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Chapter 1, Solution 2

(a) i = dq/dt = 3 mA(b) i = dq/dt = (16t + 4) A

(c) i = dq/dt = (-3e

-t

+ 10e-2t

) nA(d) i=dq/dt = 1200 120 cos t pA(e) i =dq/dt = e t t t 4 80 50 1000 50( cos sin ) A

Chapter 1, Solution 8

C15 μ1102

110idtq

Chapter 1, Solution 12

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

q t idt q tdt t t t

( ) ( ) . 0 3 0 1 50

2

0

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

q t idt q dt t t t

( ) ( ) 6 18 54 18 546 6

At t=10, q(10) = 180 – 54 = 126For 10

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The plot of the charge is shown below.

Chapter 1, Solution 18

Calculate the power absorbed or supplied by each element in Fig. 1.29

4A + 6V – I o = 3A + 10 V –

4A+ 13V 3I o +

– –2

3A – 5V +(a) (b)

Figure 1.29For Prob. 1.18

0 5 1 0 1 5 2 00

2 0

4 0

6 0

8 0

1 0 0

1 2 0

1 4 0

t

q ( t )

24V9V2

1

+

–

+

–

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Solution: Note, we will express absorbed power as positive terms and delivered power asnegative power absorbed.

(a) For the 9-V voltage source, p = -4 (9) = –36 W For element 1, p = 4 (6) = 24W

For element 2, p = 4 (3) = 12W

(b) For the 24-V voltage source, p = 24 (-3) = –72W For the current-controlled voltage source, 0 p 3 I (3) 27W = 27 WFor element 1, p = 10 (3) = 30 W For element 2, p = 3 (5) = 15 W

Chapter 1, Solution 26

A 12-V car battery supported a current of 150mA to a bulb. Calculate: (a) the power absorbed bythe bulb, (b) the energy absorbed by the bulb over an interval of 20 minutes.

Solution:

3(a) p = vi = 12 150 10 = 1.8 W

(b) w = pt = 1.8 20 60 = 2.16 kJ