Chapter 16, Problem 1. - University of Nevada, Las Vegas

[Pages:88]Chapter 16, Problem 1. Determine i(t) in the circuit of Fig. 16.35 by means of the Laplace transform.

Figure 16.35 For Prob. 16.1.

Chapter 16, Solution 1.

Consider the s-domain form of the circuit which is shown below.

1

I(s)

1/s +

1/s

-

s

1s

1

1

I(s) = 1 + s + 1 s = s2 + s + 1 = (s + 1 2)2 + ( 3 2)2

i(t) =

2 3

e-t

2

sin

3 2

t

i(t) = 1.155e-0.5t sin (0.866t) A

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Chapter 16, Problem 2. Find v x in the circuit shown in Fig. 16.36 given v s .= 4u(t)V.

Figure 16.36 For Prob. 16.2.

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Chapter 16, Solution 2.

s

8/s

+

+

4

-

Vx 2

4

s

-

Vx

-

4 s

+

Vx

-0

+

Vx

-0

=

0

s

2

4+ 8

s

Vx (4s + 8) -

(16s + 32) s

+ (2s2

+

4s)Vx

+ s2Vx

=

0

Vx

(3s 2

+

8s

+

8)

=

16s + s

32

Vx

=

-16 s + 2 s(3s2 + 8s + 8)

=

-16

0.25 s

+

- 0.125 s+4+ j

8

+

- 0.125 s+ 4 - j

8

33

3 3

v x = (-4 + 2e-(1.3333+ j0.9428)t + 2e-(1.3333- j0.9428)t )u(t) V

vx

=

4u(t) - e-4t / 3

cos

2

2 3

t -

6 2

e-4t

/

3

sin

2

2 3

t

V

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Chapter 16, Problem 3. Find i(t) for t > 0 for the circuit in Fig. 16.37. Assume i s = 4u(t) + 2 (t)mA. (Hint: Can we use superposition to help solve this problem?)

Figure 16.37 For Prob. 16.3.

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Chapter 16, Solution 3.

In the s-domain, the circuit becomes that shown below.

4 +2 s

1 I

2

0.2s

We transform the current source to a voltage source and obtain the circuit shown below.

2 1

I

8 +4

+

s

_

0.2s

I=

8+4 s

= 20s + 40 = A +

B

3 + 0.2s s(s +15) s s +15

A = 40 = 8 , B = -15x20 + 40 = 52

15 3

-15

3

I = 8 / 3 + 52 / 3 s s +15

i(t)

=

8 3

+

52 3

e-15t

u(t

)

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Chapter 16, Problem 4. The capacitor in the circuit of Fig. 16.38 is initially uncharged. Find v 0 (t) for t > 0.

Figure 16.38 For Prob. 16.4.

Chapter 16, Solution 4.

The circuit in the s-domain is shown below.

I

2

4I

+

+

5 _

Vo 1/s

1

?

I + 4I = Vo 1/ s

But I = 5 -Vo 2

5I = sVo

5

5

- Vo 2

=

sVo

Vo

=

12.5 s+5/2

vo (t) = 12.5e-2.5t V

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 16, Problem 5. If i s (t) = e -2t u(t) A in the circuit shown in Fig. 16.39, find the value of i 0 (t).

Figure 16.39 For Prob. 16.5. Chapter 16, Solution 5.

1 s+ 2

Io

s

2

2

s

V

=

s

1 +

2

1 s

+

1 1 2

+

s 2

=

s

1 +

2

s2

2s +s

+

2

=

(s

+

2)(s

+

0.5

+

2s j1.3229)(s

+

0.5

-

j1.3229)

Io

=

Vs 2

=

(s

+

2)(s +

0.5

+

s2 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

( ) io (t) = e-2t + 0.3779e-90?e- t / 2e- j1.3229t + 0.3779e90?e- t / 2e j1.3229t u(t) A

or

( ) = e-2t - 0.7559sin1.3229t u(t) A

or

io(t)

=

e

-

2t

-

2 7

sin

7 2

t u(t)A

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 16, Problem 6. Find v(t), t > 0 in the circuit of Fig. 16.40. Let v s = 20 V.

Figure 16.40 For Prob. 16.6.

Chapter 16, Solution 6.

For t0, the circuit in the s-domain is as shown below.

I

10

+

s

+

20

_

v

s

_

10

100mF = 0.1F

I

20 =s

10 +10 s

=

2 s +1

V = 10I = 20 s +1

1 = 10 sC s

v(t) = 20e-tu(t)

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.

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