( t + 1) = sin πt cos π + cos πt sin π = − sin πt g(t) = -u(t+1) sin (t+1) Let x(t) = u(t)sin t, then X(ω) =

1 ( jω) 2 + 1

=

1 1 − ω2

Using the time shifting property,

G (ω) = −

1 1 − ω2

e jω =

e jω ω2 − 1

(c ) Let y(t) = 1 + Asin at, then Y(ω) = 2πδ(ω) + jπA[δ(ω + a ) − δ(ω − a )] h(t) = y(t) cos bt Using the modulation property, 1 H(ω) = [Y(ω + b) + Y(ω − b)] 2 H(ω) = π[δ(ω + b) + δ(ω − b)] +

jπA [δ(ω + a + b) − δ(ω − a + b) + δ(ω + a − b) − δ(ω − a − b)] 2

4

(d) I(ω) = ∫ (1 − t )e − jωt dt = 0

e − j ωt e − j ωt 1 e − j4ω e − j4ω 4 − (− jωt − 1) 0 = − − ( j4ω + 1) − jω − ω 2 jω ω2 ω2

Chapter 18, Solution 14.

(a)

cos(3t + π) = cos 3t cos π − sin 3t sin π = cos 3t (−1) − sin 3t (0) = − cos(3t ) f ( t ) = −e − t cos 3t u ( t ) F(ω) =

− (1 + jω )

(1 + jω)2 + 9

(b) g(t) 1 -1 1

t

-1

g’(t)

π -1

1 t

-π g ' ( t ) = π cos πt[u ( t − 1) − u ( t − 1)]

g" ( t ) = −π 2 g( t ) − πδ( t + 1) + πδ( t − 1) − ω 2 G (ω) = − π 2 G (ω) − πe jω + πe − jω (π 2 − ω2 )G(ω) = −π(e jω − e − jω ) = −2 jπ sin ω G(ω) =

2 jπ sin ω ω2 − π 2

Alternatively, we compare this with Prob. 17.7 f(t) = g(t - 1) F(ω) = G(ω)e-jω

G (ω) = F(ω)e jω =

− j2π sin ω ω2 − π 2

=

G(ω) = (c)

π (e − jω − e jω ) 2 ω −π 2

2 jπ sin ω π 2 − ω2

cos π( t − 1) = cos πt cos π + sin πt sin π = cos πt (−1) + sin πt (0) = − cos πt Let x ( t ) = e −2( t −1) cos π( t − 1)u ( t − 1) = −e 2 h ( t ) and y( t ) = e −2 t cos(πt )u ( t ) Y(ω) =

2 + jω (2 + jω) 2 + π 2

y( t ) = x ( t − 1) Y(ω) = X(ω)e − jω

X(ω) =

(2 + jω)e jω (2 + jω)2 + π 2

X(ω) = −e 2 H(ω) H(ω) = −e −2 X(ω) =

(d)

− (2 + jω)e jω− 2

(2 + jω)2 + π 2

Let x ( t ) = e −2 t sin( −4t )u (− t ) = y(− t ) p( t ) = − x ( t ) where y( t ) = e 2 t sin 4t u ( t ) Y (ω) =

2 + jω

(2 + jω)2 + 4 2

X(ω) = Y(−ω) =

2 − jω

(2 − jω)2 + 16

p(ω) = −X(ω) =

(e)

Q(ω) =

jω − 2

(jω − 2 )2 + 16

 8 − jω 2 1  e + 3 − 2 πδ(ω) + e − jω2 jω jω  

6 jω 2 e + 3 − 2πδ(ω)e − jω 2 jω

Q(ω) =

Chapter 18, Solution 15. (a)

F(ω) = e j3ω − e − jω3 = 2 j sin 3ω

(b)

Let g( t ) = 2δ( t − 1), G (ω) = 2e − jω t F(ω) = F  ∫ g ( t ) dt   −∞ 

=

G (ω) + πF(0)δ(ω) jω

2e − j ω = + 2πδ(−1)δ(ω) jω

=

(c)

F [δ(2t )] =

F(ω) =

2e − jω jω

1 ⋅1 2

1 1 1 jω ⋅ 1 − jω = − 3 2 3 2

Chapter 18, Solution 16. (a) Using duality properly

t →

−2 ω2

−2 → 2π ω t2 4 → − 4π ω t2

or

4 F(ω) = F  2  = − 4π ω t  (b)

e

−at

2a a + ω2 2

2a a + t2

2π e

2

8 a + t2 2

4π e

−a ω

−2 ω

 8  −2 ω = 4π e G(ω) = F  2  4+t  Chapter 18, Solution 17.

(a)

Since H(ω) = F (cos ω0 t f ( t ) ) = where F(ω) = F [u (t )] = πδ(ω) +

H(ω) =

1 [F(ω + ω0 ) + F(ω − ω0 )] 2

1 , ω0 = 2 jω

1 1 1  + πδ(ω − 2) + πδ (ω + 2) + (  2 j ω + 2) j (ω − 2) 

=

π [δ(ω + 2) + δ(ω − 2)] − j  ω + 2 + ω − 2  2 2  (ω + 2)(ω − 2) 

H(ω) =

(b)

π [δ(ω + 2) + δ(ω − 2 )] − 2jω 2 ω −4

G(ω) = F [sin ω0 t f ( t )] =

j [F(ω + ω0 ) − F(ω − ω0 )] 2

where F(ω) = F [u (t )] = πδ (ω) + G (ω) =

=

=

1 jω

 j 1 1 πδ(ω + 10) + − πδ(ω − 10) −  2 j(ω + 10) j(ω − 10 )  jπ [δ(ω + 10) − δ (ω − 10)] + j  j − j  2 2  ω − 10 ω + 10 

jπ [δ(ω + 10) − δ(ω − 10 )] − 2 10 2 ω − 100

Chapter 18, Solution 18.

Let f (t ) = e − t u (t ) f (t ) cos t

Hence Y(ω) =

F(ω) =

1 j + jω

1 [F(ω − 1) + F(ω + 1)] 2

 1 1 1 +   2 1 + j (ω − 1) 1 + j (ω + 1) 

=

1  1 + jω + j + 1 + jω − j    2  [1 + j(ω − 1)][1 + j (ω + 1)]

=

1 + jω 1 + jω + j + jω − j − ω 2 + 1 =

1 + jω 2 jω − ω 2 + 2

Chapter 18, Solution 19.

∞

F(ω) = ∫ f ( t )e jωt dt = −∞

F(ω) =

1 1 j2 πt ( e + e − j2 πt )e − jωt dt ∫ 0 2

[

]

1 1 − j( ω + 2 π ) t e + e − j(ω− 2 π )t dt ∫ 0 2 1

 1 1 1 e − j( ω − 2 π ) t  e − j( ω + 2 π ) t + =  2  − j (ω + 2π ) − j(ω − 2π ) 0 1  e − j( ω+ 2 π ) − 1 e − j( ω− 2 π ) − 1  =−  +  2  j (ω + 2π) j(ω − 2π ) 

But

e j2 π = cos 2π + j sin 2π = 1 = e − j2 π 1  e − jω − 1  1 1   + F(ω) = −   2 j  ω + 2π ω − 2π 

=

(

)

jω e − jω − 1 2 ω − 4π 2

Chapter 18, Solution 20. (a)

F (cn) = cnδ(ω)

(

)

F c n e jnωo t = c n δ(ω − nωo )

(b)

 ∞  F  ∑ c n e jnωo t  =  n = −∞ 

n = −∞

T = 2π

ωo =

cn =

∞

∑ c δ(ω − nω ) n

o

2π =1 T

1  π 1 T − jnt f (t ) e − jnωo t dt =  ∫0 1⋅ e dt + 0  ∫ 0   2π T

=

1  1 jnt − e 2π  jn

π 0

 j  = ( e − jnπ − 1)  2πn

But e − jnπ = cos nπ + j sin nπ = cos nπ = (−1) n cn =

[

]

 j (− 1)n − 1 =  0−,j , 2nπ  nπ

n = even n = odd , n ≠ 0

for n = 0 cn =

1 π 1 1 dt = ∫ 0 2π 2

Hence f (t) =

∞ 1 j jnt − ∑ e 2 n = −∞ nπ n ≠0 n = odd

F(ω) =

∞ 1 j δω − ∑ δ(ω − n ) 2 n = −∞ nπ n≠0 n = odd

Chapter 18, Solution 21. Using Parseval’s theorem, ∞

∫− ∞ f

2

( t )dt =

1 ∞ | F(ω) | 2 dω ∫ − ∞ 2π

If f(t) = u(t+a) – u(t+a), then ∞

∫−∞

a

f 2 ( t )dt = ∫ (1) 2 dt = 2a = −a

or 2

4πa π  sin aω  ∫− ∞  aω  dω = 4a 2 = a as required. ∞

2

1 ∞  sin aω  4a 2   dω ∫ 2 π −∞  aω 

Chapter 18, Solution 22.

F [f ( t ) sin ωo t ] = ∫ f ( t ) ∞

(e

−∞

=

jω o t

)

− e − j ω o t − j ωt e dt 2j

∞ 1 ∞ f ( t )e − j(ω− ωo )t dt − ∫ e − j(ω+ ωo )t dt  ∫   − ∞ − ∞ 2j 

1 [F(ω − ω o ) − F(ω + ωo )] 2j

=

Chapter 18, Solution 23.

(a) f(3t) leads to F [f (− 3t )] =

1 10 30 ⋅ = 3 (2 + jω / 3)(5 + jω / 3) (6 + jω)(15 + jω)

30 (6 − jω)(15 − jω)

(b) f(2t)

1 10 20 ⋅ = 2 (2 + jω / 2)(15 + jω / 2) (4 + jω)(10 + jω) 20e − jω / 2 (4 + jω)(10 + jω)

f(2t-1) = f [2(t-1/2)]

1 1 F(ω + 2) + F(ω + 2 ) 2 2

(c) f(t) cos 2t

5

=

[2 + j(ω + 2)][5 + j(ω + 2)]

(d) F [f ' (t )] = jω F(ω) =

(e)

∫ f (t ) dt t

−∞

+

5 [2 + j(ω − 2 )[5 + j(ω − 2)]]

jω10 (2 + jω)(5 + jω)

F(ω) + πF(0 )δ(ω) j(ω)

=

10 x10 + πδ(ω) jω(2 + jω)(5 + jω) 2x5

=

10 + πδ(ω) jω(2 + jω)(5 + jω)

Chapter 18, Solution 24.

(a) X (ω) = F(ω) + F [3] = 6πδ(ω) +

(

)

j − jω e −1 ω

(b) y(t ) = f (t − 2 ) Y(ω) = e − jω2 F(ω) =

je − j2ω − jω e −1 ω

(c) If h(t) = f '(t) H(ω) = jωF(ω) = jω

(

(

)

)

j − jω e − 1 = 1 − e − jω ω

3 3  3 3  2  5  (d) g(t ) = 4f  t  + 10f  t , G (ω) = 4 x F ω  + 10x F ω  2 2  5 5  3  3  = 6⋅

=

j 3 ω 2

(

(e

− j3ω / 2

)

−1 +

)

(

)

6 j − j3ω / 5 e −1 3 ω 5

(

)

j4 − j3ω / 2 j10 − j3ω / 5 e −1 + e −1 ω ω

Chapter 18, Solution 25.

(a) F(s ) =

A=

10 A B = + , s = jω s(s + 2) s s + 2

10 10 = 5, B = = −5 2 −2

F(ω) =

5 5 − jω jω + 2

f(t) =

5 sgn(t ) − 5e −2 t u(t ) 2 jω − 4 A B = + ( jω + 1)( jω + 2) jω + 1 jω + 2

(b) F(ω) = F(s ) =

s−4 A B = + , s = jω (s + 1)(s + 2) s + 1 s + 2

A = 5, B = 6 F(ω) =

6 −5 + 1 + jω 2 + jω

(

)

f(t) = − 5e − t + 6e −2 t u(t )

Chapter 18, Solution 26. (a) f ( t ) = e −( t −2) u ( t ) (b) h ( t ) = te −4 t u ( t ) (c) If x ( t ) = u ( t + 1) − u ( t − 1)

By using duality property,

 →

X(ω) = 2

sin ω ω

G (ω) = 2u (ω + 1) − 2u (ω − 1)

 →

g( t ) =

2 sin t πt

Chapter 18, Solution 27.

(a) Let F(s ) =

100 A B = + , s = jω s (s + 10) s s + 10

A=

100 100 = 10, B = = −10 − 10 10

F(ω) =

10 10 − jω jω + 10

f(t) = 5 sgn(t ) − 10e −10 t u(t ) (b) G (s ) =

A=

10s A B = + , s = jω (2 − s )(3 + s ) 2 − s s + 3

20 − 30 = 4, B = = −6 5 5

G (ω) =

4 6 − = − jω + 2 jω + 3

g(t) = 4e 2 t u(− t ) − 6e −3 t u(t ) (c) H (ω) =

( j ω)

60 2

+ j40ω + 1300

=

60

( jω + 20)2 + 900

h(t) = 2e −20 t sin( 30t ) u(t ) 1 ∞ δ(ω)e jωt dω 1 1 1 y (t ) = = π⋅ = π ∫ − ∞ (2 + jω)( jω + 1) 2 2 4 2π

Chapter 18, Solution 28.

(a)

f (t) =

=

(b)

(c)

1 ∞ 10δ(ω + 2) jωt 10 e − j2 t e d ω = 2π ∫−∞ jω( jω + 1) 2π (− j2)(− j2 + 1)

j5 e − j2 t ( −2 + j)e − j2 t = 2π 2π 1 − j2

1 ∞ 20δ(ω − 1)e jωt 20 e jt f (t) = dω = 2π ∫−∞ (2 + jω)(3 + 5ω) 2π (2 + j)(3 + j) =

(d)

1 1 1 = = 0.05 2 (5)(2) 20

f (t) =

=

πδ(ω) e jωt 1 ∞ 1 ∞ j ωt F ( ω ) e d ω = dω 2π ∫−∞ 2π ∫−∞ (5 + jω)(2 + jω)

20e jt (1 − j)e jt = 2π(5 + 5 j) π

Let

F(ω) =

5πδ(ω) 5 + = F1 (ω) + F2 (ω) (5 + jω) jω(5 + jω)

f1 ( t ) =

1 ∞ 5πδ(ω) jωt 5π 1 e dω = ⋅ = 0.5 ∫ − ∞ 2π 5 + jω 2π 5

F2 (s) =

5 A B = + , A = 1, B = −1 s(5 + s) s s + 5

F2 (ω) =

f 2 (t) =

1 1 − jω jω + 5

1 1 sgn( t ) − e −5 t = − + u ( t ) − e 5 t 2 2

f ( t ) = f 1 ( t ) + f 2 ( t ) = u( t ) − e − 5 t

Chapter 18, Solution 29.

(a)

f(t) = F -1 [δ(ω)] + F -1 [4δ(ω + 3) + 4δ(ω − 3)] =

(b)

1 4 cos 3t 1 (1 + 8 cos 3t ) + = π 2π 2π

If h ( t ) = u ( t + 2) − u ( t − 2) H(ω) =

2 sin 2ω ω g( t ) =

G (ω) = 4H(ω)

g(t) = (c)

1 8 sin 2 t ⋅ 2π t

4 sin 2t πt

Since cos(at) ↔πδ(ω + a ) + πδ(ω − a ) Using the reversal property, 2π cos 2ω ↔ πδ( t + 2) + πδ( t − 2) or F -1 [6 cos 2ω] = 3δ(t + 2) + 3δ(t − 2)

Chapter 18, Solution 30.

(a)

y( t ) = sgn( t )

H(ω) =

(b) X(ω) =

Y(ω) =

2 , jω

Y(ω) 2(a + jω) 2a = = 2+ X(ω) jω jω

1 , 1 + jω

H(ω) =

 →

Y(ω) =

X(ω) =

 →

1 a + jω h ( t ) = 2δ( t ) + a[u ( t ) − u (− t )]

1 2 + jω

1 + jω 1 = 1− 2 + jω 2 + jω

(c ) In this case, by definition, h ( t ) =

 →

h ( t ) = δ( t ) − e − 2 t u ( t )

y( t ) = e −at sin bt u ( t )

Chapter 18, Solution 31.

Y(ω) =

(a)

X(ω) =

1 (a + jω)

2

H(ω) =

,

Y(ω) 1 = H(ω) a + jω

1 a + jω

 →

x ( t ) = e − at u ( t )

(b)

By definition, x ( t ) = y( t ) = u ( t + 1) − u ( t − 1)

(c )

Y(ω) =

X(ω) =

Y(ω) jω 1 a = − = H(ω) 2(a + jω) 2 2(a + jω)

1

H(ω) =

,

(a + jω)

2 jω  →

x(t) =

Chapter 18, Solution 32.

(a)

e − jω jω + 1 and F(− ω)

e − ( t −1) u ( t − 1)

Since

F1 (ω) =

f(-t)

e jω − jω + 1

f 1 (t ) = e − (− t −1) u (− t − 1)

f1(t) = e (t +1 )u(− t − 1) (b)

From Section 17.3, 2 t +1

2πe

2

If F2 (ω) = 2e f2(t) =

−ω

, then

2 π t +1

(

2

−ω

)

a 1 δ( t ) − e − at u ( t ) 2 2

(b)

By partial fractions F3 (ω) =

1

( jω + 1)2 ( jω − 1)2

Hence f 3 (t ) =

=

(d)

1 1 1 1 4 4 = + 4 + − 4 2 2 ( jω + 1) ( jω + 1) ( jω − 1) jω − 1

(

)

1 −t te + e − t + te t − e t u (t ) 4

1 (t + 1)e −t u(t ) + 1 (t − 1)e t u(t ) 4 4

f 4 (t ) =

1 1 ∞ 1 ∞ δ(ω)e jωt jωt ( ) F ω e d ω = = 1 ∫ ∫ 2π − ∞ 2π − ∞ 1 + j2ω 2π

Chapter 18, Solution 33. (a)

Let x (t ) = 2 sin πt[u (t + 1) − u (t − 1)] From Problem 17.9(b), 4 jπ sin ω π 2 − ω2 Applying duality property, X(ω) =

f (t ) =

(b)

1 2 j sin (− t ) X(− t ) = 2 2 2π π −t

f(t) =

2 j sin t t 2 − π2

F(ω) =

j (cos 2ω − j sin 2ω) − j (cos ω − j sin ω) ω ω

j j2 ω e − jω e j 2 ω − jω = (e − e ) = − ω jω jω

f (t ) =

1 1 sgn (t − 1) − sgn (t − 2) 2 2

But sgn( t ) = 2u ( t ) − 1

f (t ) = u (t − 1) −

1 1 − u (t − 2 ) + 2 2

= u(t − 1) − u(t − 2 )

Chapter 18, Solution 34. First, we find G(ω) for g(t) shown below. g (t ) = 10[u (t + 2 ) − u (t − 2 )] + 10[u (t + 1) − u (t − 1)] g ' (t ) = 10[δ(t + 2 ) − δ(t − 2 )] + 10[δ(t + 1) − δ(t − 1)]

The Fourier transform of each term gives g(t) 20 10

–2

–1

0

1

t

2

g ‘(t) 10δ(t+2)

10δ(t+1)

–2

–1

0

1

–10δ(t-1)

2

–10δ(t-2)

jωG (ω) = 10(e jω2 − e − jω2 ) + 10(e jω − e − jω ) = 20 j sin 2ω + 20 j sin ω G (ω) =

20 sin 2ω 20 sin ω + = 40 sinc(2ω) + 20 sinc(ω) ω ω

Note that G(ω) = G(-ω).

t

F(ω) = 2πG (− ω)

f (t ) =

1 G (t ) 2π

= (20/π)sinc(2t) + (10/π)sinc(t)

Chapter 18, Solution 35. (a)

x(t) = f[3(t-1/3)]. Using the scaling and time shifting properties,’’ X(ω) =

(b)

1 1 e − jω / 3 e − jω / 3 = 3 2 + jω / 3 (6 + jω)

Using the modulation property,  1 1 1 1 1 1 1  =  + Y(ω) = [F(ω + 5) + F(ω − 5)] =  +  2 2  2 + j(ω + 5) 2 + j(ω − 5)  2  jω + 7 jω − 3  jω 2 + jω

(c )

Z(ω) = jωF(ω) =

(d)

H(ω) = F(ω)F(ω) =

(e)

I(ω) = j

1 (2 + jω) 2

d (0 − j) 1 F(ω) = j = dω (2 + jω) 2 (2 + jω) 2

Chapter 18, Solution 36.

