Solution Manual - Fundamentals of Electric Circuits 3rd Edition Chapter 07 PDF

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Chapter 7, Problem 1. In the circuit shown in Fig. 7.81 v(t ) = 56e −200t V, t > 0 i (t ) = 8e −200t mA,

t>0

(a) Find the values of R and C. (b) Calculate the time constant τ . (c) Determine the time required for the voltage to decay half its initial value at t = 0.

Figure 7.81 For Prob. 7.1 Chapter 7, Solution 1. τ=RC = 1/200

(a)

For the resistor, V=iR= 56e −200t = 8Re −200t x10 −3 C=

1 200 R

=

(b) (c) 1 2

x 56 = 56e

200 to = ln 2

1 200 X7 X10

3

⎯⎯ →

R=

56 8

= 7 kΩ

= 0.7143 µ F

τ =1/200= 5 ms If value of the voltage at = 0 is 56 . −200 t

⎯⎯ →

⎯⎯ →

to =

e 200 t = 2 1 200

ln 2 = 3.466 m s

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Chapter 7, Problem 2. Find the time constant for the RC circuit in Fig. 7.82.

Figure 7.82 For Prob. 7.2. 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, Problem 3. Determine the time constant for the circuit in Fig. 7.83.

Figure 7.83 For Prob. 7.3. Chapter 7, Solution 3. R = 10 +20//(20+30) =10 + 40x50/(40 + 50)=32.22 kΩ

τ = RC = 32.22X10 3 X100 X10 −12 = 3.222 µ S PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 7, Problem 4. The switch in Fig. 7.84 moves instantaneously from A to B at t = 0. Find v for t > 0.

Figure 7.84 For Prob. 7.4.

Chapter 7, Solution 4. For t0. we have a source-free RC circuit.

τ = RC = 2x103 x10 x10 −6 = 0.02 −t − t v( t) = v(0)e / τ = 40 e 50 V

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Chapter 7, Problem 5. For the circuit shown in Fig. 7.85, find i(t), t > 0.

Figure 7.85 For Prob. 7.5. Chapter 7, Solution 5. Let v be the voltage across the capacitor. For t 0, we have a source-free RC circuit as shown below.

i 5Ω 4Ω

+ v –

1/3 F

1 3 = 16 e − t / 3

τ = RC = (4 + 5) = 3 s

v( t) = v(0)e −t / τ 1 1 dv i( t) = −C = − ( − )16 e − t / 3 = 1.778 e − t / 3 A 3 3 dt

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Chapter 7, Problem 6. The switch in Fig. 7.86 has been closed for a long time, and it opens at t = 0. Find v(t) for t ≥ 0.

Figure 7.86 For Prob. 7.6.

Chapter 7, Solution 6.

vo = v(0) =

2 (24) = 4 V 10 + 2

v( t ) = vo e −t / τ , τ = RC = 40x10−6 x 2x103 =

2 25

v( t ) = 4e −12.5t V

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Chapter 7, Problem 7. Assuming that the switch in Fig. 7.87 has been in position A for a long time and is moved to position B at t =0, find v 0 (t) for t ≥ 0.

Figure 7.87 For Prob. 7.7.

Chapter 7, Solution 7. When the switch is at position A, the circuit reaches steady state. By voltage division, vo (0) =

40 (12V ) = 8V 40 + 20

When the switch is at position B, the circuit reaches steady state. By voltage division, vo (∞) =

30

(12V ) = 7.2V 30 + 20 20 x30 RTh = 20 k / / 30 k = = 12 kΩ 50 τ = RThC = 12x103 x 2 x10 −3 = 24 s

vo (t) = vo (∞) + [v o (0) − v o (∞ )]e

− t/ τ

= 7.2 + (8 − 7.2)e − t/ 24 = 7.2 + 0.8e − t / 24 V

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Chapter 7, Problem 8. For the circuit in Fig. 7.88, if v = 10e −4t V and

i = 0.2e − 4t A, t > 0

(a) Find R and C. (b) Determine the time constant. (c) Calculate the initial energy in the capacitor. (d) -Obtain the time it takes to dissipate 50 percent of the initial energy.

