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18

CIRCUIT CONCEPTS

[CHAP. 2

The element cannot be a resistor since v and i are not proportional. v is an integral of i. For 2 ms < t < 4 ms, i 6¼ 0 but v is constant (zero); hence the element cannot be a capacitor. For 0 < t < 2 ms, di ¼ 5  103 A=s dt

and

v ¼ 15 V

Consequently,  L¼v

di ¼ 3 mH dt

(Examine the interval 4 ms < t < 6 ms; L must be the same.)

2.9

Obtain the voltage v in the branch shown in Fig. 2-16 for (c) i2 ¼ 0 A.

(a) i2 ¼ 1 A, (b) i2 ¼ 2 A,

Voltage v is the sum of the current-independent 10-V source and the current-dependent voltage source vx . Note that the factor 15 multiplying the control current carries the units . ðaÞ

v ¼ 10 þ vx ¼ 10 þ 15ð1Þ ¼ 25 V

ðbÞ

v ¼ 10 þ vx ¼ 10 þ 15ð2Þ ¼ 20 V

ðcÞ

v ¼ 10 þ 15ð0Þ ¼ 10 V

Fig. 2-16

2.10

Find the power absorbed by the generalized circuit element in Fig. 2-17, for (b) v ¼ 50 V.

Fig. 2-17 Since the current enters the element at the negative terminal, ðaÞ ðbÞ

2.11

p ¼ vi ¼ ð50Þð8:5Þ ¼ 425 W p ¼ vi ¼ ð50Þð8:5Þ ¼ 425 W

Find the power delivered by the sources in the circuit of Fig. 2-18. i¼ The powers absorbed by the sources are:

20  50 ¼ 10 A 3

(a) v ¼ 50 V,

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