H(ω) =

Vo (ω) Vi (ω)

Vo (ω) = H(ω)Vi (ω) =

10Vi (ω) 2 + jω

(a)

vi = 4δ(t) Vo (ω) =

Vi(ω) = 4

40 2 + jω

v o ( t ) = 40e −2 t u (t ) vo(2) = 40e–4 = 0.7326 V (b)

v i = 6e − t u (t )

Vi (ω) =

6 1 + jω

Vo (ω) =

60 (2 + jω)(1 + jω)

Vo (s ) =

60 A B = + , s = jω (s + 2)(s + 1) s + 1 s + 2

A=

60 60 = 60, B = = −60 −1 1

Vo (ω) =

60 60 − 1 + jω 2 + jω

[ (2) = 60[e

] ] = 60 (0.13533 − 0.01831)

v o ( t ) = 60 e − t − e −2 t u ( t ) vo = 7.021 V (c)

−2

− e −4

vi(t) = 3 cos 2t Vi(ω) = π[δ(ω + 2) + δ(ω- 2)] Vo =

10π[δ(ω + 2) + δ(ω − 2 )] 2 + jω

v o (t) =

1 ∞ Vo (ω)e jωt dω 2π ∫−∞ jωt ∞ δ(ω − 2 )e δ(ω + 2 ) jωt e dω + 5 ∫ = 5∫ dω −∞ − ∞ 2 + jω 2 + jω ∞

=

[

5e − j2 t 5e + j2 t 5 + = e − j(2 t − 45° ) + e j(2 t − 45° ) 2 − j2 2 + j2 2 2 5 cos(2 t − 45°) 2

=

v o (2 ) =

5 5 cos(229.18° − 45°) cos(4 − 45°) = 2 2

vo(2) = –3.526 V

Chapter 18, Solution 37.

2 jω =

j2ω 2 + jω

By current division, j2ω I (ω) j2ω 2 + jω H(ω) = o = = j2ω I s (ω) j2ω + 8 + j4ω 4+ 2 + jω H(ω) =

jω 4 + j3ω

Chapter 18, Solution 38.

Vi (ω) = πδ(ω) + Vo (ω) =

]

1 jω

5  1  10  πδ (ω) +  Vi (ω) = 5 + jω  jω  10 + jω2

Let Vo (ω) = V1 (ω) + V2 (ω) =

5πδ(ω) 5 + 5 + jω jω(5 + jω)

V2 (ω) =

5 A B = + s(s + 5) s s + 5

V2 (ω) =

1 1 − jω 5 + j ω

V1 =

5πδ(ω) 5 + jω

A = 1, B = -1, s = jω 1 v 2 (t ) = sgn( t ) − e − 5 t 2 v 1 (t ) =

1 ∞ 5πδ (ω) jωt e dω 2π ∫−∞ 5 + jω

5π 1 ⋅ = 0.5 2π 5

v1(t) =

v 0 (t ) = v1 (t ) + v 2 (t ) = 0.5 + 0.5 sgn (t ) − e −5 t sgn (t ) = −1 + 2u (t )

But

v o (t ) = +0.5 − 0.5 + u (t ) − e −5 t u (t ) = u(t ) − e −5t u(t )

Chapter 18, Solution 39. ∞

Vs (ω) =

∫ (1 − t )e

− jωt

dt =

−∞

I(ω) =

Vs (ω) 3

10 + jωx10

−3

=

1 1 1 − jω + − e 2 jω ω ω2

10 3

 1 1 1 − jω   +  e − 2 10 + jω  jω ω ω2  6

Chapter 18, Solution 40. v( t ) = δ( t ) − 2δ( t − 1) + δ( t − 2) − ω 2 V(ω) = 1 − 2e − jω + e jω2

V(ω) =

Now

1 − 2e − jω + e − jω2 − ω2

Z(ω) = 2 +

1 1 + j2ω = jω jω

I=

=

V(ω) 2e jω − e jω2 − 1 jω = ⋅ 2 Z(ω) 1 + j2ω ω

1 ( 0.5 + 0.5e − jω2 −e − jω ) jω(0.5 + jω)

1 A B = + s(s + 0.5) s s + 0.5

But

I(ω) =

i(t) =

A = 2, B = -2

2 2 ( (0.5 + 0.5e− jω2 − e− jω ) 0.5 + 0.5e jω2 − e − jω ) − jω 0.5 + jω

1 1 sgn( t ) + sgn(t − 2) − sgn( t − 1) − e − 0.5t u(t ) − e − 0.5( t − 2 ) u(t − 2) − 2e − 0.5( t −1) u(t − 1) 2 2

Chapter 18, Solution 41.

2 +

+ −

1 2 + jω

V

1/s

0.5s

−

V−

1 2V 2 + jω + jω V + −2=0 2 jω

(

)

jω − 4ω 2 + j9ω jω − 2ω + 4 V = j4ω + = 2 + jω 2 + jω

V(ω) =

2

2 jω(4.5 + j2ω) (2 + jω)(4 − 2ω 2 + jω)

2

Chapter 18, Solution 42.

By current division, I o = (a)

2 ⋅ I(ω) 2 + jω

For i(t) = 5 sgn (t), 10 jω 2 10 20 Io = ⋅ = 2 + jω jω jω(2 + jω)

I(ω) =

Let I o =

20 A B = + , A = 10, B = −10 s(s + 2) s s + 2

I o (ω) =

10 10 − j ω 2 + jω

io(t) = 5 sgn( t ) − 10e −2 t u(t )A i(t)

(b)

i’(t)

4

4δ(t)

1 1

t

i' ( t ) = 4δ( t ) − 4δ( t − 1) jω I(ω) = 4 − 4e − jω

(

4 1 − e − jω I(ω) = jω

Io =

)

 1 8(1 − e − jω ) 1  (1 − e − jω ) = 4 − ω + ω jω(2 + jω) j 2 j   =

4 4 4e − j ω 4e − j ω − − + jω 2 + jω jω 2 + jω

t –4δ(t–1)

io(t) = 2 sgn( t ) − 2 sgn( t − 1) − 4e −2 t u(t ) + 4e −2( t −1) u(t − 1)A

Chapter 18, Solution 43.

20 mF

1 1 50 = = , − 3 jωC j20x10 ω jω

 →

Vo =

Vo =

50 50 40 = Is • , 50 jω (s + 1.25)(s + 5) 40 + jω

i s = 5e − t

 →

Is =

1 5 + jω

s = jω

A B 40  1 1  + = −  s + 1.25 s + 5 3  jω + 1.25 jω + 5  v o (t) =

40 −1.25t (e − e −5 t ) u ( t ) 3

Chapter 18, Solution 44.

1H

jω

We transform the voltage source to a current source as shown in Fig. (a) and then combine the two parallel 2Ω resistors, as shown in Fig. (b). Io +

Vs/2 2

2 Vo

Io +

Vs/2

1 Vo

jω

− (a) 2 2 = 1Ω, I o =

Vo = jω I o =

V 1 ⋅ s 1 + jω 2

jω Vs 2(1 + jω)

− (b)

jω

v s ( t ) = 10δ(t ) − 10δ( t − 2)

jω Vs (ω) = 10 − 10e − j2 ω

Vs (ω) =

(

10 1 − e − j2ω jω

)

(

)

5 1 − e − j2 ω 5 5 Hence Vo = = − e − j2 ω 1 + jω 1 + jω 1 + jω

v o ( t ) = 5e − t u ( t ) − 5e − ( t − 2) u ( t − 2) v o (1) = 5e −1 − 1 − 0 = 1.839 V Chapter 18, Solution 45.

Vo =

1 jω 1 2 + jω + jω

(2) =

2 ( jω + 1)

2

 →

v o ( t ) = 2te − t u ( t )

Chapter 18, Solution 46.

1 F 4

1 jω

2H 3δ( t )

1 4

=

− j4 ω

jω2 3 1 1 + jω

e − t u(t)

The circuit in the frequency domain is shown below: 2Ω

Vo Io(ω) –j4/ω

1/(1+jω)

+ −

3

+ −

j2ω

At node Vo, KCL gives 1 − Vo 3 − Vo V 1 + jω + = o − j4 2 j2ω ω

2 j2Vo − 2Vo + jω3 − jωVo = − ω 1 + jω 2 + jω3 1 + jω Vo = j2 2 + jω − ω 2 + jω3 − 3ω 2 V 1 + jω I o (ω) = o = j2  j2ω  j2ω 2 + jω −  ω 

Io(ω) =

2 + jω 2 − 3ω 2 4 − 6ω 2 + j(8ω − 2ω 3 )

Chapter 18, Solution 47.

Transferring the current source to a voltage source gives the circuit below: 1/(jω)

2Ω +

8V

+ −

Vo

−

1Ω

jω/2

jω 4 + j3ω jω Let Z in = 2 + 1 = 2+ 2 = jω 2 + jω 2 1+ 2 By voltage division, 1 8 8 jω Vo (ω) = ⋅8 = = 1 jω(4 + j3ω) 1 + jωZ in + Z in 1+ jω 2 + jω =

8(2 + jω) 2 + jω + jω4 − 3ω 2

=

8(2 + jω) 2 + jω5 − 3ω 2

Chapter 18, Solution 48.

0.2F

1 j5 =− jωC ω

As an integrator, RC = 20 x 10 3 x 20 x 10 −6 = 0.4 vo = −

1 t v i dt RC ∫o

Vo = −

 1  Vi + πVi (0)δ(ω)  RC  jω  =−

Io =

 1  2 + πδ (ω)  ( 0 .4  j ω 2 + j ω ) 

  Vo 2 mA = −0.125  + π δ (ω) 20  jω (2 + jω) 

=−

0.125 0.125 + − 0.125πδ (ω) jω 2 + jω

i o ( t ) = −0.125 sgn( t ) + 0.125e − 2 t u (t ) −

0.125 πδ (ω)e jωt dt ∫ 2π

= 0.125 + 0.25u ( t ) + 0.125e −2 t u ( t ) −

0.125 2

io(t) = 0.625 − 0.25u(t ) + 0.125e −2t u(t ) mA

Chapter 18, Solution 49. Consider the circuit shown below:

jω j2ω

jω +

VS

+ −

i1

i2 2Ω

1 Ω vo

−

Vs = π[δ (ω + 1) + δ (ω − 2)] For mesh 1, − Vs + (2 + j2ω)I1 − 2I 2 − jωI 2 = 0 Vs = 2 (1 + jω) I1 − (2 + jω)I 2 0 = (3 + jω)I 2 − 2I1 − jωI1

For mesh 2, I1 =

(3 + ω)I 2 (2 + ω)

(2)

Substituting (2) into (1) gives Vs = 2

(1)

2 (1 + jω)(3 + jω)I 2 − (2 + jω)I 2 2 + jω

[(

) (

)]

Vs (2 + ω) = 2 3 + j4ω − ω 2 − 4 + j4ω − ω 2 I 2 = I 2 (2 + j4ω − ω

2

)

I2 =

(s + 2)Vs s 2 + 4s + 2

Vo = I 2 =

( jω + 2) π [δ (ω + 1) + δ (ω − 1)] ( jω)2 + jω4 + 2

1 ∞ v o (ω)e jωt dω ∫ − ∞ 2π

v o (t) =

=∫

, s = jω

∞

−∞

1 ( jω + 2) e jωt δ (ω + 1)dω 1 ( jω + 2)e jωt δ(ω − 1)dω 2 +2 2 ( jω) + jω4 + 2 ( jω)2 + jω4 + 2

1 (− j + 2)e jt 1 ( j + 2)e jt = 2 + 2 − 1 − j4 + 2 − 1 + j4 + 2

1 1 (2 − j)(1 + j4) (2 − j)(1 − j4)e jt v o (t) = 2 e jt + 2 17 17 =

1 (6 + j7 )e jt + 1 (6 − j7 )e jt 34 34

= 0.271e − j ( t −13.64° ) + 0.271e j ( t −13.64° ) vo(t) = 0.542 cos(t − 13.64°)V

Chapter 18, Solution 50. Consider the circuit shown below:

j0.5ω 1Ω

VS

+ −

i1

jω

i2 jω

+ 1Ω

vo −

For loop 1, For loop 2,

− 2 + (1 + jω)I1 + j0.5ωI 2 = 0

(1)

(1 + jω)I 2 + j0.5ωI1 = 0

(2)

From (2), I1 =

(1 + jω)I 2 − j0.5ω

= −2

(1 + jω)I 2 jω

Substituting this into (1), − 2(1 + jω)I 2 jω 2= + I2 jω 2 3   2 jω = − 4 + j4ω − ω 2 I 2 2   I2 =

2 jω 4 + j4ω − 1.5ω 2

Vo = I 2 =

− 2 jω

4 + j4ω + 1.5( jω)

2

4 jω 3 Vo = 8 8ω 2 +j + ( jω ) 3 3

=

 4 − 4 + jω   3   8  4  + jω  +    3  3 2

2

+

16 3 2   8  4  + jω  +    3  3

2

 8   8  t  u(t ) + 5.657e − 4t / 3 sin t u(t ) V Vo ( t ) = − 4e − 4t / 3 cos 3 3    

Chapter 18, Solution 51.

1 1 1 jω Z = 1 // = = 1 jω 1 + 1 + jω jω

1 Z 2 1 2 1 + jω ∗ = Vo = Vo = 1 Z+2 1 + jω 3 + 2 jω 1 + jω 2+ 1 + jω 1 = , s = jω (s + 1)(s + 1.5) Vo =

A B 2 2 + = − s + 1 s + 1.5 s + 1 s + 1.5 ∞

W=

→

v o ( t ) = 2(e − t − e −1.5t )u ( t )

∞

∫

−∞

f 2 ( t )dt = 4 ∫ (e − t − e −1.5t ) 2 dt

∞

= 4 ∫ (e

0

∞

− 2t

− 2e

0

− 2.5 t

+e

− 3t

 e − 2t e − 2.5t e − 3t  +2 − )dt = 4  −2 2 . 5 3   0

1 2 1 W = 4( − + ) = 0.1332 J 2 2.5 3 Chapter 18, Solution 52.

∞

J = 2 ∫ f 2 ( t ) dt = 0

1 ∞ 2 F(ω) dω ∫ 0 π ∞

1 ∞ 1 1 1 π dω = tan −1 (ω / 3) = = (1/6) = 2 2 ∫ 0 π 9 +ω 3π 3π 2 0

Chapter 18, Solution 53.

J =

∫

∞

0

∞

F(ω) dω = 2π∫ f 2 ( t ) dt 2

−∞

f(t) =

e 2t ,

t<0

e −2 t ,

t>0

 e 4t 0 ∞ J = 2π  ∫ e 4 t dt + ∫ e − 4 t dt  = 2π  −∞  0  4

0

−∞

e −4 t + −4

  = 2π[(1/4) + (1/4)] = π 0   ∞

Chapter 18, Solution 54.

W1Ω =

∫

∞

−∞

∞

∞

0

0

f 2 ( t ) dt = 16 ∫ e − 2 t dt = − 8e − 2 t

= 8J

Chapter 18, Solution 55.

f(t) = 5e2e–tu(t) F(ω) = 5e2/(1 + jω), |F(ω)|2 = 25e4/(1 + ω2) W1Ω

1 ∞ 25e 4 2 = F ( ω ) d ω = π ∫0 π

∫

∞

0

= 12.5e4 = 682.5 J or

W1Ω =

∫

∞

−∞

∞

f 2 ( t ) dt = 25e 4 ∫ e − 2 t dt = 12.5e4 = 682.5 J 0

Chapter 18, Solution 56.

W1Ω = But,

∫

∞

−∞

∞

f 2 ( t ) dt = ∫ e − 2 t sin 2 (2 t ) dt 0

sin2(A) = 0.5(1 – cos(2A))

∞

1 25e 4 d ω = tan −1 (ω) 2 π 1+ ω 0

W1Ω =

∫

∞

0

e

−2 t

1 e −2 t 0.5[1 − cos(4 t )]dt = 2 −2

∞

0

∞

e −2 t − [−2 cos(4 t ) + 4 sin(4t )] 4 + 16 0

= (1/4) + (1/20)(–2) = 0.15 J Chapter 18, Solution 57.

W1Ω = or

∫

∞

−∞

0

0

−∞

−∞

i 2 ( t ) dt = ∫ 4e 2 t dt = 2e 2 t

= 2J

I(ω) = 2/(1 – jω), |I(ω)|2 = 4/(1 + ω2) ∞

W1Ω =

4 4π 1 ∞ 4 ∞ 1 2 = 2J dω = tan −1 (ω) = ω ω = I ( ) d 2 ∫ ∫ − ∞ − ∞ 2 π π 2π 2π (1 + ω ) 0

In the frequency range, –5 < ω < 5, 5

4 4 4 W = tan −1 ω = tan −1 (5) = (1.373) = 1.7487 π π π 0

W/ W1Ω = 1.7487/2 = 0.8743 or 87.43% Chapter 18, Solution 58.

ωm = 200π = 2πfm which leads to fm = 100 Hz (a)

ωc = πx104 = 2πfc which leads to fc = 104/2 = 5 kHz

(b)

lsb = fc – fm = 5,000 – 100 = 4,900 Hz

(c)

usb = fc + fm = 5,000 + 100 = 5,100 Hz

Chapter 18, Solution 59.

