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Using Thevenin's Theorem:

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Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is d. v f (t )  K 1 2 r 2  4r  4  0 r 2  2r  2  0 2 48 r  1  j1 2 v n (t )  K 2 e  t cos t  K 3 e  t sin t

v(t )  K 1  K 2 e  t cos t  K 3 e  t sin t

Chapter 14: Application of the Laplace Transform To Circuit Analysis

Problem 14.FE-1

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is c. 4 I ( s)  2 sI ( s )  2 I ( s ) s 4  V s ( s )    2 s  2 I ( s ) s  V0 ( s )  2 I ( s ) V (s) I (s)  0 2 4  V ( s)  Vs ( s )    2 s  2  0  s  2  V0 ( s ) s  2 Vs ( s ) s  s  2 Vs ( s ) 

Chapter 14: Application of the Laplace Transform To Circuit Analysis

Problem 14.FE-2

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is a.    s4   I s (s) I 0 ( s)   s4 3   s 

I 0 (s) s( s  4)  2 I s (s) s  4s  3

Chapter 14: Application of the Laplace Transform To Circuit Analysis

Problem 14.FE-3

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is b. KVL:

4s 4     s   2  I (s) s s 1  

I (s) 

4s 2 ( s 2  1)( s 2  2 s  4)

2

Chapter 14: Application of the Laplace Transform To Circuit Analysis

Problem 14.FE-4

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is d. KVL: Vs ( s )  4 I ( s)  2 sI ( s ) Vs ( s )  (2 s  4) I ( s ) V0 ( s)  2 sI ( s) V (s) I (s)  0 2s V s ( s )  ( 2 s  4)

V0 ( s ) 2s

V0 ( s ) s  Vs ( s ) s  2

Chapter 14: Application of the Laplace Transform To Circuit Analysis

Problem 14.FE-5