Figure 7.88 For Prob. 7.8. 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 0.5 = 1 − e -8t 0 ⎯ ⎯→ e -8t 0 = 2 8t 0 or e =2 1 t 0 = ln (2) = 86.6 ms 8 R=

(b) (c) (d)

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Chapter 7, Problem 9. The switch in Fig. 7.89 opens at t = 0. Find v 0 for t > 0

Figure 7.89 For Prob. 7.9. Chapter 7, Solution 9. For t < 0, the switch is closed so that vo (0) =

4 2+4

(6) = 4 V

For t >0, we have a source-free RC circuit. τ = RC = 3 x10−3 x4 x103 = 12 s vo (t) = vo (0)e

−t / τ

= 4e −t/ 12 V

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Chapter 7, Problem 10. For the circuit in Fig. 7.90, find v 0 (t) for t > 0. Determine the time necessary for the capacitor voltage to decay to one-third of its value at t = 0.

Figure 7.90 For Prob. 7.10. Chapter 7, Solution 10. For t0, we have a source-free RC circuit τ =RC = 3 x10 3 x 20 x10 − 6 =0.06 s vo(t) = 9e–16.667t V Let the time be to. 3 = 9e–16.667to or e16.667to = 9/3 = 3 to = ln(3)/16.667 = 65.92 ms.

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Chapter 7, Problem 11. For the circuit in Fig. 7.91, find i 0 for t > 0.

Figure 7.91 For Prob. 7.11. Chapter 7, Solution 11. For t0, we have a source-free RL circuit. 4 L τ= = = 1/ 3 R 4+ 8 −t / −3 t A io (t) = io (0)e τ = 1.4118 e PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 7, Problem 12. The switch in the circuit of Fig. 7.92 has been closed for a long time. At t = 0 the switch is opened. Calculate i(t) for t > 0.

Figure 7.92 For Prob. 7.12. 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)

(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

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Chapter 7, Problem 13. In the circuit of Fig. 7.93,

v(t) = 20e −10 t V,

t>0

i(t) = 4e −10 t mA,

t>0

3

3

(a) Find R, L, and τ . (b) Calculate the energy dissipated in the resistance for 0 < t < 0.5 ms.

Figure 7.93 For Prob. 7.13. Chapter 7, Solution 13. (a) τ =

1 1 3 = ms 10

→ 20 e −1000 t = Rx4 e −1000 t x10 −3 v = iR ⎯⎯ From this, R = 20/4 kΩ= 5 kΩ 5 x1000 L L= ⎯⎯→ = 5H But τ = = 1 3 10 R 1000 (b) The energy dissipated in the resistor is 80 x10 − − 2x 103 t w = ∫ p d t = ∫ 80x 10 −3e −2 x10 d t = − e 2x103 0 0 t

t

3

3

0.5 x10 −3 0

= 40(1 − e −1)µ J = 25.28 µ J

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Chapter 7, Problem 14. Calculate the time constant of the circuit in Fig. 7.94.

Figure 7.94 For Prob. 7.14. Chapter 7, Solution 14. RTh = (40 + 20)/ / (10 + 30) =

τ = L/ R =

60 x40 = 24 kΩ 100

5 x10 −3 = 0.2083µ s 24 x10 3

Chapter 7, Problem 15. Find the time constant for each of the circuits in Fig. 7.95.

Figure 7.95 For Prob. 7.15. Chapter 7, Solution 15 (a) RTh = 12 + 10 // 40 = 20Ω , (b) RTh = 40 // 160 + 8 = 40Ω ,

L = 5 / 20 = 0.25s R Th L τ= = (20 x10− 3 ) / 40 = 0.5 ms RTh

τ=

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Chapter 7, Problem 16. Determine the time constant for each of the circuits in Fig. 7.96.

Figure 7.96 For Prob. 7.16.

Chapter 7, Solution 16.