10 6 − V (ω) 2 + jω 4 + jω 5 3 H(ω) = o = = − Vi (ω) 2 2 + jω 4 + jω

 5 3  4  Vo (ω) = H(ω)Vi (ω) =  −  2 + jω 4 + jω  1 + jω 20 12 = − , s = jω (s + 1)(s + 2) (s + 1)(s + 4) Using partial fraction, A B C D 16 20 4 + + + = − + s + 1 s + 2 s + 1 s + 4 1 + jω 2 + jω 4 + jω

Vo (ω) = Thus,

(

)

v o ( t ) = 16e − t − 20e −2 t + 4e −4 t u ( t ) V

Chapter 18, Solution 60. 2 + Is

1/jω

jω

V

−

V = jωI s

1 jω 1 + 2 + jω jω

=

jωI s 1 − ω 2 + j2ω

Since the voltage appears across the inductor, there is no DC component. V1 =

2π∠90°8 1 − 4π 2 + j4π

V2 =

=

50.27∠90° = 1.2418∠ − 71.92° − 38.48 + j12.566

4π∠90°5 2

1 − 16π + j8π

=

62.83∠90° = 0.3954∠ − 80.9° − 156.91 + j25.13

v( t ) = 1.2418 cos( 2πt − 41.92°) + 0.3954 cos( 4πt + 129.1°) mV

Chapter 18, Solution 61. lsb = 8,000,000 – 5,000 = 7,995,000 Hz usb = 8,000,000 + 5,000 = 8,005,000 Hz Chapter 18, Solution 62. For the lower sideband, the frequencies range from 10,000 – 3,500 Hz = 6,500 Hz to 10,000 – 400 Hz = 9,600 Hz For the upper sideband, the frequencies range from 10,000 + 400 Hz = 10,400 Hz to 10,000 + 3,500 Hz = 13,500 Hz Chapter 18, Solution 63. Since fn = 5 kHz, 2fn = 10 kHz i.e. the stations must be spaced 10 kHz apart to avoid interference. ∆f = 1600 – 540 = 1060 kHz The number of stations = ∆f /10 kHz = 106 stations Chapter 18, Solution 64. ∆f = 108 – 88 MHz = 20 MHz The number of stations = 20 MHz/0.2 MHz = 100 stations Chapter 18, Solution 65. ω = 3.4 kHz fs = 2ω = 6.8 kHz

Chapter 18, Solution 66. ω = 4.5 MHz fc = 2ω = 9 MHz Ts = 1/fc = 1/(9x106) = 1.11x10–7 = 111 ns Chapter 18, Solution 67. We first find the Fourier transform of g(t). We use the results of Example 17.2 in conjunction with the duality property. Let Arect(t) be a rectangular pulse of height A and width T as shown below. Arect(t) transforms to Atsinc(ω2/2) f(t)

F(ω)

A ω

t –T/2

T/2 G(ω)

ω –ωm/2 According to the duality property, Aτsinc(τt/2)

becomes 2πArect(τ)

g(t) = sinc(200πt) becomes 2πArect(τ) where Aτ = 1 and τ/2 = 200π or T = 400π i.e. the upper frequency ωu = 400π = 2πfu or fu = 200 Hz The Nyquist rate = fs = 200 Hz The Nyquist interval = 1/fs = 1/200 = 5 ms

ωm/2

Chapter 18, Solution 68. The total energy is WT =

∫

∞

−∞

v 2 ( t ) dt

Since v(t) is an even function, WT =

∫

∞

0

2500e

−4 t

e −4 t dt = 5000 −4

∞

= 1250 J 0

V(ω) = 50x4/(4 + ω2) 1 5 1 5 (200) 2 2 | V ( ω ) | d ω = dω 2π ∫1 2π ∫1 (4 + ω 2 ) 2

W =

But

∫ (a

2

1  x 1 1  + tan −1 ( x / a ) + C dx = 2  2 2 2 2 a +x ) 2a  x + a  5

2x10 4 1  ω 1  W = + tan −1 (ω / 2) 2  π 8 4 + ω 2 1

= (2500/π)[(5/29) + 0.5tan-1(5/2) – (1/5) – 0.5tan–1(1/2) = 267.19 W/WT = 267.19/1250 = 0.2137 or 21.37% Chapter 18, Solution 69.

The total energy is WT =

=

W =

1 ∞ 1 ∞ 400 2 F(ω) dω = dω ∫ 2π − ∞ 2π ∫−∞ 4 2 + ω 2

[

400 (1 / 4) tan −1 (ω / 4) π

]

∞ 0

=

100 π = 50 π 2

[

1 2 400 2 F(ω) dω = (1 / 4) tan −1 (ω / 4) ∫ 0 2π 2π

]

2 0

= [100/(2π)]tan–1(2) = (50/π)(1.107) = 17.6187 W/WT = 17.6187/50 = 0.3524 or 35.24%

Chapter 19, Solution 1. To get z 11 and z 21 , consider the circuit in Fig. (a).

1Ω

4Ω

I2 = 0

+ I1

6Ω

V1

Io

+

2Ω

V2

−

−

(a) z 11 =

V1 = 1 + 6 || (4 + 2) = 4 Ω I1

Io =

1 I , 2 1

z 21 =

V2 = 2 I o = I 1

V2 = 1Ω I1

To get z 22 and z 12 , consider the circuit in Fig. (b). I1 = 0

1Ω

4Ω Io '

+

+

6Ω

V1

2Ω

V2

−

−

(b) z 22 =

V2 = 2 || (4 + 6) = 1.667 Ω I2

Io' =

2 1 I2 = I2 , 2 + 10 6

z 12 =

V1 = 1Ω I2

V1 = 6 I o ' = I 2

I2

Hence, 4 1  [z ] =  Ω  1 1.667 

Chapter 19, Solution 2.

Consider the circuit in Fig. (a) to get z 11 and z 21 . 1Ω

Io '

1Ω

1Ω

+ I1

1Ω

V1

1Ω

1Ω Io

+

1Ω

V2 −

− 1Ω

1Ω

1Ω (a)

z 11 =

V1 = 2 + 1 || [ 2 + 1 || (2 + 1) ] I1

(1)(11 4)  11 3 = 2 + = 2.733 z 11 = 2 + 1 ||  2 +  = 2 +  15 4 1 + 11 4

Io =

1 1 Io' = Io' 1+ 3 4

Io' =

1 4 I1 = I1 1 + 11 4 15

Io =

1 4 1 ⋅ I1 = I1 4 15 15

V2 = I o = z 21 =

I2 = 0

1 I 15 1

V2 1 = = z 12 = 0.06667 I 1 15

1Ω

To get z 22 , consider the circuit in Fig. (b). I1 = 0

1Ω

1Ω

1Ω

1Ω

+

+ 1Ω

V1

1Ω

1Ω

V2 −

− 1Ω

1Ω

1Ω

1Ω

(b) z 22 =

V2 = 2 + 1 || (2 + 1 || 3) = z 11 = 2.733 I2

Thus,  2.733 0.06667  [z ] =  Ω  0.06667 2.733 

Chapter 19, Solution 3. (a)

To find z 11 and z 21 , consider the circuit in Fig. (a). -j Ω Io

+ I1

V1

I2 = 0

jΩ

1Ω

−

V1 j (1 − j) = j || (1 − j) = = 1+ j I1 j +1− j

By current division, j Io = I = j I1 j +1− j 1

V2 −

(a) z 11 =

+

I2

V2 = I o = jI 1 z 21 =

V2 =j I1

To get z 22 and z 12 , consider the circuit in Fig. (b). -j Ω

I1 = 0

+

+ jΩ

V1

1Ω

−

I2

V2 −

(b) z 22 =

V2 = 1 || ( j − j) = 0 I2

V1 = j I 2 z 12 =

V1 =j I2

Thus,  1+ j j  [z ] =  Ω 0   j (b)

To find z 11 and z 21 , consider the circuit in Fig. (c). jΩ

-j Ω

+ I1

+

1Ω

V1 −

I2 = 0

V2 -j Ω

1Ω (c)

−

z 11 =

V1 -j = j + 1 + 1 || (-j) = 1 + j + = 1.5 + j0.5 1− j I1

V2 = (1.5 − j0.5) I 1 z 21 =

V2 = 1.5 − j0.5 I1

To get z 22 and z 12 , consider the circuit in Fig. (d).

I1 = 0

jΩ

-j Ω

+

+

1Ω

V1 −

V2 -j Ω

1Ω (d)

z 22 =

V2 = -j + 1 + 1 || (-j) = 1.5 - j1.5 I2

V1 = (1.5 − j0.5) I 2 z 12 =

V1 = 1.5 − j0.5 I2

Thus,

 1.5 + j0.5 1.5 − j0.5  [z ] =  Ω  1.5 − j0.5 1.5 − j1.5 

Chapter 19, Solution 4. Transform the Π network to a T network.

Z1

Z3

Z2

−

I2

Z1 =

(12)( j10) j120 = 12 + j10 − j5 12 + j5

Z2 =

- j60 12 + j5

Z3 =

50 12 + j5

The z parameters are

z 12 = z 21 = Z 2 =

(-j60)(12 - j5) = -1.775 - j4.26 144 + 25

z 11 = Z1 + z 12 =

( j120)(12 − j5) + z 12 = 1.775 + j4.26 169

z 22 = Z 3 + z 21 =

(50)(12 − j5) + z 21 = 1.7758 − j5.739 169

Thus,

 1.775 + j4.26 - 1.775 − j4.26  [z ] =  Ω  - 1.775 − j4.26 1.775 − j5.739 

Chapter 19, Solution 5. Consider the circuit in Fig. (a).

1

s

I2 = 0 Io

+ I1

V1

1

1/s

1/s

+ V2 −

− (a)  1  1  1 + s +   1  1 1   s + 1  s z 11 = 1 || || 1 + s +  = || 1 + s +  = 1  1 s  s s  1  1+   +1+ s + s  s + 1 s 1 s

s2 + s +1 s 3 + 2s 2 + 3s + 1

z 11 =

Io =

Io =

1 ||

1 s

1 1 1 || + 1 + s + s s

I1 =

s s + 2s + 3s + 1 3

2

1 s +1

1 1 +1+ s + s +1 s

I1 =

s s +1

s + s2 + s +1 s +1

I1

I1

I1 1 V2 = I o = 3 s s + 2s 2 + 3s + 1 z 21 =

V2 1 = 3 2 I 1 s + 2s + 3s + 1

Consider the circuit in Fig. (b).

s

1

I1 = 0 +

+ 1

V1

1/s

1/s

−

− (b) z 22 =

z 22

V2 1  1 1  1   = || 1 + s + 1 ||  = || 1 + s + s s  s + 1 I2 s 

 1  1  1  1 + s +  1 s + +  s  s + 1 s +1 = = s 1 1 1+ s + s2 + +1+ s + s +1 s s +1

z 22 =

s 2 + 2s + 2 s 3 + 2s 2 + 3s + 1

z 12 = z 21

Hence,

V2

I2

  s2 + s + 1 1  s 3 + 2s 2 + 3s + 1 s 3 + 2s 2 + 3s + 1   [z ] =  1 s 2 + 2s + 2    s 3 + 2s 2 + 3s + 1 s 3 + 2s 2 + 3s + 1 

Chapter 19, Solution 6. To find z 11 and z 21 , connect a voltage source V1 to the input and leave the output open as in Fig. (a).

I1

10 Ω

Vo

20 Ω +

V1

+ −

30 Ω

0.5 V2

V2 −

(a) V1 − Vo Vo , = 0.5 V2 + 10 50

where V2 =

30 3 Vo = Vo 20 + 30 5

3  V V1 = Vo + 5  Vo  + o = 4.2 Vo 5  5 I1 =

V1 − Vo 3.2 = V = 0.32 Vo 10 10 o

z 11 =

4.2 Vo V1 = = 13.125 Ω I 1 0.32 Vo

z 21 =

0.6 Vo V2 = = 1.875 Ω I 1 0.32 Vo

To obtain z 22 and z 12 , use the circuit in Fig. (b).

10 Ω

20 Ω

I2

+ V1

0.5 V2

− (b)

30 Ω

+ −

V2

V2 = 0.5333 V2 30 V2 1 = = = 1.875 Ω I 2 0.5333

I 2 = 0.5 V2 + z 22

V1 = V2 − (20)(0.5 V2 ) = -9 V2 z 12 =

- 9 V2 V1 = = -16.875 Ω I 2 0.5333 V2

Thus,

 13.125 - 16.875  [z ] =  Ω 1.875   1.875

Chapter 19, Solution 7. To get z11 and z21, we consider the circuit below. I1

20 Ω

I2=0

100 Ω +

+ V1

vx

50 Ω

60 Ω

-

-

12vx +

V1 − Vx Vx Vx + 12Vx = + 20 50 160 V − Vx 81 V1 I1 = 1 ( ) = 20 121 20

 → →

Vx = z11 =

40 V1 121

V1 = 29.88 I1

+ V2 -

13Vx 57 57 40 57 40 20x121 ) − 12Vx = − Vx = − ( )V1 = − ( ) I1 160 8 8 121 8 121 81 V = −70.37 I1  → z 21 = 2 = −70.37 I1

V2 = 60(

To get z12 and z22, we consider the circuit below. I1=0

20 Ω

I2

100 Ω +

+ V1

50 Ω

vx

60 Ω

-

-

12vx

+ V2 -

+ Vx =

50 1 V2 = V2 , 100 + 50 3

z 22 =

V2 = 1 / 0.09 = 11.11 I2

I2 =

V2 V2 + 12Vx + = 0.09V2 150 60

11.11 1 I 2 = 3.704I 2 V1 = Vx = V2 = 3 3

 →

Thus,  29.88 3.704 [z] =  Ω − 70.37 11.11

V z12 = 1 = 3.704 I2

Chapter 19, Solution 8.

To get z11 and z21, consider the circuit below. j4 Ω

I1 -j2 Ω

•

5Ω

I2 =0

•

j6 Ω

+

j8 Ω

+ V2

V1 10 Ω

-

-

V1 = (10 − j2 + j6)I1

V2 = −10I1 − j4I1

V z11 = 1 = 10 + j4 I1

 →

 →

z 21 =

V2 = −(10 + j4) I1

To get z22 and z12, consider the circuit below. j4 Ω

I1=0 -j2 Ω

•

j6 Ω

5Ω •

j8 Ω

I2

+

+ V2

V1 10 Ω -

-

V2 = (5 + 10 + j8)I 2

V1 = −(10 + j4)I 2

 →

 →

z 22 =

V2 = 15 + j8 I2

V z12 = 1 = −(10 + j4) I2

Thus,  (10 + j4) − (10 + j4) [z] =  Ω − (10 + j4) (15 + j8) 

Chapter 19, Solution 9. It is evident from Fig. 19.5 that a T network is appropriate for realizing the z parameters.

6Ω

R2

R1

2Ω

4Ω

R3

R 1 = z 11 − z 12 = 10 − 4 = 6 Ω R 2 = z 22 − z 12 = 6 − 4 = 2 Ω

R 3 = z 12 = z 21 = 4 Ω

Chapter 19, Solution 10. (a)

This is a non-reciprocal circuit so that the two-port looks like the one shown in Figs. (a) and (b).

I1

z11

z22

+

+ V1

I2

z12 I2

+ −

+ −

−

z21 I1

V2 −

(a)

(b)

This is a reciprocal network and the two-port look like the one shown in Figs. (c) and (d).

z11 – z12

I1

z22 – z12

I2 +

+ z12

V1

V2

−

−

(c) 25 Ω

I1

10 Ω

I2 +

+ + −

20 I2

V1

+ −

5 I1

V2

−

−

(b)

z 11 − z 12 = 1 +

2 1 = 1+ s 0.5 s

z 22 − z 12 = 2s

z 12 = I1

1 s

1Ω

0.5 F

2H

+

I2 +

1F

V1 −

V2 −

(d)

Chapter 19, Solution 11. This is a reciprocal network, as shown below. 1+j5 3+j

1Ω

j5 Ω

3Ω

j1 Ω

5Ω

5-j2

-j2 Ω

Chapter 19, Solution 12. This is a reciprocal two-port so that it can be represented by the circuit in Figs. (a) and (b). I1

z11 – z12

z22 – z12

I2

+

+ z12

V1

V2

−

2Ω

− (a)

I1

8Ω

2Ω

+ V1

+ 4Ω

(b)

V1 = (8 + 4 || 4) I 1 = 10 I 1

V2 −

−

From Fig. (b),

I2 Io 2Ω

By current division,

Io =

1 I , 2 1

V2 = 2 I o = I 1

V2 I1 = = 0 .1 V1 10 I 1

Chapter 19, Solution 13. This is a reciprocal two-port so that the circuit can be represented by the circuit below.

40 Ω

120∠0° V rms

+ −

I1

50 Ω

10 I2

20 Ω

+ −

+ −

30 I1

We apply mesh analysis. For mesh 1, - 120 + 90 I 1 + 10 I 2 = 0  → 12 = 9 I 1 + I 2 For mesh 2, 30 I 1 + 120 I 2 = 0  → I 1 = -4 I 2 Substituting (2) into (1), - 12 12 = -36 I 2 + I 2 = -35 I 2  → I 2 = 35 2

1 1  12  2 P = I 2 R =   (100) = 5.877 W 2 2  35 

I2

100 Ω

(1) (2)

Chapter 19, Solution 14. To find Z Th , consider the circuit in Fig. (a). I2

I1 + ZS

+ −

V1

Vo = 1

− (a) V1 = z 11 I 1 + z 12 I 2 V2 = z 21 I 1 + z 22 I 2

(1) (2)

But V1 = - Z s I 1

V2 = 1 ,

0 = (z 11 + Z s ) I 1 + z 12 I 2

Hence,

 → I 1 =

- z 12 I z 11 + Z s 2

 - z 21 z 12  1= + z 22  I 2  z 11 + Z s  Z Th =

V2 z z 1 = = z 22 − 21 12 z 11 + Z s I2 I2

To find VTh , consider the circuit in Fig. (b). ZS

VS

+ −

I1

I2 = 0

+

+

V1

V2 = VTh

−

− (b)

I2 = 0 ,

V1 = Vs − I 1 Z s

Substituting these into (1) and (2), Vs − I 1 Z s = z 11 I 1

V2 = z 21 I 1 =

VTh = V2 =

 → I 1 =

Vs z 11 + Z s

z 21 Vs z 11 + Z s

z 21 Vs z 11 + Z s

Chapter 19, Solution 15. (a) From Prob. 18.12, ZTh = z 22 −

z12z 21 80x 60 = 120 − = 24 z11 + Zs 40 + 10 ZL = ZTh = 24Ω

(b) VTh =

z 21 80 Vs = (120) = 192 z11 + Zs 40 + 10 Pmax =

V 2Th 192 2 = = 192 W 8R Th 8x 24

Chapter 19, Solution 16. As a reciprocal two-port, the given circuit can be represented as shown in Fig. (a).

5Ω

15∠0° V

10 – j6 Ω

+ −

4 – j6 Ω

a

j6 Ω

(a)

j4 Ω b

At terminals a-b, Z Th = (4 − j6) + j6 || (5 + 10 − j6)

Z Th = 4 − j6 +

j6 (15 − j6) = 4 − j6 + 2.4 + j6 15

Z Th = 6.4 Ω

VTh =

j6 (15∠0°) = j6 = 6∠90° V j6 + 5 + 10 − j6

The Thevenin equivalent circuit is shown in Fig. (b).

6.4 Ω + 6∠90° V

+ −

Vo

j4 Ω

− (b) From this,

Vo =

j4 ( j6) = 3.18∠148° 6.4 + j4

v o ( t ) = 3.18 cos( 2t + 148°) V

Chapter 19, Solution 17. To obtain z 11 and z 21 , consider the circuit in Fig. (a).

4Ω + I1

V1

Io '

2Ω

Io

I2 = 0 + V2

8Ω

−

− 6Ω (a)

In this case, the 4-Ω and 8-Ω resistors are in series, since the same current, I o , passes through them. Similarly, the 2-Ω and 6-Ω resistors are in series, since the same current, I o ' , passes through them. V1 (12)(8) = (4 + 8) || (2 + 6) = 12 || 8 = = 4 .8 Ω I1 20

z 11 =

Io = But

8 2 I1 = I1 8 + 12 5

Io' =

3 I 5 1

- V2 − 4 I o + 2 I o ' = 0 V2 = -4 I o + 2 I o ' = z 21 =

-8 6 -2 I1 + I1 = I 5 5 5 1

V2 - 2 = = -0.4 Ω I1 5

To get z 22 and z 12 , consider the circuit in Fig. (b). 4Ω

I1 = 0 + V1

2Ω

+ V2

8Ω

−

− 6Ω (b)

z 22 =

V2 (6)(14) = (4 + 2) || (8 + 6) = 6 || 14 = = 4 .2 Ω I2 20

z12 = z 21 = -0.4 Ω

Thus,  4.8 - 0.4  [z ] =  Ω  - 0.4 4.2 

We may take advantage of Table 18.1 to get [y] from [z]. ∆ z = (4.8)(4.2) − (0.4) 2 = 20

I2

y 11 =

z 22 4.2 = = 0.21 20 ∆z

y 12 =

- z 12 0.4 = = 0.02 20 ∆z

y 21 =

- z 21 0.4 = = 0.02 20 ∆z

y 22 =

z 11 4.8 = = 0.24 20 ∆z

Thus,  0.21 0.02  [y ] =  S  0.02 0.24 

Chapter 19, Solution 18.

