Page 29 - Schaum's Outline of Theory and Problems of Electric Circuits

[CHAP. 2
CIRCUIT CONCEPTS
18
The element cannot be a resistor since v and i are not proportional. v is an integral of i. For
2ms < t < 4 ms, i 6¼ 0 but v is constant (zero); hence the element cannot be a capacitor. For 0 < t < 2 ms,
di 3
¼ 5 10 A=s and v ¼ 15 V
dt
Consequently,

di
L ¼ v ¼ 3mH
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 (a) i 2 ¼ 1A, (b) i 2 ¼ 2A,
(c) i 2 ¼ 0A.
Voltage v is the sum of the current-independent 10-V source and the current-dependent voltage source
v x . Note that the factor 15 multiplying the control current carries the units
.
ðaÞ v ¼ 10 þ v x ¼ 10 þ 15ð1Þ¼ 25 V
ðbÞ v ¼ 10 þ v x ¼ 10 þ 15ð 2Þ¼ 20 V

ðcÞ v ¼ 10 þ 15ð0Þ¼ 10 V

Fig. 2-16

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

Fig. 2-17

Since the current enters the element at the negative terminal,
ðaÞ p ¼ vi ¼ ð50Þð8:5Þ¼ 425 W
ðbÞ p ¼ vi ¼ ð 50Þð8:5Þ¼ 425 W

2.11 Find the power delivered by the sources in the circuit of Fig. 2-18.

20 50
i ¼ ¼ 10 A
3
The powers absorbed by the sources are: