Basic Engineering Circuit Analysis Chapter 6 Solution

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S S

s s

S S

t

2.24

/

c

3

t

t

t

3

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is a. Yes. The capacitors should be connected as shown.

6 F 2 F

4 F

C eq 

6  (6  )  3 F 6  6

Chapter 6: Capacitance and Inductance

Ceq

Problem 6.FE-1

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is c. q (t )   i (t ) dt

0 C , t  0  q (t )  6t C , 0  t  1 s 6  C , t  1 s  q (t ) C 0 V , t  0  v(t )  6 x10 6 t V , 0  t  1 s 6 V , t  1 s  v(t ) 

1 C v 2 (t ) 2 0 J , t  0  w(t )  18 x10 6 t 2 J , 0  t  1 s 18 J , t  1 s 

w(t ) 

Chapter 6: Capacitance and Inductance

Problem 6.FE-2

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is b. The voltage across the unknown capacitor Cx is (using KVL): 24  8  V x V x  16V q  Cv The capacitors are connected in series and the charge is the same. q  60  (8)  480  C q 480   30  F Cx   v 16

Chapter 6: Capacitance and Inductance

Problem 6.FE-3

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: 2 mH

Leq

3mH

3mH

6mH

12mH

9mH

The correct answer is d. Leq  [ [(3m  9m) 12m] 6m 3m]  2m Leq  [ [(12m) 12m] 6m 3m]  2m

Leq  [ 6m 6m 3m]  2m Leq  [ 3m 3m]  2m Leq  1.5m  2m Leq  3.5mH

Chapter 6: Capacitance and Inductance

Problem 6.FE-4

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is a. di(t ) v(t )  L dt di(t )  20te  2t (2)  20e  2t  20e  2t  40te  2t dt





v(t )  10m 20e  2t  40te  2t 0 V , t  0 v(t )   2t  2t 0.2e  0.4te V , t  0

Chapter 6: Capacitance and Inductance

Problem 6.FE-5