To get y 11 and y 21 , consider the circuit in Fig.(a). 6Ω

I1

3Ω

I2 +

+ −

V1

6Ω

3Ω

V2 = 0 −

(a) V1 = (6 + 6 || 3) I 1 = 8 I 1 I1 1 = V1 8

y 11 =

I2 =

-6 - 2 V1 - V1 I1 = = 6+3 3 8 12

y 21 =

I 2 -1 = V1 12

To get y 22 and y 12 , consider the circuit in Fig.(b). I1

6Ω

Io

3Ω

I2

+ V1 = 0

6Ω

3Ω

− (b)

+ −

V2

I2 1 1 1 = = = V2 3 || (3 + 6 || 6) 3 || 6 2

y 22 =

I1 =

- Io , 2

I1 =

- I 2  - 1  1  - V2 =   V2  =  6  2  12 6

y 12 =

Io =

3 1 I2 = I2 3+ 6 3

I1 -1 = = y 21 V2 12

Thus,  1 -1   8 12  [y ] =  S  -1 1   12 2 

Chapter 19, Solution 19.

Consider the circuit in Fig.(a) for calculating y 11 and y 21 . 1

I1

I2 +

V1

+ −

1/s

s 1 (a)

2s 1  2 V1 =  || 2  I 1 = I1 = I s  2 + (1 s) 2s + 1 1 y 11 =

I2 =

I 1 2s + 1 = = s + 0 .5 V1 2 - I1 - V1 (- 1 s ) I1 = = (1 s) + 2 2s + 1 2

V2 = 0

−

y 21 =

I2 = -0.5 V1

To get y 22 and y 12 , refer to the circuit in Fig.(b). 1

I1

I2

+ V1 = 0

1/s

s 1

−

(b) V2 = (s || 2) I 2 = y 22 =

I1 =

2s I s+2 2

I2 s + 2 1 = = 0 .5 + V2 s 2s

- V2 -s -s s+ 2 I2 = ⋅ V2 = s+2 s + 2 2s 2

y 12 =

I1 = -0.5 V2

Thus,

 s + 0.5 - 0.5  [y ] =  S  - 0.5 0.5 + 1 s 

Chapter 19, Solution 20.

To get y11 and y21, consider the circuit below.

+ −

V2

3ix

2Ω

I1

I2

+ V1

I1

+

ix 4Ω

6Ω

V2 =0 -

-

Since 6-ohm resistor is short-circuited, ix = 0 V1 = I1(4 // 2) =

I2 = −

8 I1 6

→

I y11 = 1 = 0.75 V1

4 2 6 1 I1 = − ( V1) = − V1 4+2 3 8 2

→

I y 21 = 2 = −0.5 V1

To get y22 and y12, consider the circuit below. 3ix

2Ω

I1 + V1=0

ix 4Ω -

+

6 Ω V2 -

I2

ix =

V2 , 6

V V I 2 = i x − 3i x + 2 = 2 2 6

V I1 = 3i x − 2 = 0 2

I 1 y 22 = 2 = = 0.1667 V2 6

→

I y12 = 1 = 0 V2

 →

Thus, 0   0.75 [ y] =  S − 0.5 0.1667 

Chapter 19, Solution 21.

To get y 11 and y 21 , refer to Fig. (a). I1

0.2 V1

V1

I2 +

+ −

V1

5Ω

10 Ω

V2 = 0

− (a)

At node 1, I1 =

V1 + 0.2 V1 = 0.4 V1 5

I 2 = -0.2 V1

 → y 11 =

I1 = 0 .4 V1

I2 = -0.2 V1

 → y 21 =

To get y 22 and y 12 , refer to the circuit in Fig. (b). I1

0.2 V1

V1

I2

+ V1 = 0

5Ω

10 Ω

+ −

V2

− (b)

Since V1 = 0 , the dependent current source can be replaced with an open circuit.

V2 = 10 I 2

y 12 =

 → y 22 =

I2 1 = = 0 .1 V2 10

I1 =0 V2

Thus,  0.4 0 [y ] =  S  - 0.2 0.1 Consequently, the y parameter equivalent circuit is shown in Fig. (c). I1

I2

+

+ 0.2 V1

0.4 S

V1

0.1 S

V2

−

− (c)

Chapter 19, Solution 22.

(a)

To get y 11 and y 21 refer to the circuit in Fig. (a). I1

2Ω

V1

Vo

3Ω

+

+ V1

+ −

Vx

I2

1Ω

Vx/2

−

V2 = 0

− (a)

At node 1, I1 =

V1 V1 − Vo + 1 2

At node 2, V1 − Vo V1 Vo + = 2 2 3

 → I 1 = 1.5 V1 − 0.5 Vo  → 1.2 V1 = Vo

Substituting (2) into (1) gives,

(1)

(2)

I 1 = 1.5 V1 − 0.6 V1 = 0.9 V1

I2 =

- Vo = -0.4 V1 3

 → y 11 =

 → y 21 =

I1 = 0 .9 V1

I2 = -0.4 V1

To get y 22 and y 12 refer to the circuit in Fig. (b). I1

2Ω

V1

+

+

V1 = 0

Vx

−

−

3Ω

1Ω

Vx/2

I2

+ −

V2

(b) Vx = V1 = 0 so that the dependent current source can be replaced by an open circuit. I2 1 V2 = (3 + 2 + 0) I 2 = 5 I 2  → y 22 = = = 0 .2 V2 5 I1 I 1 = - I 2 = -0.2 V2  → y 12 = = -0.2 V2 Thus,  0.9 - 0.2  [y ] =  S  - 0.4 0.2 

(b)

To get y 11 and y 21 refer to Fig. (c). jΩ

Io ' I1

Io

1Ω

1Ω

Io''

I2 +

V1

+ −

-j Ω

V2 = 0 −

Zin

(c)

 -j   = j || (1.5 − j0.5) Z in = j || (1 + 1 || -j) = j || 1 +  1− j

=

j (1.5 − j0.5) = 0.6 + j0.8 1.5 + j0.5

V1 = Z in I 1

Io =

 → y 11 =

I1 1 1 = = = 0.6 − j0.8 V1 Z in 0.6 + j0.8

j I , 1.5 + j0.5 1

I o '' =

Io' =

1.5 − j0.5 I 1.5 + j0.5 1

I1 I -j Io = = 1 1− j (1 − j)(1.5 + j0.5) 2 − j

- I2 = Io + Io '

''

(1.5 − j0.5) 2 2+ j = I1 + I = (1.2 − j0.4) I 1 2 .5 5 1

- I 2 = (1.2 − j0.4)(0.6 − j0.8) V1 = (0.4 − j1.2) V1 y 21 =

I2 = -0.4 + j1.2 = y 12 V1

To get y 22 refer to the circuit in Fig.(d). jΩ I1

1Ω

1Ω

I2

+ + −

-j Ω

V1 = 0 −

(d) Z out = j || (1 + 1 || - j) = 0.6 + j0.8 y 22 =

1 = 0.6 − j0.8 Z out

Thus,  0.6 − j0.8 - 0.4 + j1.2  [y ] =  S  - 0.4 + j1.2 0.6 − j0.8 

Zout

V2

Chapter 19, Solution 23. (a)

1 1 − y12 = 1 // = s s +1

y11 + y12 = 1

 →

 →

y 22 + y12 = s

y12 = −

1 s +1

y11 = 1 − y12 = 1 +

 →

1 s+2 = s +1 s +1

y 22 = s − y12 = s +

s + 2  s +1 [ y] =   −1  s + 1

1 s2 + s +1 = s +1 s +1

−1  s +1   s 2 + s + 1 s + 1 

(b) Consider the network below. I1

1 +

+

Vs -

V1 -

I2 + [y]

V2 -

2

Vs = I1 + V1

(1)

V2 = −2I 2

(2)

I1 = y11V1 + y12 V2

(3)

I 2 = y 21V1 + y 22 V2

(4)

From (1) and (3)

Vs − V1 = y11V1 + y12 V2

 →

Vs = (1 + y11 )V1 + y12 V2

(5)

From (2) and (4), − 0.5V2 = y 21V1 + y 22 V2

→

V1 = −

1 (0.5 + y 22 )V2 y 21

(6)

Substituting (6) into (5), Vs = − =

2 s

(1 + y11)(0.5 + y 22 ) V2 + y12V2 y 21  →

V2 =

2/s   1 (1 + y11)(0.5 + y 22 )  y12 − y 21  

2/s

V2 = −

2 1  2s + 3  1 s + s + 1  + (s + 1)  + s +1 s + 1   s + 1  2

=

2(s + 1) s(2s3 + 6s 2 + 7.5s + 3.5)

Chapter 19, Solution 24.

Since this is a reciprocal network, a Π network is appropriate, as shown below. Y2

Y1

Y3 (a) 4Ω

1/4 S

1/4 S

1/8 S (b)

4Ω

8Ω (c)

Y1 = y 11 + y 12 =

Y2 = - y 12 =

1 1 1 − = S, 2 4 4

Z1 = 4 Ω

1 S, 4

Y3 = y 22 + y 21 =

Z2 = 4 Ω

3 1 1 − = S, 8 4 8

Z3 = 8 Ω

Chapter 19, Solution 25. This is a reciprocal network and is shown below. 0.5 S

0.5S

1S

Chapter 19, Solution 26. To get y 11 and y 21 , consider the circuit in Fig. (a).

4Ω 2Ω

V1

+ −

1

2

+ Vx

1Ω

2 Vx

I2 + V2 = 0 −

−

(a)

At node 1, V1 − Vx V V + 2 Vx = x + x 2 1 4

 → 2 V1 = -Vx

But

I1 =

V1 − Vx V1 + 2 V1 = = 1.5 V1 2 2

Also,

I2 +

Vx = 2 Vx 4

y 21 =

(1)

 → y 11 =

I1 = 1 .5 V1

 → I 2 = 1.75 Vx = -3.5 V1

I2 = -3.5 V1

To get y 22 and y 12 , consider the circuit in Fig.(b).

4Ω 2Ω

1

2

+

I1

1Ω

Vx

2 Vx

I2

+ −

V2

−

(b) At node 2, I 2 = 2 Vx +

V2 − Vx 4

(2)

At node 1, 2 Vx +

V2 − Vx Vx Vx 3 = + = Vx 4 2 1 2

 → V2 = -Vx

Substituting (3) into (2) gives 1 I 2 = 2 Vx − Vx = 1.5 Vx = -1.5 V2 2 y 22 =

I1 =

I2 = -1.5 V2

- Vx V2 = 2 2

 → y 12 =

I1 = 0 .5 V2

(3)

Thus,

 1.5 0.5  [y ] =  S  - 3.5 - 1.5 

Chapter 19, Solution 27.

Consider the circuit in Fig. (a). 4Ω

I1

I2 +

+ −

V1

0.1 V2

− +

10 Ω

20 I1

V2 = 0 −

(a) V1 = 4 I 1

 → y 11 =

I 2 = 20 I 1 = 5 V1

I1 I1 = = 0.25 V1 4 I 1

 → y 21 =

I2 =5 V1

Consider the circuit in Fig. (b). I1

4Ω

I2

+ V1 = 0

0.1 V2

− +

10 Ω

20 I1

+ −

V2

− (b) 4 I 1 = 0.1 V2

I 2 = 20 I 1 +

 → y 12 =

I 1 0 .1 = = 0.025 4 V2

V2 = 0.5 V2 + 0.1 V2 = 0.6 V2 10

 → y 22 =

I2 = 0 .6 V2

Thus,  0.25 0.025  [y ] =  S 0.6   5

Alternatively, from the given circuit, V1 = 4 I 1 − 0.1 V2 I 2 = 20 I 1 + 0.1 V2 Comparing these with the equations for the h parameters show that h 11 = 4 , h 12 = -0.1, h 21 = 20 , h 22 = 0.1 Using Table 18.1,

y 11 =

1 1 = = 0.25 , h11 4

y 12 =

- h 12 0.1 = = 0.025 4 h 11

y 21 =

h 21 20 = = 5, 4 h 11

y 22 =

∆ h 0 .4 + 2 = = 0 .6 4 h 11

as above. Chapter 19, Solution 28. We obtain y 11 and y 21 by considering the circuit in Fig.(a). 1Ω

4Ω

I2

+ I1

V1

+ 6Ω

(a) Z in = 1 + 6 || 4 = 3.4

I2 =

V2 = 0 −

−

y 11 =

2Ω

I1 1 = = 0.2941 V1 Z in

 - 6  V1  - 6 -6 I 1 =    = V  10  3.4  34 1 10

y 21 =

I2 - 6 = = -0.1765 V1 34

To get y 22 and y 12 , consider the circuit in Fig. (b). 1Ω

I1

4Ω

Io

+

+ 6Ω

V1 = 0

2Ω

−

V2

I2

− (b) 1 6  (2)(34 7) 34 V2  = 2 || (4 + 6 || 1) = 2 ||  4 +  = = = y 22 7  2 + (34 7) 24 I 2 

y 22 =

24 = 0.7059 34

I1 =

-6 I 7 o

Io =

I1 =

-6 V 34 2

 → y 12 =

2 14 7 I2 = I2 = V 2 + (34 7) 48 34 2 I1 - 6 = = -0.1765 V2 34

Thus,  0.2941 - 0.1765 [y ] =  S  - 0.1765 0.7059  The equivalent circuit is shown in Fig. (c). After transforming the current source to a voltage source, we have the circuit in Fig. (d).

6/34 S

1A

4/34 S

(c)

18/34 S

2Ω

8.5 Ω

5.667 Ω +

8.5 V

+ −

1.889 Ω

V

2Ω

− (d) V=

(2 || 1.889)(8.5) (0.9714)(8.5) = = 0.5454 2 || 1.889 + 8.5 + 5.667 0.9714 + 14.167

P=

V 2 (0.5454) 2 = = 0.1487 W R 2

Chapter 19, Solution 29. (a)

Transforming the ∆ subnetwork to Y gives the circuit in Fig. (a).

1Ω

1Ω

Vo

+ 10 A

+ 2Ω

V1

−

-4 A

V2

− (a)

It is easy to get the z parameters z 12 = z 21 = 2 , z 11 = 1 + 2 = 3 ,

z 22 = 3

∆ z = z 11 z 22 − z 12 z 21 = 9 − 4 = 5 y 11 =

z 22 3 = = y 22 , ∆z 5

y 12 = y 21 =

- z 12 - 2 = ∆z 5

Thus, the equivalent circuit is as shown in Fig. (b).

2/5 S

I1

I2 +

+ 10 A

V1

1/5 S

1/5 S

−

V2

− (b)

-4 A

I 1 = 10 =

3 2 V1 − V2 5 5

I 2 = -4 =

-2 3 V1 + V2 5 5

10 = V1 − 1.5 V2

 → 50 = 3 V1 − 2 V2  → - 20 = -2 V1 + 3 V2

 → V1 = 10 + 1.5 V2

Substituting (2) into (1), 50 = 30 + 4.5 V2 − 2 V2

(1)

(2)

 → V2 = 8 V

V1 = 10 + 1.5 V2 = 22 V

(b)

For direct circuit analysis, consider the circuit in Fig. (a). For the main non-reference node, Vo 10 − 4 =  → Vo = 12 2 10 =

V1 − Vo 1

 → V1 = 10 + Vo = 22 V

-4=

V2 − Vo 1

 → V2 = Vo − 4 = 8 V

Chapter 19, Solution 30. (a)

Convert to z parameters; then, convert to h parameters using Table 18.1. z 11 = z 12 = z 21 = 60 Ω , z 22 = 100 Ω ∆ z = z 11 z 22 − z 12 z 21 = 6000 − 3600 = 2400 h 11 =

∆ z 2400 = = 24 , 100 z 22

h 12 =

z 12 60 = = 0 .6 z 22 100

h 21 =

- z 21 = -0.6 , z 22

h 22 =

1 = 0.01 z 22

Thus,  24 Ω 0.6  [h] =    - 0.6 0.01 S  (b)

Similarly, z 11 = 30 Ω

z 12 = z 21 = z 22 = 20 Ω

∆ z = 600 − 400 = 200

h11 =

200 = 10 20

h 21 = -1

h12 =

20 =1 20

h 22 =

1 = 0.05 20

Thus,  10 Ω 1  [h] =    - 1 0.05 S 

Chapter 19, Solution 31. We get h11 and h 21 by considering the circuit in Fig. (a). 1Ω

2Ω

V3

V4

1Ω

I2

+ I1

2Ω

V1

4 I1

− (a) At node 1, I1 =

V3 V3 − V4 + 2 2

 → 2 I 1 = 2 V3 − V4

(1)

 → 16 I 1 = -2 V3 + 6 V4

(2)

At node 2, V3 − V4 V + 4 I1 = 4 2 1

8 I 1 = -V3 + 3 V4

Adding (1) and (2), 18 I 1 = 5 V4  → V4 = 3.6 I 1 V3 = 3 V4 − 8 I 1 = 2.8 I 1 V1 = V3 + I 1 = 3.8 I 1 h11 =

V1 = 3 .8 Ω I1

I2 =

- V4 = -3.6 I 1 1

 → h 21 =

I2 = -3.6 I1

To get h 22 and h12 , refer to the circuit in Fig. (b). The dependent current source can be replaced by an open circuit since 4 I 1 = 0 .

1Ω

I1

1Ω

2Ω

I2

+ 2Ω

V1

+ −

4 I1 = 0

V2

− (b) V1 =

2 2 V2 = V2 2 + 2 +1 5

I2 =

V2 V2 = 2 + 2 +1 5

 → h12 =

 → h 22 =

V1 = 0 .4 V2

I2 1 = = 0 .2 S V2 5

Thus,

 38 Ω 0.4  [h] =    - 3.6 0.2 S 

Chapter 19, Solution 32. (a)

We obtain h11 and h 21 by referring to the circuit in Fig. (a).

1

s

s

+ I1

I2 +

1/s

V1

V2 = 0 −

− (a)

  s  1 I V1 = 1 + s + s ||  I 1 = 1 + s + 2   s + 1 1 s h11 =

V1 s = s +1+ 2 I1 s +1

By current division, - I1 I2 -1 s -1 I2 = I1 =  → h 21 = = 2 s +1 s s +1 I1 s + 1 To get h 22 and h12 , refer to Fig. (b).

I1 = 0

1

s

s

I2

+ + −

1/s

V1

V2

− (b) V1 =

V2 V1 1s 1 V2 = 2  → h12 = = 2 s +1 s s +1 V2 s + 1

 1 V2 = s +  I 2  s

 → h 22 =

I2 1 s = = 2 V2 s + 1 s s + 1

Thus,  s  s + 1 + s2 + 1 [h] =  -1  2 s +1

(b)

1  s +1  s   s2 + 1  2

To get g11 and g 21 , refer to Fig. (c).