τ= (a)

L eq R eq L eq = L and R eq = R 2 +

τ=

(b)

R1 R 3 R (R + R 3 ) + R 1 R 3 = 2 1 R1 + R3 R1 + R 3

L ( R1 + R 3 ) R2 (R1 + R 3 ) + R1 R3 L 1L 2 R1 R 2 R 3 (R 1 + R 2 ) + R 1R 2 = and R eq = R 3 + L1 + L 2 R1 + R 2 R1 + R 2 L1 L2 (R1 + R 2 ) τ= ( L 1 + L 2 ) ( R 3 ( R 1 + R 2 ) + R 1R 2 )

where L eq =

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Chapter 7, Problem 17. Consider the circuit of Fig. 7.97. Find v 0 (t) if i(0) = 2 A and v(t) = 0.

Figure 7.97 For Prob. 7.17.

Chapter 7, Solution 17. i( t ) = i(0) e -t τ ,

τ=

L 14 1 = = Req 4 16

i( t ) = 2 e-16t v o ( t ) = 3i + L

di = 6 e -16t + (1 4)(-16) 2 e -16t dt

vo ( t ) = - 2 e -16t u( t )V

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Chapter 7, Problem 18. For the circuit in Fig. 7.98, determine v 0 (t) when i(0) = 1 A and v(t) = 0.

Figure 7.98 For Prob. 7.18. Chapter 7, Solution 18. If v( t ) = 0 , the circuit can be redrawn as shown below.

6 L 2 5 1 , τ= = × = 5 R 5 6 3 -t τ -3t i( t ) = i(0) e = e di - 2 (-3) e -3t = 1.2 e -3t V vo ( t ) = -L = dt 5 R eq = 2 || 3 =

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Chapter 7, Problem 19. In the circuit of Fig. 7.99, find i(t) for t > 0 if i(0) = 2 A.

Figure 7.99 For Prob. 7.19. Chapter 7, Solution 19. i

1V

i1

i2

− +

10 Ω

i1

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 But i = i2 + i 2 and i = i1 i.e. i1 = 2 i 2 = i 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 u( t ) A

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Chapter 7, Problem 20.

For the circuit in Fig. 7.100, v = 120e −50t V

and i = 30e −50t A, t > 0

(a) Find L and R. (b) Determine the time constant. (c) Calculate the initial energy in the inductor. (d) What fraction of the initial energy is dissipated in 10 ms?

Figure 7.100 For Prob. 7.20. Chapter 7, Solution 20.

(a)

L 1 = R 50 di -v= L dt τ=

⎯ ⎯→ R = 50 L

- 120 e- 50t = L(30)(-50) e- 50t ⎯ ⎯→ L = 80 mH R = 50 L = 4 Ω L 1 (b) τ= = = 20 ms R 50 1 1 (c) w = L i 2 (0) = (0.08)(30) 2 = 36J 2 2 The value of the energy remaining at 10 ms is given by: w10 = 0.04(30e–0.5)2 = 0.04(18.196)2 = 13.24J. So, the fraction of the energy dissipated in the first 10 ms is given by: (36–13.24)/36 = 0.6322 or 63.2%.

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Chapter 7, Problem 21. In the circuit of Fig. 7.101, find the value of R stored in the inductor will be 1 J.

for which the steady-state energy

Figure 7.101 For Prob. 7.21. Chapter 7, Solution 21. The circuit can be replaced by its Thevenin equivalent shown below.

Rth

Vth

+ −

2H

80 (60) = 40 V 80 + 40 80 R th = 40 || 80 + R = +R 3 V 40 I = i(0) = i(∞ ) = th = R th 80 3 + R Vth =

1 1 ⎛ 40 ⎞ ⎟ =1 w = L I 2 = (2)⎜ 2 2 ⎝ R + 80 3 ⎠ 40 40 =1 ⎯ ⎯→ R = R + 80 3 3 R = 13.333 Ω 2

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Chapter 7, Problem 22. Find i(t) and v(t) for t > 0 in the circuit of Fig. 7.102 if i(0) = 10 A.

Figure 7.102 For Prob. 7.22.

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

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Chapter 7, Problem 23. Consider the circuit in Fig. 7.10...