Page 30 - Schaum's Outline of Theory and Problems of Electric Circuits

CIRCUIT CONCEPTS
CHAP. 2]

Fig. 2-18 19

p a ¼ v a i ¼ ð20Þð 10Þ¼ 200 W
p b ¼ v b i ¼ð50Þð 10Þ¼ 500 W
Since power delivered is the negative of power absorbed, source v b delivers 500 W and source v a absorbs
200 W. The power in the two resistors is 300 W.

2.12 A 25.0-
resistance has a voltage v ¼ 150:0 sin 377t (V). Find the power p and the average power
p avg over one cycle.
i ¼ v=R ¼ 6:0 sin 377t ðAÞ
2
p ¼ vi ¼ 900:0 sin 377t ðWÞ
The end of one period of the voltage and current functions occurs at 377t ¼ 2 . For P avg the
integration is taken over one-half cycle, 377t ¼ . Thus,
ð
1
2
P avg ¼ 900:0 sin ð377tÞdð377tÞ¼ 450:0 ðWÞ
0

2.13 Find the voltage across the 10.0-
resistor in Fig. 2-19 if the control current i x in the dependent
source is (a) 2 A and (b) 1A.
i ¼ 4i x 4:0; v R ¼ iR ¼ 40:0i x 40:0 ðVÞ
i x ¼ 2; v R ¼ 40:0V
i x ¼ 1; v R ¼ 80:0V

Fig. 2-19

Supplementary Problems

2.14 A resistor has a voltage of V ¼ 1:5 mV. Obtain the current if the power absorbed is (a) 27.75 nW and
(b) 1.20 mW. Ans. 18.5 mA, 0.8 mA