I1

1

s

s

I2 = 0 +

V1

+ −

1/s

V2 −

(c)

 1 V1 = 1 + s +  I 1  s V2 =

 → g 11 =

I1 1 s = = 2 V1 1 + s + 1 s s + s + 1

V1 V2 1s 1 V1 = 2  → g 21 = = 2 1+ s +1 s s + s +1 V1 s + s + 1

To get g 22 and g 12 , refer to Fig. (d).

1

I1

s

s

+

I2 +

1/s

V1 = 0 −

V2

I2

− (d)

 (s + 1) s   1  I2 V2 = s + || (s + 1)  I 2 = s +  s   1+ s +1 s  g 22 =

I1 =

V2 s +1 =s+ 2 s + s +1 I2

- I2 I1 -1 s -1 I2 = 2  → g 12 = = 2 1+ s +1 s s + s +1 I2 s + s +1

Thus,   2 [g ] =  s  2 s

 s -1 2 +s+1 s +s+1  s+1  1 s+ 2  +s+1 s +s+1 

Chapter 19, Solution 33.

To get h11 and h21, consider the circuit below.

4Ω

j6 Ω

+ I1

V1 = 5 //( 4 + j6)I1 =

Also, I 2 = −

-j3 Ω

5Ω

V1 -

5 I1 9 + j6

I2

5(4 + j6)I1 9 + j6

+ V2=0 -

V h11 = 1 = 3.0769 + j1.2821 I1

I h 21 = 2 = −0.3846 + j0.2564 I1

 →

To get h22 and h12, consider the circuit below. 4Ω

j6 Ω

I2

I1 +

V1 =

-j3 Ω

5Ω

V1

-

5 V2 9 + j6

+

-

 →

V2 = − j3 //(9 + j6)I 2

h12 =

 →

+ V2

V1 5 = = 0.3846 − j0.2564 V2 9 + j6 I 1 9 + j3 h 22 = 2 = = V2 − j3 //(9 + j6) − j3(9 + j6) = 0.0769 + j0.2821

Thus,  3.0769 + j1.2821 0.3846 − j0.2564 [h ] =   − 0.3846 + j0.2564 0.0769 + j0.2821

Chapter 19, Solution 34. Refer to Fig. (a) to get h11 and h 21 .

300 Ω

10 Ω

50 Ω

2

1 + I1

V1 −

− +

100 Ω

Vx

I2

+

+

10 Vx

V2 = 0 −

−

(a) At node 1, I1 =

Vx Vx − 0 + 100 300

 → 300 I 1 = 4 Vx

300 I = 75 I 1 4 1

Vx =

V1 = 10 I 1 + Vx = 85 I 1

But

(1)

 → h11 =

V1 = 85 Ω I1

At node 2, I2 =

0 + 10 Vx Vx Vx Vx 75 75 − = − = I1 − I = 14.75 I 1 50 300 5 300 5 300 1

h 21 =

I2 = 14.75 I1

To get h 22 and h 12 , refer to Fig. (b). 300 Ω I1 = 0 10 Ω

50 Ω 1

+ V1 −

+ Vx

− +

100 Ω

−

(b)

10 Vx

2

I2

+ −

V2

At node 2, I2 =

V2 V2 + 10 Vx + 400 50

But

Vx =

V2 100 V2 = 400 4

Hence,

400 I 2 = 9 V2 + 20 V2 = 29 V2 h 22 =

 → 400 I 2 = 9 V2 + 80 Vx

I2 29 = = 0.0725 S V2 400

V1 = Vx =

V2 4

 → h 12 =

V1 1 = = 0.25 V2 4

 85 Ω 0.25  [h] =    14.75 0.0725 S 

To get g 11 and g 21 , refer to Fig. (c). 300 Ω

I1

10 Ω

50 Ω 1

V1

Vx

I2 = 0 +

+

+ −

2

100 Ω

− +

10 Vx

V2 −

− (c)

At node 1, I1 =

Vx Vx + 10 Vx + 100 350

But

I1 =

V1 − Vx 10

or

Vx = V1 − 10 I 1

 → 350 I 1 = 14.5 Vx

(2)

 → 10 I 1 = V1 − Vx

(3)

Substituting (3) into (2) gives 350 I 1 = 14.5 V1 − 145 I 1 g 11 =

 → 495 I 1 = 14.5 V1

I 1 14.5 = = 0.02929 S V1 495

At node 2,  11  V2 = (50)  Vx  − 10 Vx = -8.4286 Vx  350 

 14.5  = -8.4286 V1 + 84.286 I 1 = -8.4286 V1 + (84.286)   V1  495  V2 = -5.96 V1

 → g 21 =

V2 = -5.96 V1

To get g 22 and g 12 , refer to Fig. (d). 300 Ω

I1

Io

10 Ω

+ V1 = 0

+

+ Vx

−

Io

50 Ω

− +

100 Ω

10 Vx

I2

V2 −

− (d)

10 || 100 = 9.091 I2 =

But

V2 + 10 Vx V2 + 50 300 + 9.091

309.091 I 2 = 7.1818 V2 + 61.818 Vx

(4)

9.091 V = 0.02941 V2 309.091 2

(5)

Vx =

Substituting (5) into (4) gives 309.091 I 2 = 9 V2

g 22 =

V2 = 34.34 Ω I2

Io =

34.34 I 2 V2 = 309.091 309.091

I1 =

- 34.34 I 2 - 100 Io = 110 (1.1)(309.091)

g 12 =

I1 = -0.101 I2

Thus,  0.02929 S - 0.101  [g ] =  34.34 Ω   - 5.96

Chapter 19, Solution 35.

To get h11 and h 21 consider the circuit in Fig. (a). 1Ω

I1

1:2

4Ω

+

+

V1

V2 = 0

−

− (a)

ZR =

I2

4 4 =1 2 = n 4

V1 = (1 + 1) I 1 = 2 I 1

 → h 11 =

V1 = 2Ω I1

I1 - N 2 I 2 -1 = = -2  → h 21 = = = -0.5 I2 N1 I1 2

To get h 22 and h 12 , refer to Fig. (b). 1Ω

I1 = 0

4Ω

1:2

I2

+

− +

V1

V2

− (b)

Since I 1 = 0 , I 2 = 0 . Hence, h 22 = 0 . At the terminals of the transformer, we have V1 and V2 which are related as V2 N 2 V1 1 = =n=2  → h12 = = = 0 .5 V1 N 1 V2 2

Thus,  2 Ω 0.5  [h] =    - 0.5 0 

Chapter 19, Solution 36.

We replace the two-port by its equivalent circuit as shown below. 4Ω

I1

16 Ω

2 I1

+ 10 V

+ −

V1

I2 +

3 V2

+ −

−

-2 I1

100 Ω V2 −

100 || 25 = 20 Ω V2 = (20)(2 I 1 ) = 40 I 1 - 10 + 20 I 1 + 3 V2 = 0 10 = 20 I 1 + (3)(40 I 1 ) = 140 I 1

(1)

25 Ω

I1 =

1 , 14

V2 =

V1 = 16 I 1 + 3 V2 =

40 14

136 14

-8  100  I2 =   (2 I 1 ) = 70  125  (a)

V2 40 = = 0.2941 V1 136

(b)

I2 = - 1.6 I1

(c)

I1 1 = = 7.353 × 10 -3 S V1 136

(d)

V2 40 = = 40 Ω 1 I1

Chapter 19, Solution 37. (a)

We first obtain the h parameters. To get h11 and h 21 refer to Fig. (a). 6Ω

3Ω

I2

+ I1

V1

+ 6Ω

3Ω

−

− (a)

3 || 6 = 2 V1 = (6 + 2) I 1 = 8 I 1

V2 = 0

 → h11 =

V1 =8Ω I1

I2 =

-6 -2 I1 = I 3+ 6 3 1

 → h 21 =

I2 - 2 = I1 3

To get h 22 and h12 , refer to the circuit in Fig. (b). 6Ω

I1 = 0

3Ω

I2

+ 6Ω

V1

+ −

3Ω

V2

− (b) 3 || 9 =

9 4

V2 =

9 I 4 2

V1 =

6 2 V2 = V2 6+3 3

 → h 22 =

I2 4 = V2 9

 → h12 =

V1 2 = V2 3

 2  8 Ω 3  [h] =  - 2 4  S  3 9 

The equivalent circuit of the given circuit is shown in Fig. (c). I1

8Ω

I2 +

10 V

+ −

2/3 V2

+ −

-2/3 I1

9/4 Ω V2 −

(c) 2 8 I 1 + V2 = 10 3

(1)

5Ω

V2 =

I1 =

2  9  2  45  30 I 5 ||  = I   = I 3 1  4  3 1  29  29 1 29 V 30 2

(2)

Substituting (2) into (1),  29  2 (8)   V2 + V2 = 10  30  3 V2 = (b)

300 = 1.19 V 252

By direct analysis, refer to Fig.(d). 6Ω

3Ω +

10 V

+ −

6Ω

3Ω

V2

5Ω

− (d) 10 -A current source. Since 6 6 || 6 = 3 Ω , we combine the two 6-Ω resistors in parallel and transform 10 × 3 = 5 V voltage source shown in Fig. (e). the current source back to 6 Transform the 10-V voltage source to a

3Ω

3Ω +

5V

+ −

V2 − (e)

3 || 5 =

(3)(5) 15 = 8 8

3 || 5 Ω

V2 =

15 8 75 (5) = = 1.19 V 6 + 15 8 63

Chapter 19, Solution 38. We replace the two-port by its equivalent circuit as shown below. 200 Ω

I1 800 Ω

I2

+ 10 V

+ −

V1

+ 10-4 V2

+ −

50 I1

200 kΩ V2

−

Z in =

−

Vs , I1

200 || 50 = 40 kΩ

V2 = -50 I 1 (40 × 10 3 ) = (-2 × 10 6 ) I 1 For the left loop, Vs − 10 -4 V2 = I1 1000 Vs − 10 -4 (-2 × 10 6 I 1 ) = 1000 I 1 Vs = 1000 I 1 − 200 I 1 = 800I 1 Z in =

Vs = 800 Ω I1

Alternatively, Z in = Z s + h11 −

h12 h 21 Z L 1 + h 22 Z L

(10 -4 )(50)(50 × 10 3 ) Z in = 200 + 800 − = 800 Ω 1 + (0.5 × 10 -5 )(50 × 10 3 )

50 kΩ

Chapter 19, Solution 39. To get g11 and g21, consider the circuit below which is partly obtained by converting the delta to wye subnetwork. I1

R1

R2

I2

+ +

R3 V2

V1

10 Ω -

8x8 = 3.2 20

R1 =

4 x8 = 1.6 = R 2 , 8+8+ 4

V2 =

13.2 V1 = 0.8919V1 13.2 + 1.6

→

V1 = I1(1.6 + 3.2 + 10) = 14.8I1

 →

R3 =

g 21 =

V2 = 0.8919 V1

I 1 = 0.06757 g11 = 1 = V1 14.8

To get g22 and g12, consider the circuit below. 1.6 Ω

1.6 Ω

I1 + V1=0

V2 13.2 Ω -

I2

I1 = −

13.2 I2 = −0.8919I 2 13.2 + 1.6

V2 = I 2 (1.6 + 13.2 // 1.6) = 3.027I 2

I g12 = 1 = −0.8919 I2

 →

 →

g 22 =

V2 = 3.027 I2

0.06757 − 0.8919 [g ] =  3.027   0.8919

Chapter 19, Solution 40. To get g 11 and g 21 , consider the circuit in Fig. (a). -j6 Ω

I1

j10 Ω

I2 = 0 +

+ −

V1

12 Ω

V2 −

(a) V1 = (12 − j6) I 1

g 21 =

 → g 11 =

I1 1 = = 0.0667 + j0.0333 S V1 12 − j6

12 I 1 V2 2 = = = 0.8 + j0.4 V1 (12 − j6) I 1 2 − j

To get g 12 and g 22 , consider the circuit in Fig. (b). I1

-j6 Ω

j10 Ω

I2

+ V1 = 0

12 Ω

−

(b)

I2

I1 =

- 12 I 12 - j6 2

 → g 12 =

I1 - 12 = = - g 21 = -0.8 − j0.4 I 2 12 - j6

V2 = ( j10 + 12 || -j6) I 2 g 22 =

V2 (12)(-j6) = j10 + = 2.4 + j5.2 Ω 12 - j6 I2

 0.0667 + j0.0333 S - 0.8 − j0.4  [g ] =  0.8 + j0.4 2.4 + j5.2 Ω  

Chapter 19, Solution 41. For the g parameters I 1 = g 11 V1 + g 12 I 2 V2 = g 21 V1 + g 22 I 2 V1 = Vs − I 1 Z s and But V2 = - I 2 Z L = g 21 V1 + g 22 I 2 0 = g 21 V1 + (g 22 + Z L ) I 2 or

V1 =

- (g 22 + Z L ) I2 g 21

Substituting this into (1), (g 22 g 11 + Z L g 11 − g 21 g 12 ) I1 = I2 - g 21 or

I2 - g 21 = I 1 g 11 Z L + ∆ g

Also,

V2 = g 21 (Vs − I 1 Z s ) + g 22 I 2 = g 21 Vs − g 21 Z s I 1 + g 22 I 2 = g 21 Vs + Z s (g 11 Z L + ∆ g ) I 2 + g 22 I 2

But

I2 =

- V2 ZL

(1) (2)

 V2  V2 = g 21 Vs − [ g 11 Z s Z L + ∆ g Z s + g 22 ]   ZL 

V2 [ Z L + g 11 Z s Z L + ∆ g Z s + g 22 ] ZL

= g 21 Vs

V2 g 21 Z L = Vs Z L + g 11 Z s Z L + ∆ g Z s + g 22 V2 g 21 Z L = Vs Z L + g 11 Z s Z L + g 11 g 22 Z s − g 21 g 12 Z s + g 22 V2 g 21 Z L = Vs (1 + g 11 Z s )(g 22 + Z L ) − g 12 g 21 Z s

Chapter 19, Solution 42. (a)

The network is shown in Fig. (a). 20 Ω

I1

I2

+

+ 100 Ω

V1

+ −

-0.5 I2

0.5 I1

V2

−

−

(a) (b)

The network is shown in Fig. (b). 2Ω

I1 + V1

s

I2 +

10 Ω

+ −

−

12 V1

V2 −

(b)

Chapter 19, Solution 43. (a)

To find A and C , consider the network in Fig. (a).

Z

I1

I2 +

V1

+ −

V2 −

(a) V1 = V2

 → A =

I1 = 0  → C =

V1 =1 V2

I1 =0 V2

To get B and D , consider the circuit in Fig. (b).

Z

I1

I2 +

V1

+ −

V2 = 0 −

(b) V1 = Z I 1 , B=

- V1 - Z I 1 = =Z I2 - I1

D=

- I1 =1 I2

Hence,

1 Z [T] =   0 1 

I 2 = - I1

(b)

To find A and C , consider the circuit in Fig. (c).

I1

I2 +

V1

+ −

Z

V2

− (c) V1 = V2

 → A =

V1 = Z I 1 = V2

V1 =1 V2

 → C =

I1 1 = =Y V2 Z

To get B and D , refer to the circuit in Fig.(d).

I2 + I1

+ Y

V1

−

− (d)

V1 = V2 = 0 B=

- V1 = 0, I2

V2 = 0

I 2 = - I1 D=

- I1 =1 I2

Thus,

 1 0 [T] =   Y 1

Chapter 19, Solution 44. To determine A and C , consider the circuit in Fig.(a).

j15 Ω

Io -j10 Ω

I1

-j20 Ω Io '

V1

+ −

I2 = 0 Io

20 Ω

+ V2 −

(a) V1 = [ 20 + (- j10) || ( j15 − j20) ] I 1

  (-j10)(-j5)  10  V1 =  20 + I 1 =  20 − j  I 1   - j15  3  '

I o = I1

 - j10  2  I 1 =   I 1 I o =  3  - j10 − j5 

V2 = (-j20) I o + 20 I o ' = − j

A=

C=

40  40  I1 + 20I1 =  20 − j I1 3  3 

V1 (20 − j10 3) I 1 = = 0.7692 + j0.3461 40  V2   20 − j I1 3   I1 = V2

1 40 20 − j 3

= 0.03461 + j0.023

To find B and D , consider the circuit in Fig. (b). j15 Ω

I1

-j10 Ω

-j20 Ω

I2 +

+ −

V1

20 Ω

V2 = 0 −

(b) We may transform the ∆ subnetwork to a T as shown in Fig. (c).

Z1 =

( j15)(-j10) = j10 j15 − j10 − j20

Z2 =

40 (-j10)(-j20) = -j 3 - j15

Z3 =

( j15)(-j20) = j20 - j15 I1

j10 Ω

j20 Ω

I2 +

V1

+ −

20 – j40/3 Ω

V2 = 0 −

(c) - I2 =

D=

20 − j40 3 3 − j2 I1 = I 20 − j40 3 + j20 3+ j 1

- I1 3+ j = 0.5385 + j0.6923 = 3 − j2 I2

 ( j20)(20 − j40 3)  V1 =  j10 + I 20 − j40 3 + j20  1 

V1 = [ j10 + 2 (9 + j7) ] I 1 = j I 1 (24 − j18)

B=

- V1 - j I 1 (24 − j18) 6 = = (-15 + j55) - (3 - j2) I2 13 I1 3+ j

B = -6.923 + j25.385 Ω  0.7692 + j0.3461 - 6.923 + j25.385 Ω  [T] =    0.03461 + j0.023 S 0.5385 + j0.6923 

Chapter 19, Solution 45. To obtain A and C, consider the circuit below. I1

sL

1/sC

I2 =0 +

+ V1

R1

V2

-

R2

V2 =

R1 V1 R1 + R 2 + sL

V2 = I1R1

→

→ I 1 C= 1 = V2 R1

To obtain B and D, consider the circuit below.

A=

V1 R1 + R 2 + sL = V2 R1

I1

sL

1/sC

I2 +

+ V1

R1

V2=0

-

R2

I2 = −

R1 R1 +

1 sC

I1 = −

sR1C I1 1 + sR1C

 →

I 1 + sR1C D=− 1 = I2 sR1C

R1     sC  I1 = − [(1 + sR1C)(R 2 + sL) + R1 ] (1 + sR1C) I 2 V1 = R 2 + sL + 1  1 + sR1C sR1C  R1 +   sC   V 1 [R1 + (1 + sR1C)(R 2 + sL)] B=− 1 = I 2 sR1C

Chapter 19, Solution 46. To get A and C , refer to the circuit in Fig.(a).

I1

V1

1Ω

+ −

1Ω

1 + Vo

I2 = 0

2

Ix 2Ω

+ 4 Ix

V2 −

−

(a) At node 1, I1 =

Vo Vo − V2 + 2 1

 → 2 I 1 = 3 Vo − 2 V2

(1)

At node 2,

Vo − V2 4 Vo = 4Ix = = 2 Vo 1 2

 → Vo = -V2

(2)

From (1) and (2), 2 I 1 = -5 V2

But

I1 =

 → C =

V1 − Vo = V1 + V2 1

- 2.5 V2 = V1 + V2 A=

I1 - 5 = = -2.5 S V2 2

 → V1 = -3.5 V2

V1 = -3.5 V2

To get B and D , consider the circuit in Fig. (b).

I1

V1

1Ω +

+ −

1Ω

1

Vo

I2

2

Ix 2Ω

+ 4 Ix

V2 = 0 −

− (b) At node 1, I1 =

Vo Vo + 2 1

I2 +

Vo + 4Ix = 0 1

 → 2 I 1 = 3 Vo

(3)

At node 2,

→ I 2 = -3 Vo – I 2 = Vo + 2 Vo = 0 

Adding (3) and (4), 2 I1 + I 2 = 0  → I 1 = -0.5 I 2 D=

- I1 = 0 .5 I2

(4)

(5)

But

I1 =

V1 − Vo 1

 → V1 = I 1 + Vo

(6)

Substituting (5) and (4) into (6), -1 -1 -5 V1 = I 2 + I 2 = I 2 3 6 2 B=

- V1 5 = = 0.8333 Ω I2 6

Thus,  - 3.5 0.8333 Ω  [T] =  - 0.5   - 2.5 S

Chapter 19, Solution 47.

To get A and C, consider the circuit below. 6Ω I1

1Ω

+ V1 -

V1 − Vx Vx Vx − 5Vx = + 1 2 10

4Ω + Vx

2Ω

-

 →

V2 = 4(−0.4Vx ) + 5Vx = 3.4Vx V − Vx I1 = 1 = 1.1Vx − Vx = 0.1Vx 1

I2=0

+ 5Vx

+ V2

-

-

V1 = 1.1Vx

 →

→

A=

V1 = 1.1 / 3.4 = 0.3235 V2

I C = 1 = 0.1 / 3.4 = 0.02941 V2

Chapter 19, Solution 48.

(a)

Refer to the circuit below. I2

I1

+ V1

+ −

[T]

V2

ZL

−

V1 = 4 V2 − 30 I 2 I 1 = 0.1 V2 − I 2

(1) (2)

When the output terminals are shorted, V2 = 0 . So, (1) and (2) become V1 = -30 I 2 and I1 = - I 2 Hence, V1 = 30 Ω Z in = I1 (b)

When the output terminals are open-circuited, I 2 = 0 . So, (1) and (2) become V1 = 4 V2 I 1 = 0.1 V2 or V2 = 10 I 1 V1 = 40 I 1 Z in =

(c)

V1 = 40 Ω I1

When the output port is terminated by a 10-Ω load, V2 = -10 I 2 . So, (1) and (2) become V1 = -40 I 2 − 30 I 2 = -70 I 2 I 1 = - I 2 − I 2 = -2 I 2 V1 = 35 I 1 Z in =

V1 = 35 Ω I1

Alternatively, we may use Z in =

A ZL + B CZL + D

Chapter 19, Solution 49.

To get A and C , refer to the circuit in Fig.(a). 1/s

I1

I2 = 0 +

V1

+ −

1Ω

1/s

1/s

1Ω

V2 −

(a)

1s 1 1 1 || = = s 1+1 s s +1 V2 =

1 || 1 s V 1 s + 1 || 1 s 1

1 V2 s s +1 A= = = 1 V1 1 2s + 1 + s s +1  1  1  1   2s + 1  1    ||  +  = I1  ||  V1 = I 1   s + 1  s s + 1  s + 1   s (s + 1) 

 1   2s + 1   ⋅  V1  s + 1   s (s + 1)  2s + 1 = = 1 2s + 1 (s + 1)(3s + 1) I1 + s + 1 s (s + 1) But

V1 = V2 ⋅

2s + 1 s

V2 2s + 1 2s + 1 ⋅ = s (s + 1)(3s + 1) I1

Hence,

C=

V2 (s + 1)(3s + 1) = I1 s

To get B and D , consider the circuit in Fig. (b). 1/s

I1

I2 +

V1

+ −

1Ω

1/s

1/s

1Ω

V2 = 0 −

(b) I  1 1  1 V1 = I 1 1 || ||  = I 1 1 ||  = 1  s s  2s  2s + 1 -1 I -s s +1 1 I2 = = I 1 1 2s + 1 1 + s +1 s D=

- I 1 2s + 1 1 = = 2+ I2 s s

I  1  2s + 1   I2 = 2 V1 =   2s + 1  - s  -s

Thus,  2  2s + 1 [T] =  (s + 1)( 3s + 1)  s

  1 2+  s 1 s

Chapter 19, Solution 50. To get a and c, consider the circuit below.

 → B =

- V1 1 = I2 s

I1=0

2

s

I2

+

+ 4/s

V1

V2

-

V1 =

-

4/s 4 V2 = V2 2 s + 4/s s +4

 →

a = V2

V1

= 1 + 0.25s 2

V2 = (s + 4 / s)I 2 or V2 (1 + 0.25s 2 )V1 I2 = = s + 4/s s + 4/s

 →

I 2 s + 0.25s3 c= = V1 s2 + 4

To get b and d, consider the circuit below. I1

2

s

I2

+

+

V1=0

4/s

V2

-

I1 =

− 4/s 2I I2 = − 2 2 + 4/s s+2

-

 →

I d = − 2 = 1 + 0.5s I1

4 (s 2 + 2s + 4) V2 = (s + 2 // )I2 = I2 s s+2 =−

(s 2 + 2s + 4)( s + 2) I1 s+2 2

 →

V b = − 2 = 0.5s 2 + s + 2 I1

 0.25s 2 + 1 0.5s 2 + s + 2   [ t ] =  0.25s 2 + s 0.5s + 1    s2 + 4  

Chapter 19, Solution 51. To get a and c , consider the circuit in Fig. (a).

jΩ I1 = 0

1Ω

-j3 Ω

I2

+ j2 Ω

V1

+ −

jΩ

V2

−

(a) V2 = I 2 ( j − j3) = -j2 I 2 V1 = -jI 2

a=

V2 - j2 I 2 = =2 V1 - jI 2

c=

I2 1 = =j V1 - j

To get b and d , consider the circuit in Fig. (b).

jΩ I1

1Ω

-j3 Ω

I2

+ V1 = 0

j2 Ω

jΩ

−

(b) For mesh 1, 0 = (1 + j2) I1 − j I 2

or

I 2 1 + j2 = = 2− j I1 j

+ −

V2

d=

- I2 = -2 + j I1

For mesh 2, V2 = I 2 ( j − j3) − j I 1 V2 = I 1 (2 − j)(- j2) − j I 1 = (-2 − j5) I 1 b=

- V2 = 2 + j5 I1

Thus,  2 2 + j5  [t ] =    j -2+ j

Chapter 19, Solution 52.

It is easy to find the z parameters and then transform these to h parameters and T parameters.  R1 + R 2 [z ] =   R2

 R 2 + R 3  R2

∆ z = (R 1 + R 2 )(R 2 + R 3 ) − R 22 = R 1R 2 + R 2 R 3 + R 3 R 1

(a)

 ∆z z [h] =  22 -z  21  z 22

z 12   R 1 R 2 + R 2 R 3 + R 3 R 1 z 22   R2 + R3 = - R2 1   z 22   R2 + R3

R2  R2 + R3   1  R2 + R3 

Thus, h 11 = R 1 +

R 2R 3 , R2 + R3

h 12 =

R2 = - h 21 , R2 + R3

as required.

(b)

 z 11 z [T] =  21 1   z 21

∆ z   R1 + R 2 z 21   R 2   z 22 = 1   z 21   R 2

R 1R 2 + R 2 R 3 + R 3 R 1   R2  R2 + R3  R2 

h 22 =

1 R2 + R3

Hence, A = 1+

R3 R1 R1 1 , B = R3 + (R 2 + R 3 ) , C = , D = 1+ R2 R2 R2 R2

as required. Chapter 19, Solution 53.

For the z parameters, V1 = z11 I1 + z12 I 2 V2 = z12 I1 + z 22 I 2

(1) (2)

For ABCD parameters, V1 = A V2 − B I 2 I1 = C V2 − D I 2 From (4), I D V2 = 1 + I 2 C C Comparing (2) and (5), 1 z 21 = , C

(3) (4) (5)

z 22 =

D C

Substituting (5) into (3),  AD  A V1 = I1 +  − B I 2  C  C =

A AD − BC I1 + I2 C C

Comparing (6) and (1), A z11 = C Thus, A  [Z] =  C 1  C

(6)

z 12 = ∆T  C D  C

AD − BC ∆ T = C C

Chapter 19, Solution 54. For the y parameters I 1 = y 11 V1 + y 12 V2 I 2 = y 21 V1 + y 22 V2 From (2), I 2 y 22 V1 = − V y 21 y 21 2 or

V1 =

(1) (2)

- y 22 1 V2 + I y 12 y 21 2

(3)

Substituting (3) into (1) gives - y 11 y 22 y 11 I1 = V2 + y 12 V2 + I y 21 y 21 2 or

I1 =

- ∆y y 21

V2 +

y 11 I y 21 2

(4)

Comparing (3) and (4) with the following equations V1 = A V2 − B I 2 I 1 = C V2 − D I 2 clearly shows that A=

- y 22 , y 21

B=

-1 , y 21

C=

- ∆y y 21

,

D=

- y 11 y 21

as required. Chapter 19, Solution 55. For the z parameters V1 = z11 I1 + z12 I 2 V2 = z 21 I1 + z 22 I 2 From (1), z 1 I1 = V1 − 12 I 2 z11 z11 Substituting (3) into (2) gives

(1) (2) (3)

or

V2 =

 z z  z 21 V1 +  z 22 − 21 12  I 2 z11  z11 

V2 =

∆ z 21 V1 + z I 2 z11 z11

Comparing (3) and (4) with the following equations I1 = g11 V1 + g12 I 2 V2 = g 21 V1 + g 22 I 2 indicates that -z z 1 , g 12 = 12 , g 21 = 21 , g 11 = z 11 z 11 z 11

(4)

g 22 =

∆z z 11

as required. Chapter 19, Solution 56. (a) ∆ y = (2 + j)(3 − j) + j4 = 7 + j5  y 22 / ∆ y [z] =  − y 21 / ∆ y

− y12 / ∆ y  0.2162 − j0.2973 − 0.2703 − j0.3784 = Ω y11 / ∆ y  0.0946 − j0.0676 0.2568 − j0.0405 

− y12 / y11   0.4 − j0.2 − 0.8 − j1.6  1 / y11 (b) [h ] =  =   y 21 / y11 ∆ y / y11  − 0.4 + j0.2 3.8 + j0.6  − y11 / y12 (c ) [ t ] =   − ∆ y / y12

− 1 / y12   − 0.25 + j0.5 j0.25  =  − y 22 / y12  − 1.25 + j1.75 0.25 + j0.75

Chapter 19, Solution 57. ∆ T = (3)(7) − (20)(1) = 1

A  [z ] =  C 1  C

∆T  C =  3 1Ω D   1 7   C

D  [y ] =  B -1  B

- ∆T   B =  A     B  

B  [h] =  -D1  D

1  ∆ T   20 Ω  D = 7 7  C   -1 1  S   D 7 7 

C  [g ] =  A 1  A

- ∆T  A = B   A 

D ∆ [t ] =  CT   ∆T

7 20 -1 20

-1 20 3 20

1 3S  1  3

  S 

-1  3  20  Ω  3

B  ∆ T   7 20 Ω  A  =  1 S 3   ∆T 

Chapter 19, Solution 58. The given set of equations is for the h parameters. 1 Ω 2  ∆ h = (1)(0.4) − (2)(-2) = 4.4 [h] =    - 2 0.4 S

(a)

 1 h [y ] =  11 h  21  h11

(b)

  [T] =   

- ∆h h 21 - h 22 h 21

- h12 h11 ∆h h11

   1 -2  = S   - 2 4.4  

- h11 h 21 -1 h 21

   2.2 0.5 Ω  =     0.2 S 0.5  

Chapter 19, Solution 59. ∆ g = (0.06)(2) − (-0.4)(2) = 0.12 + 0.08 = 0.2

(a)

  [z] =   

1 g 11 g 21 g 11 ∆g

(b)

  [y ] =   

(c)

 g 22  ∆ g [h] =  - g  21  ∆ g

- g 12  2  ∆ g   10 Ω = g 11   - 1 0.3 S  ∆ g 

(d)

  [T] =   

g 22  10 Ω  g 21   5 =  ∆g 1    0.3 S g 21 

g 22 - g 21 g 22

1 g 21 g 11 g 21

- g 12  g 11   16.667 6.667   ∆ g =  3.333 3.333  Ω  g 11  g 12 g 22 1 g 22

   0.1 - 0.2  = S   - 0.1 0.5  

Chapter 19, Solution 60. ∆ y = y 11 y 22 − y 12 y 21 = 0.3 − 0.02 = 0.28

(a)

  [z ] =   

y 22 ∆y - y 21 ∆y

- y 12 ∆y y 11 ∆y

   1.786 0.7143  =  Ω   0.3571 2.143  

- y 12  y 11   1.667 Ω 0.3333   ∆ y =  - 0.1667 0.4667 S   y 11 

(b)

  [h] =   

1 y 11 y 21 y 11

(c)

  [t ] =   

- y 11 y 12 - ∆y

y 12

-1  5Ω y 12   3 = - y 22   1.4 S 2.5  y 12 

Chapter 19, Solution 61. (a)

To obtain z 11 and z 21 , consider the circuit in Fig. (a).

1Ω

Io 1Ω

1Ω

+ I1

+ 1Ω

V1 −

 2 5 V1 = I 1 [1 + 1 || (1 + 1) ] = I 1 1 +  = I 1  3 3

Io =

V1 5 = I1 3

1 1 I1 = I1 1+ 2 3

- V2 + I o + I 1 = 0 1 4 V2 = I 1 + I 1 = I 1 3 3

V2 −

(a)

z 11 =

I2 = 0

z 21 =

V2 4 = I1 3

To obtain z 22 and z 12 , consider the circuit in Fig. (b).

1Ω 1Ω

I1

1Ω

+

+ 1Ω

V1 −

V2

I2

−

(b) Due to symmetry, this is similar to the circuit in Fig. (a). 5 4 z 22 = z 11 = , z 21 = z 12 = 3 3   [z ] =   

5 3 4 3

4 3 Ω 5 3

(b)

  [h] =   

∆z z 22 - z 21 z 22

z 12 z 22 1 z 22

    =     

(c)

  [T] =   

z 11 z 21 1 z 21

∆z z 21 z 22 z 21

    =    

Chapter 19, Solution 62. Consider the circuit shown below.

3 4  Ω 5 5  -4 3  S 5 5 

5 4 3 S 4

3  Ω 4  5   4 

I1

10 kΩ

a

40 kΩ

+ −

+

I2 +

50 kΩ b

30 kΩ

V1

Ib 20 kΩ

−

Since no current enters the input terminals of the op amp, V1 = (10 + 30) × 10 3 I 1 But

Va = Vb =

V2

−

(1)

30 3 V1 = V1 40 4

Vb 3 V 3 = 20 × 10 80 × 10 3 1 which is the same current that flows through the 50-kΩ resistor. Ib =

Thus,

V2 = 40 × 10 3 I 2 + (50 + 20) × 10 3 I b V2 = 40 × 10 3 I 2 + 70 × 10 3 ⋅ V2 =

3 V 80 × 10 3 1

21 V + 40 × 10 3 I 2 8 1

V2 = 105 × 10 3 I 1 + 40 × 10 3 I 2 From (1) and (2),  40 0  [z ] =   kΩ  105 40  ∆ z = z 11 z 22 − z 12 z 21 = 16 × 10 8

(2)

 A B  [T] =  = C D  

∆z z 21 z 22 z 21

z 11 z 21 1 z 21

   0.381 15.24 kΩ  = 0.381    9.52 µS 

Chapter 19, Solution 63. To get z11 and z21, consider the circuit below. 1:3

I1

• + 4Ω

V1

I2=0

• +

+ V’1

V’2

-

-

+ 9Ω

-

V2 -

ZR =

9 n2

= 1,

V1 = (4 // ZR )I1 =

n = 3 4 I1 5

 →

V2 = V2 ' = nV1' = nV1 = 3(4 / 5)I1

V z11 = 1 = 0.8 I1

 →

z 21 =

V2 = 2.4 I1

To get z21 and z22, consider the circuit below. I1=0

1:3 •

+ V1

4Ω

• +

+ V’1

V’2

-

-

-

Z R ' = n 2 ( 4 ) = 36 ,

I2 + 9Ω

V2 -

n =3

V2 = (9 // ZR ' )I 2 =

V1 =

9x36 I2 45

V2 V2 = = 2.4I 2 n 3

→

→

z 22 =

V2 = 7.2 I2

V z 21 = 1 = 2.4 I2

Thus, 0.8 2.4 [z] =  Ω 2.4 7.2

Chapter 19, Solution 64.

1 -j = = - j kΩ 3 jωC (10 )(10 -6 )

1 µF  →

Consider the op amp circuit below. 40 kΩ I1

20 kΩ

Vx

10 kΩ

1

+

− +

2 -j kΩ

V1

I2 + V2

−

−

At node 1, V1 − Vx Vx Vx − 0 = + 20 -j 10 V1 = (3 + j20) Vx

(1)

At node 2, Vx − 0 0 − V2 = 10 40

But

I1 =

V1 − Vx 20 × 10 3

Substituting (2) into (3) gives

 → Vx =

-1 V 4 2

(2)

(3)

I1 =

V1 + 0.25 V2 = 50 × 10 -6 V1 + 12.5 × 10 -6 V2 20 × 10 3

(4)

Substituting (2) into (1) yields -1 V1 = (3 + j20) V2 4 or

0 = V1 + (0.75 + j5) V2

(5)

Comparing (4) and (5) with the following equations I 1 = y 11 V1 + y 12 V2 I 2 = y 21 V1 + y 22 V2 indicates that I 2 = 0 and that  50 × 10 -6 [y ] =  1 

12.5 × 10 -6  S 0.75 + j5 

∆ y = (77.5 + j25. − 12.5) × 10 -6 = (65 + j250) × 10 -6

  [h] =   

1 y 11 y 21 y 11

- y 12  - 0.25  y 11   2 × 10 4 Ω  ∆ y =  2 × 10 4 1.3 + j5 S   y 11 

Chapter 19, Solution 65.

The network consists of two two-ports in series. It is better to work with z parameters and then convert to y parameters. 4 2 For N a , [z a ] =   2 2 For N b ,

 2 1 [z b ] =    1 1 6 3 [z ] = [z a ] + [z b ] =   3 3

∆ z = 18 − 9 = 9  z 22  ∆ [y ] =  z -z  21  ∆z

- z 12   ∆z   = z 11   ∆z  

1 3 -1 3

-1  3 S 2   3 

Chapter 19, Solution 66. Since we have two two-ports in series, it is better to convert the given y parameters to z parameters. ∆ y = y 11 y 22 − y 12 y 21 = (2 × 10 -3 )(10 × 10 -3 ) − 0 = 20 × 10 -6   [z a ] =   

y 22 ∆y - y 21 ∆y

- y 12 ∆y y 11 ∆y

   500 Ω 0  = 100 Ω    0 

 500 0   100 100   600 100  [z ] =  + =   0 100   100 100   100 200  i.e.

V1 = z 11 I 1 + z 12 I 2 V2 = z 21 I 1 + z 22 I 2

or

V1 = 600 I 1 + 100 I 2 V2 = 100 I 1 + 200 I 2

(1) (2)

But, at the input port, Vs = V1 + 60 I 1

(3)

and at the output port, V2 = Vo = -300 I 2

(4)

From (2) and (4), 100 I 1 + 200 I 2 = -300 I 2 I 1 = -5I 2

(5)

Substituting (1) and (5) into (3), Vs = 600 I 1 + 100 I 2 + 60 I 1

= (660)(-5) I 2 + 100 I 2 = -3200I 2

(6)

From (4) and (6), Vo - 300 I 2 = = 0.09375 V2 - 3200 I 2

Chapter 19, Solution 67. The y parameters for the upper network is  2 -1  [y ] =  ,  -1 2    [z a ] =   

- y 12 ∆y y 11 ∆y

y 22 ∆y - y 21 ∆y

 2   = 3  1   3

∆ y = 4 −1 = 3 1 3  2 3

1 1 [z b ] =   1 1 5 3 4 3 [z ] = [z a ] + [z b ] =   4 3 5 3 ∆z =

25 16 − =1 9 9

  [T] =   

z 11 z 21 1 z 21

∆z z 21 z 22 z 21

   1.25 0.75 Ω  =    0.75 S 1.25  

Chapter 19, Solution 68.

 4 -2 For the upper network N a , [y a ] =   -2 4 

 2 -1  and for the lower network N b , [y b ] =   1 2  For the overall network,

 6 -3 [y ] = [y a ] + [y b ] =    -3 6  ∆ y = 36 − 9 = 27   [h] =   

1 y 11 y 21 y 11

- y 12   1 -1  y 11   6 Ω 2   ∆y =  -1 9    S 2  y 11   2

Chapter 19, Solution 69. We first determine the y parameters for the upper network N a . To get y 11 and y 21 , consider the circuit in Fig. (a). n=

1 , 2

ZR =

1s 4 = n2 s

 2s + 4   4 I V1 = (2 + Z R ) I 1 =  2 +  I 1 =   s  1  s I1 s = V1 2 (s + 2)

y 11 =

I2 =

- s V1 - I1 = -2 I 1 = s+2 n

y 21 =

I2 -s = V1 s + 2

To get y 22 and y 12 , consider the circuit in Fig. (b).

I1

2Ω

1/s

2:1

I2

+

+

V1 =0

V2

−

−

I2

(b) 1 1 Z R ' = (n 2 )(2) =   (2) = 4 2 s + 2 1   1 1 I V2 =  + Z R '  I 2 =  +  I 2 =   2s  2 s   s 2 y 22 =

I2 2s = V2 s + 2

 - 1  2s   -s   V2 =  V I 1 = - n I 2 =    2  s + 2  s + 2 2 y 12 =

I1 -s = V2 s + 2

 s  2 (s + 2) [y a ] =  -s   s+2

-s  s+2  2s   s+2 

For the lower network N b , we obtain y 11 and y 21 by referring to the network in Fig. (c).

I1

2

I2 +

V1

+ −

s

V2 = 0 −

(c) V1 = 2 I 1

 → y 11 =

I1 1 = V1 2

I 2 = - I1 =

- V1 2

I 2 -1 = V1 2

 → y 21 =

To get y 22 and y 12 , refer to the circuit in Fig. (d).

I1

2

I2

+

+ s

V1 = 0 −

I2

V2 −

(d) V2 = (s || 2) I 2 =

I1 = - I 2 ⋅

y 12 =

2s I s+2 2

 → y 22 =

I2 s + 2 = 2s V2

- V2  - s  s + 2  -s   V2 = = s + 2  s + 2  2s  2

I1 - 1 = 2 V2

12 -1 2  [y b ] =    - 1 2 (s + 2) 2s   s+1  s+2 [y ] = [y a ] + [y b ] =   - (3s + 2)  2 (s + 2)

- (3s + 2) 2 (s + 2) 5s 2 + 4s + 4 2s (s + 2)

    

Chapter 19, Solution 70. We may obtain the g parameters from the given z parameters.  25 20  [z a ] =  ∆ z a = 250 − 100 = 150 ,  5 10 

 50 25  [z b ] =  ,  25 30 

∆ z b = 1500 − 625 = 875

  [g ] =   

1 z 11 z 21 z 11

- z 12 z 11 ∆z z 11

    

 0.04 - 0.8  [g a ] =  , 6   0.2

 0.02 - 0.5  [g b ] =    0.5 17.5 

 0.06 S - 1.3  [g ] = [g a ] + [ g b ] =  23.5 Ω   0.7

Chapter 19, Solution 71. This is a parallel-series connection of two two-ports. We need to add their g parameters together and obtain z parameters from there. For the transformer, V1 =

1 V2 , I1 = −2I 2 2

Comparing this with V1 = AV2 − BI2 ,

I1 = CV2 − DI 2

shows that

0.5 0 [Tb1] =    0 2 To get A and C for Tb2 , consider the circuit below. I1 + V1 -

4Ω

I2 =0

5Ω 2Ω

+ V2 -

V1 = 9I1,

A=

V2 = 5I1

V1 = 9 / 5 = 1.8, V2

I C = 1 = 1 / 5 = 0.2 V2

Chapter 19, Solution 72. Consider the network shown below.

I1

Ia1 + Va1

+

V1

Ia2 Na

Ib1 + Vb1

−

+ Va2

+

V2

Ib2 Nb

Va1 = 25 I a1 + 4 Va 2 I a 2 = - 4 I a1 + Va 2 Vb1 = 16 I b1 + Vb 2 I b 2 = - I b1 + 0.5 Vb 2 V1 = Va1 + Vb1 V2 = Va 2 = Vb 2 I 2 = I a 2 + I b2 I 1 = I a1 Now, rewrite (1) to (4) in terms of I 1 and V2 Va1 = 25 I 1 + 4 V2 I a 2 = - 4 I 1 + V2 Vb1 = 16 I b1 + V2 I b 2 = - I b1 + 0.5 V2 Adding (5) and (7),

I2

+ Vb2

−

(1) (2) (3) (4)

(5) (6) (7) (8)

V1 = 25 I 1 + 16 I b1 + 5 V2

(9)

Adding (6) and (8), I 2 = - 4 I 1 − I b1 + 1.5 V2

(10)

I b1 = I a1 = I 1

(11)

Because the two networks N a and N b are independent, I 2 = - 5 I 1 + 1.5 V2 V2 = 3.333 I 1 + 0.6667 I 2 or

(12)

Substituting (11) and (12) into (9), 25 5 V1 = 41I 1 + I1 + I 1.5 1.5 2 V1 = 57.67 I 1 + 3.333 I 2

(13)

Comparing (12) and (13) with the following equations V1 = z 11 I 1 + z 12 I 2 V2 = z 21 I 1 + z 22 I 2 indicates that  57.67 3.333  [z ] =  Ω  3.333 0.6667  Alternatively,

 25 4  [h a ] =  , -4 1

 16 1  [h b ] =    - 1 0.5 

 41 5  [h] = [h a ] + [h b ] =    - 5 1.5   ∆h  h [z ] =  22 -h  21  h 22 as obtained previously.

h12 h 22 1 h 22

∆ h = 61.5 + 25 = 86.5

   57.67 3.333  = Ω 3 . 333 0 . 6667    

Chapter 19, Solution 73. From Example 18.14 and the cascade two-ports,

2 3 [Ta ] = [Tb ] =   1 2

 2 3  2 3   7 12 Ω  [T] = [Ta ][Tb ] =  =   7   1 2  1 2   4 S When the output is short-circuited, V2 = 0 and by definition V1 = - B I 2 , I1 = - D I 2 Hence, V1 B 12 Z in = = = Ω I1 D 7 Chapter 19, Solution 74. From Prob. 18.35, the transmission parameters for the circuit in Figs. (a) and (b) are 1 Z [Ta ] =  , 0 1 

 1 0 [Tb ] =   1 Z 1 

Z

Z (a)

(b)

We partition the given circuit into six subcircuits similar to those in Figs. (a) and (b) as shown in Fig. (c) and obtain [T] for each. s

s

1

1/s

T1

T2

T3

1

T4

T5

1/s

T6

1 0  [T1 ] =  , 1 1 

1 s [T2 ] =  ,  0 1

1 0  [T3 ] =   s 1

[T4 ] = [T2 ] ,

[T5 ] = [T1 ] ,

[T6 ] = [T3 ]

 1 0  1 0  [T] = [T1 ][T2 ][T3 ][T4 ][T5 ][T6 ] = [T1 ][T2 ][T3 ][T4 ]    1 1  s 1 

0 0 1 s 1  1 = [T1 ][T2 ][T3 ]  = [T1 ][T2 ][T3 ][T4 ]       0 1   s +1 1   s +1 1   1 0   s2 + s +1 s  = [T1 ][T2 ]    1  s 1   s +1

s   1 s   s2 + s +1 = [T1 ]   3   2 2  0 1   s + s + 2s + 1 s + 1   1 0   s 4 + s 3 + 3s 2 + 2s + 1 s 3 + 2s  =   3 2 s2 +1   1 1   s + s + 2s + 1  s 4 + s 3 + 3s 2 + 2s + 1  s 3 + 2s [T] =  4  3 2 3 2  s + 2s + 4s + 4s + 2 s + s + 2s + 1 

Note that AB − CD = 1 as expected. Chapter 19, Solution 75. (a) We convert [za] and [zb] to T-parameters. For Na, ∆ z = 40 − 24 = 16 . ∆ z / z 21   2 4  z / z = [Ta ] =  11 21    1 / z 21 z 22 / z 21  0.25 1.25

For Nb, ∆ y = 80 + 8 = 88 . − y 22 / y 21 − 1 / y 21   − 5 − 0.5 [Tb ] =  =   − ∆ y / y 21 − y11 / y 21  − 44 − 4 

− 17   − 186 [T] = [Ta ][Tb ] =   − 56.25 − 5.125 We convert this to y-parameters. ∆ T = AD − BC = −3.  D / B − ∆ T / B 0.3015 − 0.1765 = [ y] =   A / B  0.0588 10.94  − 1 / B (b)

The equivalent z-parameters are A / C ∆ T / C  3.3067 0.0533 [z] =  =   1 / C D / C  − 0.0178 0.0911

Consider the equivalent circuit below. I1

z11

z22

+

I2

+

+ +

Vi

z12 I2

-

-

z21 I1

ZL

Vo -

-

Vi = z11I1 + z12 I 2

(1)

Vo = z 21I1 + z 22 I 2

(2)

But Vo = −I 2 ZL

→

I 2 = −Vo / ZL

(3)

From (2) and (3) , V Vo = z 21I1 − z 22 o ZL

→

  1 z I1 = Vo  + 22   z 21 ZL z 21 

(4)

Substituting (3) and (4) into (1) gives Vi  z11 z11z 22  z12 − = + = −194.3 Vo  z 21 z 21ZL  ZL

→

Vo. = −0.0051 Vi

Chapter 19, Solution 76. To get z11 and z21, we open circuit the output port and let I1 = 1A so that V V z11 = 1 = V1, z 21 = 2 = V2 I1 I1 The schematic is shown below. After it is saved and run, we obtain z11 = V1 = 3.849,

z 21 = V2 = 1.122

Similarly, to get z22 and z12, we open circuit the input port and let I2 = 1A so that V z12 = 1 = V1, I2

z 22 =

V2 = V2 I2

The schematic is shown below. After it is saved and run, we obtain z12 = V1 = 1.122,

z 22 = V2 = 3.849

Thus, 3.949 1.122  [z] =  Ω 1.122 3.849

Chapter 19, Solution 77. We follow Example 19.15 except that this is an AC circuit. (a) We set V2 = 0 and I1 = 1 A. The schematic is shown below. In the AC Sweep Box, set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, the output file includes FREQ

IM(V_PRINT2)

IP(V_PRINT2)

1.592 E–01

3.163 E–.01

–1.616 E+02

FREQ

VM($N_0001)

VP($N_0001)

1.592 E–01

9.488 E–01

–1.616 E+02

From this we obtain h11 = V1/1 = 0.9488∠–161.6° h21 = I2/1 = 0.3163∠–161.6°.

(b) In this case, we set I1 = 0 and V2 = 1V. The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain an output file which includes FREQ

VM($N_0001)

VP($N_0001)

1.592 E–01

3.163 E–.01

1.842 E+01

FREQ

IM(V_PRINT2)

IP(V_PRINT2)

1.592 E–01

9.488 E–01

–1.616 E+02

From this, h12 = V1/1 = 0.3163∠18.42° h21 = I2/1 = 0.9488∠–161.6°. Thus,

0.9488∠ − 161.6° 0.3163∠18.42°  [h] =   0.3163∠ − 161.6° 0.9488∠ − 161.6°

Chapter 19, Solution 78 For h11 and h21, short-circuit the output port and let I1 = 1A. f = ω / 2π = 0.6366 . The schematic is shown below. When it is saved and run, the output file contains the following: FREQ

IM(V_PRINT1)IP(V_PRINT1)

6.366E-01 FREQ

1.202E+00

1.463E+02

VM($N_0003) VP($N_0003)

6.366E-01

3.771E+00

-1.350E+02

From the output file, we obtain I 2 = 1.202∠146.3o ,

V1 = 3.771∠ − 135o

so that V h11 = 1 = 3.771∠ − 135o , 1

I h 21 = 2 = 1.202∠146.3o 1

For h12 and h22, open-circuit the input port and let V2 = 1V. The schematic is shown below. When it is saved and run, the output file includes: FREQ

VM($N_0003) VP($N_0003)

6.366E-01 FREQ

1.202E+00

-3.369E+01

IM(V_PRINT1)IP(V_PRINT1)

6.366E-01

3.727E-01

-1.534E+02

From the output file, we obtain I 2 = 0.3727∠ − 153.4o ,

V1 = 1.202∠ − 33.69o

so that V h12 = 1 = 1.202∠ − 33.69o , 1

I h 22 = 2 = 0.3727∠ − 153.4o 1

Thus, 3.771∠ − 135o [h ] =   1.202∠146.3

1.202∠ − 33.69o   0.3727∠ − 153.4o 

Chapter 19, Solution 79 We follow Example 19.16. (a) We set I1 = 1 A and open-circuit the output-port so that I2 = 0. The schematic is shown below with two VPRINT1s to measure V1 and V2. In the AC Sweep box, we enter Total Pts = 1, Start Freq = 0.3183, and End Freq = 0.3183. After simulation, the output file includes FREQ

VM(1)

VP(1)

3.183 E–01

4.669 E+00

–1.367 E+02

FREQ

VM(4)

VP(4)

3.183 E–01

2.530 E+00

–1.084 E+02

From this, z11 = V1/I1 = 4.669∠–136.7°/1 = 4.669∠–136.7° z21 = V2/I1 = 2.53∠–108.4°/1 = 2.53∠–108.4°.

(b) In this case, we let I2 = 1 A and open-circuit the input port. The schematic is shown below. In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.3183, and End Freq = 0.3183. After simulation, the output file includes FREQ

VM(1)

VP(1)

3.183 E–01

2.530 E+00

–1.084 E+02

FREQ

VM(2)

VP(2)

3.183 E–01

1.789 E+00

–1.534 E+02

From this, z12 = V1/I2 = 2.53∠–108.4°/1 = 2.53∠–108..4° Thus,

z22 = V2/I2 = 1.789∠–153.4°/1 = 1.789∠–153.4°. 4.669∠ − 136.7° 2.53∠ − 108.4°  [z] =    2.53∠ − 108.4° 1.789∠ − 153.4°

Chapter 19, Solution 80 To get z11 and z21, we open circuit the output port and let I1 = 1A so that V z11 = 1 = V1, I1

z 21 =

V2 = V2 I1

The schematic is shown below. After it is saved and run, we obtain z11 = V1 = 29.88,

z 21 = V2 = −70.37

Similarly, to get z22 and z12, we open circuit the input port and let I2 = 1A so that V z12 = 1 = V1, I2

z 22 =

V2 = V2 I2

The schematic is shown below. After it is saved and run, we obtain z12 = V1 = 3.704,

z 22 = V2 = 11.11

Thus,  29.88 3.704 [z] =  Ω − 70.37 11.11

Chapter 19, Solution 81 (a) We set V1 = 1 and short circuit the output port. The schematic is shown below. After simulation we obtain y11 = I1 = 1.5, y21 = I2 = 3.5

(b) We set V2 = 1 and short-circuit the input port. The schematic is shown below. Upon simulating the circuit, we obtain y12 = I1 = –0.5, y22 = I2 = 1.5 1.5 − 0.5 [Y] =    3.5 1.5 

Chapter 19, Solution 82 We follow Example 19.15. (a) Set V2 = 0 and I1 = 1A. The schematic is shown below. After simulation, we obtain h11 = V1/1 = 3.8, h21 = I2/1 = 3.6

(b) Set V1 = 1 V and I1 = 0. The schematic is shown below. After simulation, we obtain h12 = V1/1 = 0.4, h22 = I2/1 = 0.25 Hence,

 3.8 0.4  [h] =    3.6 0.25

Chapter 19, Solution 83 To get A and C, we open-circuit the output and let I1 = 1A. The schematic is shown below. When the circuit is saved and simulated, we obtain V1 = 11 and V2 = 34.

A=

V1 = 0.3235, V2

I 1 C= 1 = = 0.02941 V2 34

Similarly, to get B and D, we open-circuit the output and let I1 = 1A. The schematic is shown below. When the circuit is saved and simulated, we obtain V1 = 2.5 and I2 = -2.125.

V 2.5 = 1.1765, B=− 1 = I 2 2.125

I 1 = 0.4706 D=− 1 = I 2 2.125

Thus,  0.3235 1.1765  [T ] =   0.02941 0.4706

Chapter 19, Solution 84

(a)

Since A =

V1 V2

and C = I 2 =0

I1 V2

, we open-circuit the output port and let V1 I 2 =0

= 1 V. The schematic is as shown below. After simulation, we obtain A = 1/V2 = 1/0.7143 = 1.4 C = I2/V2 = 1.0/0.7143 = 1.4

(b) To get B and D, we short-circuit the output port and let V1 = 1. The schematic is shown below. After simulating the circuit, we obtain B = –V1/I2 = –1/1.25 = –0.8 D = –I1/I2 = –2.25/1.25 = –1.8

A B  1.4 − 0.8  C D = 1.4 − 1.8     

Thus

Chapter 19, Solution 85

(a)

Since A =

V1 V2

and C = I 2 =0

I1 V2

, we let V1 = 1 V and openI 2 =0

circuit the output port. The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain an output file which includes FREQ 1.592 E–01

IM(V_PRINT1) 6.325 E–01

IP(V_PRINT1) 1.843 E+01

FREQ 1.592 E–01

VM($N_0002) 6.325 E–01

VP($N_0002) –7.159 E+01

From this, we obtain A =

1 1 = = 1.581∠71.59° V2 0.6325∠ − 71.59°

C =

(b)

I1 0.6325∠18.43° = 1∠90° = j = V2 0.6325∠ − 71.59°

Similarly, since B =

V1 I2

and D = − V2 = 0

I1 I2

, we let V1 = 1 V and shortV2 = 0

circuit the output port. The schematic is shown below. Again, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box. After simulation, we get an output file which includes the following results: FREQ 1.592 E–01

IM(V_PRINT1) 5.661 E–04

IP(V_PRINT1) 8.997 E+01

FREQ 1.592 E–01

IM(V_PRINT3) 9.997 E–01

IP(V_PRINT3) –9.003 E+01

From this, B = −

1 1 =− = −1∠90° = − j I2 0.9997∠ − 90°

D = −

I1 5.661x10 −4 ∠89.97° =− = 5.561x10–4 I2 0.9997∠ − 90°

−j A B  1.581∠71.59°   C D =  −4  j 5.661x10    

Chapter 19, Solution 86

(a)

By definition, g11 =

I1 V1

, g21 = I 2 =0

V1 V2

. I 2 =0

We let V1 = 1 V and open-circuit the output port. The schematic is shown below. After simulation, we obtain g11 = I1 = 2.7 g21 = V2 = 0.0

(b)

Similarly, g12 =

I1 I2

, g22 = V1 = 0

V2 I2

V1 = 0

We let I2 = 1 A and short-circuit the input port. The schematic is shown below. After simulation, g12 = I1 = 0

g22 = V2 = 0  2.727S 0 [g] =  0  0

Thus

Chapter 19, Solution 87

(a)

Since

a =

V2 V1

and c = I1 = 0

I2 V1

, I1 = 0

we open-circuit the input port and let V2 = 1 V. The schematic is shown below. In the AC Sweep box, set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain an output file which includes FREQ 1.592 E–01

IM(V_PRINT2) 5.000 E–01

IP(V_PRINT2) 1.800 E+02

FREQ 1.592 E–01

VM($N_0001) 5.664 E–04

VP($N_0001) 8.997 E+01

From this, a =

1 = 1765∠ − 89.97° 5.664x10 −4 ∠89.97°

c =

0.5∠180° = −882.28∠ − 89.97° 5.664x10 − 4 ∠89.97°

(b)

Similarly, b = −

V2 I1

and d = − V1 = 0

I2 I1

V1 = 0

We short-circuit the input port and let V2 = 1 V. The schematic is shown below. After simulation, we obtain an output file which includes FREQ 1.592 E–01

IM(V_PRINT2) 5.000 E–01

IP(V_PRINT2) 1.800 E+02

FREQ 1.592 E–01

IM(V_PRINT3) 5.664 E–04

IP(V_PRINT3) –9.010 E+01

From this, we get b = −

d = −

Thus

1 −4

5.664x10 ∠ − 90.1°

= –j1765

0.5∠180° = j888.28 5.664x10 − 4 ∠ − 90.1°

 − j1765 − j1765 [t] =    j888.2 j888.2 

Chapter 19, Solution 88

To get Z in , consider the network in Fig. (a).

Rs

I1

I2

+ Vs

+ −

+ Two-Port

V1

RL

V2

−

− Zin

(a)

I 1 = y 11 V1 + y 12 V2 I 2 = y 21 V1 + y 22 V2

But

I2 =

- V2 = y 21 V1 + y 22 V2 RL

V2 =

- y 21 V1 y 22 + 1 R L

(1) (2)

(3)

Substituting (3) into (1) yields  - y 21 V1  , I 1 = y 11 V1 + y 12 ⋅   y 22 + 1 R L   ∆ y + y 11 YL   V1 , I1 =   y 22 + YL 

or

YL =

∆ y = y 11 y 22 − y 12 y 21

Z in =

y 22 + YL V1 = I 1 ∆ y + y 11 YL

Ai =

y I2 y V + y 22 V2 = 21 1 = y 21 Z in +  22 I1 I1  I1

= y 21 Z in −

y 22 y 21 Z in  y 22 + YL = ∆ +y Y y 22 + YL 11 L  y

Ai =

y 21 YL ∆ y + y 11 YL

Av =

V2 - y 21 = V1 y 22 + YL

From (3),

1 RL

  - y 21 V1     y 22 + YL

  

  y 21 − y 22 y 21  y 22 + YL 

  

To get Z out , consider the circuit in Fig. (b). I1

I2

+ Rs

V1

+ −

Two-Port

V2

− (b) Z out =

But

V2 V2 = I 2 y 21 V1 + y 22 V2

V1 = - R s I 1

Substituting this into (1) yields I 1 = - y 11 R s I 1 + y 12 V2 (1 + y 11 R s ) I 1 = y 12 V2 I1 =

or

- V1 y 12 V2 = 1 + y 11 R s Rs

- y 12 R s V1 = V2 1 + y 11 R s

Substituting this into (4) gives 1 Z out = y 12 y 21 R s y 22 − 1 + y 11 R s =

Z out =

y 22

1 + y 11 R s + y 11 y 22 R s − y 21 y 22 R s

y 11 + Ys ∆ y + y 22 Ys

Zout

(4)

Chapter 19, Solution 89

Av =

- h fe R L h ie + (h ie h oe − h re h fe ) R L

Av =

- 72 ⋅ 10 5 2640 + (2640 × 16 × 10 -6 − 2.6 × 10 -4 × 72) ⋅ 10 5

Av =

- 72 ⋅ 10 5 = - 1613 2640 + 1824

dc gain = 20 log A v = 20 log (1613) = 64.15

Chapter 19, Solution 90

(a)

Z in = h ie −

h re h fe R L 1 + h oe R L

10 -4 × 120 R L 1500 = 2000 − 1 + 20 × 10 -6 R L 500 =

12 × 10 -3 1 + 2 × 10 -5 R L

500 + 10 -2 R L = 12 × 10 -3 R L 500 × 10 2 = 0.2 R L R L = 250 kΩ (b)

Av =

- h fe R L h ie + (h ie h oe − h re h fe ) R L

- 120 × 250 × 10 3 Av = 2000 + (2000 × 20 × 10 -6 − 120 × 10 -4 ) × 250 × 10 3 - 30 × 10 6 = - 3333 Av = 2 × 10 3 + 7 × 10 3

Ai =

(c)

h fe 120 = = 20 1 + h oe R L 1 + 20 × 10 -6 × 250 × 10 3

Z out =

R s + h ie 600 + 2000 = (R s + h ie ) h oe − h re h fe (600 + 2000) × 20 × 10 -6 − 10 -4 × 120

Z out =

2600 kΩ = 65 kΩ 40 Vc Vc = Vb Vs

Av =

 → Vc = A v Vs = -3333 × 4 × 10 -3 = - 13.33 V

Chapter 19, Solution 91

R s = 1.2 kΩ ,

R L = 4 kΩ

Av =

- h fe R L h ie + (h ie h oe − h re h fe ) R L

Av =

- 80 × 4 × 10 3 1200 + (1200 × 20 × 10 -6 − 1.5 × 10 -4 × 80) × 4 × 10 3

Av =

- 32000 = - 25.64 1248

(b)

Ai =

h fe 80 = = 74.074 1 + h oe R L 1 + 20 × 10 -6 × 4 × 10 3

(c)

Z in = h ie − h re A i

(a)

Z in = 1200 − 1.5 × 10 -4 × 74.074 ≅ 1.2 kΩ (d)

Z out =

R s + h ie (R s + h ie ) h oe − h re h fe

Z out =

1200 + 1200 2400 = = 51.282 kΩ -6 -4 2400 × 20 × 10 − 1.5 × 10 × 80 0.0468

Chapter 19, Solution 92 Due to the resistor R E = 240 Ω , we cannot use the formulas in section 18.9.1. We will need to derive our own. Consider the circuit in Fig. (a).

Rs

Ib

hie

Ic

+

+ hre Vc

Vs

+ −

+ −

hfe Ib

Vb

Vc

IE RE

−

−

(a)

Zin IE = Ib + Ic

(1)

Vb = h ie I b + h re Vc + (I b + I c ) R E

(2)

Vc RE + 1

(3)

I c = h fe I b +

But

hoe

h oe

Vc = - I c R L

(4)

Substituting (4) into (3),

I c = h fe I b −

or

Ai =

RL RE + 1

Ic h oe

I c h fe (1 + R E h oe ) = Ib 1 + h oe (R L

100(1 + 240x30 x10 −6 ) Ai = 1 + 30 × 10 -6 (4,000 + 240) A i = 79.18

From (3) and (5),

(5)

RL

Ic =

h fe (1 + R E )h oe Vc I b = h fe I b + 1 + h oe (R L + R E ) RE + 1

(6) h oe

Substituting (4) and (6) into (2), Vb = (h ie + R E ) I b + h re Vc + I c R E Vb =

Vc (h ie + R E ) V + h re Vc − c R E RL   1   h fe (1 + R E h oe )    R E + − h fe  h oe  1 + h oe (R L + R E )  

V (h ie + R E ) R 1 + h re − E = b = A v Vc  RL  1   h fe (1 + R E h oe )    R E + − h fe  h oe  1 + h oe (R L + R E )   1 = Av  1  240 + 30 x10 −6 

(7)

(4000 + 240) 240 + 10 -4 − −6 4000   100(1 + 240 x 30 x10 ) − 100  -6   1 + 30 × 10 × 4240 

1 = −6.06x10 −3 + 10 -4 − 0.06 = -0.066 Av A v = –15.15 From (5), Ic =

h fe I 1 + h oe R L b

We substitute this with (4) into (2) to get Vb = (h ie + R E ) I b + (R E − h re R L ) I c  h (1 + R E h oe )  Vb = (h ie + R E ) I b + (R E − h re R L )  fe I b   1 + h oe (R L + R E )  Z in =

Vb h (R − h re R L )(1 + R E h oe ) = h ie + R E + fe E Ib 1 + h oe (R L + R E )

(100)(240 × 10 -4 × 4 × 10 3 )(1 + 240x30x10 −6 ) Z in = 4000 + 240 + 1 + 30 × 10 -6 × 4240 Z in = 12.818 kΩ

(8)

To obtain Z out , which is the same as the Thevenin impedance at the output, we introduce a 1-V source as shown in Fig. (b). Rs

hie

Ib

Ic

+

+ + −

hre Vc Vb

hfe Ib

hoe

IE RE

−

+ −

Vc

1V

− (b)

Zout

From the input loop, I b (R s + h ie ) + h re Vc + R E (I b + I c ) = 0 But So,

Vc = 1 I b (R s + h ie + R E ) + h re + R E I c = 0

(9)

From the output loop, Ic =

or

Vc

RE +

1 h oe

+ h fe I b =

h oe + h fe I b R E h oe + 1

h oe h fe Ic Ib = − h fe 1 + R E h oe

(10)

Substituting (10) into (9) gives I (R s + R E + h ie )  c  h fe

  + h re + R E I c − 

h  (R s + R E + h ie ) oe h fe   =0 1 + R E h oe

R s + R E + h ie R + R E + h ie Ic + R E Ic = s h fe 1 + R E h oe

 h oe   h fe

  − h re 

 R + R E + h ie  (h oe h fe )  s  − h re 1 + R E h oe   Ic = R E + (R s + R E + h ie ) h fe Z out =

Z out =

Z out =

R E h fe + R s + R E + h ie 1 = I c  R s + R E + h ie    h oe − h re h fe  1 + R E h oe  240 × 100 + (1200 + 240 + 4000) 1200 + 240 + 4000  -6 -4  1 + 240 x 30 x10 −6  × 30 × 10 − 10 × 100  

24000 + 5440 = 193.7 kΩ 0.152

Chapter 19, Solution 93

We apply the same formulas derived in the previous problem. (h ie + R E ) R 1 = + h re − E Av  RL  1   h fe (1 + R E h oe )  R E +   − h fe  h oe  1 + h oe (R L + R E )   1 = Av

200 (2000 + 200) + 2.5 × 10 -4 − 3800 150(1 + 0.002)  − 150 (200 + 10 5 )   1 + 0.04 

1 = −0.004 + 2.5 × 10- 4 − 0.05263 = -0.05638 Av A v = –17.74 h fe (1 + R E h oe ) 150(1 + 200x10 −5 ) Ai = = = 144.5 1 + h oe (R L + R E ) 1 + 10 -5 × (200 + 3800) Z in = h ie + R E +

h fe (R E − h re R L )(1 + R E h oe ) 1 + h oe (R L + R E )

Z in = 2000 + 200 +

(150)(200 − 2.5 × 10 -4 × 3.8 × 10 3 )(1.002) 1.04

Z in = 2200 + 28966 Z in = 31.17 kΩ

Z out =

Z out =

R E h fe + R s + R E + h ie  R s + R E + h ie    h oe − h re h fe  1 + R E h oe  33200 200 × 150 + 1000 + 200 + 2000 = -5 - 0.0055  3200 × 10  -4  − 2.5 × 10 × 150   1.002 

Z out = –6.148 MΩ Chapter 19, Solution 94

We first obtain the ABCD parameters.  200 0  [h] =  Given ,  100 10 -6    [T] =   

∆h h 21 - h 22 h 21

- h11 h 21 -1 h 21

∆ h = h11 h 22 − h12 h 21 = 2 × 10 -4

   - 2 × 10 -6 = -8   - 10 

-2   - 10 -2 

The overall ABCD parameters for the amplifier are  - 2 × 10 -6 - 2  - 2 × 10 -6 -2 [T] =  -8 -2  -8 - 10  - 10 - 10 -2  - 10 ∆ T = 2 × 10 -12 − 2 × 10 -12 = 0 B  [h] =  D -1  D

∆T D C D

 0    200  =  - 10 -4 10 -6  

  2 × 10 -8 ≅   10 -10

2 × 10 -2   10 -4 

Thus,

h ie = 200 ,

Av =

h re = 0 ,

h fe = -10 -4 ,

h oe = 10 -6

(10 4 )(4 × 10 3 ) = 2 × 10 5 200 + (2 × 10 -4 − 0) × 4 × 10 3

Z in = h ie −

h re h fe R L = 200 − 0 = 200 Ω 1 + h oe R L

Chapter 19, Solution 95

Let Z A =

1 s 4 + 10s 2 + 8 = s 3 + 5s y 22

Using long division, 5s 2 + 8 ZA = s + 3 = s L1 + Z B s + 5s

i.e.

L1 = 1 H

and

ZB =

5s 2 + 8 s 3 + 5s

as shown in Fig (a). L1 ZB y22 = 1/ZA (a) 1 s 3 + 5s YB = = Z B 5s 2 + 8

Using long division, YB = 0.2s +

where

C 2 = 0 .2 F

3.4s = sC 2 + YC 5s 2 + 8 and

YC =

3.4s 5s 2 + 8

as shown in Fig. (b). L1 C2 Yc = 1/ZC (b) ZC =

1 5s 2 + 8 5s 8 1 = = + = s L3 + YC 3.4s 3.4 3.4s s C4

i.e. an inductor in series with a capacitor 5 L3 = = 1.471 H and 3.4

C4 =

3.4 = 0.425 F 8

Thus, the LC network is shown in Fig. (c). 0.425 F

1.471 H

1H

0.2 F

(c)

Chapter 19, Solution 96

This is a fourth order network which can be realized with the network shown in Fig. (a). L1

L3

C2

C4

(a)

∆ (s) = (s 4 + 3.414s 2 + 1) + (2.613s 3 + 2.613s)

1Ω

1 2.613s + 2.613s H(s) = s 4 + 3.414s 2 + 1 1+ 2.613s 3 + 2.613s 3

which indicates that -1 2.613s + 2.613s s 4 + 3.414s + 1 = 2.613s 3 + 2.613s

y 21 = y 22

3

We seek to realize y 22 . By long division, 2.414s 2 + 1 y 22 = 0.383s + = s C 4 + YA 2.613s 3 + 2.613s i.e.

C 4 = 0.383 F

YA =

and

2.414s 2 + 1 2.613s 3 + 2.613s

as shown in Fig. (b). L1

YA

L3

C2

C4

y22

(b) 1 2.613s 3 + 2.613s = ZA = YA 2.414s 2 + 1

By long division, Z A = 1.082s +

i.e.

L 3 = 1.082 H

1.531s = s L3 + Z B 2.414s 2 + 1 and

ZB =

1.531s 2.414s 2 + 1

as shown in Fig.(c). L1

ZB

L3

C2

C4

(c) YB =

i.e.

1 1 1 = 1.577s + = s C2 + 1.531s ZB s L1

C 2 = 1.577 F

and

L1 = 1.531 H

Thus, the network is shown in Fig. (d). 1.531 H

1.577 F

1.082 H

0.383 F

1Ω

(d)

Chapter 19, Solution 97

Hence,

s3 s3 s 3 + 12s H(s) = 3 = 6s 2 + 24 (s + 12s) + (6s 2 + 24) 1+ 3 s + 12s y 22 =

6s 2 + 24 1 = + ZA 3 s + 12s s C 3

where Z A is shown in the figure below.

(1)

C1

C3

L2

ZA

y22

We now obtain C 3 and Z A using partial fraction expansion. Let

6s 2 + 24 A Bs + C = + s (s 2 + 12) s s 2 + 12

6s 2 + 24 = A (s 2 + 12) + Bs 2 + Cs Equating coefficients : s0 : 24 = 12A  → A = 2 1 s : 0=C 2 s : 6= A+B  → B = 4 Thus, 6s 2 + 24 2 4s = + 2 2 s (s + 12) s s + 12

(2)

Comparing (1) and (2), 1 1 C3 = = F A 2

But

1 s 2 + 12 1 3 = = s+ 4s 4 ZA s

(3)

1 1 = sC1 + ZA s L2

(4)

Comparing (3) and (4), 1 1 C1 = F and L2 = H 4 3 Therefore, C1 = 0.25 F , L 2 = 0.3333 H ,

C 3 = 0.5 F

Chapter 19, Solution 98 ∆ h = 1 − 0 .8 = 0 .2  − ∆ h / h 21 − h11 / h 21   − 0.001 [Ta ] = [Tb ] =  = −6 − h 22 / h 21 − 1 / h 21  − 2.5x10

− 10  − 0.005

2.6x10−5 0.06  [T] = [Ta ][Tb ] =   −8 5x10−5  1.5x10 We now convert this to z-parameters A / C ∆ T / C 1.733x103 [z] =  = 7  1 / C D / C  6.667 x10 1000

I1

0.0267   3.33x103 

z11

+

z22 +

I2 +

+ Vs

z12 I2

-

-

z21 I1

Vo

ZL -

Vs = (1000 + z11)I1 + z12 I 2

(1)

Vo = z 22 I 2 + z 21I1

(2)

But Vo = −I 2 ZL

→

I 2 = −Vo / ZL

(3)

Substituting (3) into (2) gives   1 z I1 = Vo  + 22   z 21 z 21ZL 

We substitute (3) and (4) into (1)

(4)

  1 z z Vs = (1000 + z11) + 22  Vo − 12 Vo ZL  z11 z 21ZL  = 7.653x10− 4 − 2.136 x10−5 = 744µV

Chapter 19, Solution 99 Z ab = Z1 + Z 3 = Z c || (Z b + Z a ) Z1 + Z 3 =

Z c (Z a + Z b ) Za + Zb + Zc

(1)

Z cd = Z 2 + Z 3 = Z a || (Z b + Z c ) Z2 + Z3 =

Z a (Z b + Z c ) Za + Zb + Zc

(2)

Z ac = Z1 + Z 2 = Z b || (Z a + Z c ) Z1 + Z 2 =

Z b (Z a + Z c ) Za + Zb + Zc

(3)

Z b (Z c − Z a ) Za + Zb + Zc

(4)

Subtracting (2) from (1), Z1 − Z 2 =

Adding (3) and (4), Z1 =

ZbZc Za + Zb + Zc

(5)

Subtracting (5) from (3), Z2 =

ZaZb Za + Zb + Zc

(6)

Subtracting (5) from (1),

Z3 =

ZcZa Za + Zb + Zc

(7)

Using (5) to (7) Z1Z 2 + Z 2 Z 3 + Z 3 Z1 =

Z a Z b Z c (Z a + Z b + Z c ) (Z a + Z b + Z c ) 2

Z1Z 2 + Z 2 Z 3 + Z 3 Z1 =

Za ZbZc Za + Zb + Zc

(8)

Dividing (8) by each of (5), (6), and (7), Z1Z 2 + Z 2 Z 3 + Z 3 Z1 Za = Z1

Zb =

Z1Z 2 + Z 2 Z 3 + Z 3 Z1 Z3

Zc =

Z1Z 2 + Z 2 Z 3 + Z 3 Z1 Z2

as required. Note that the formulas above are not exactly the same as those in Chapter 9 because the locations of Z b and Z c are interchanged in Fig. 18.122.