CHAP. 10]

In this case hi1 i2 i ¼ 0 because hcos ð2000t  458Þ cos ð4000t  3:48Þi ¼ 0. applies and P ¼ P1 þ P2 ¼ 0:625 þ 1 ¼ 1:625 W. (b) The current in the coil is i ¼ 0:61 cos ð2000t  1358Þ (see Example 9.7). coil is P ¼ RI 2 =2 ¼ 5  ð0:61Þ2 ¼ 1:875 W. Note that P > P1 þ P2 . (c)

235

AC POWER

Therefore, superposition of power The average power dissipated in the

By applying superposition in the time domain we ﬁnd i1 ¼ 0:35 cos ð2000t  458Þ; P1 ¼ 0:625 W pﬃﬃﬃ i2 ¼ 0:41 cos ð1000 2t  35:38Þ; P2 ¼ 0:833 W

pﬃﬃﬃ i ¼ i1  i2 ; P ¼ hRi2 =2i ¼ P1 þ P2  1:44hcos ð2000t  458Þ cos ð1000 2t  35:38Þi pﬃﬃﬃ The term hcos ð2000t  458Þ cos ð1000 2t  35:38Þi is not determined because a common period can’t be found. The average power depends on the duration of averaging.

Solved Problems 10.1

The current plotted in Fig. 10-2(a) enters a 0.5-mF capacitor in series with a 1-k resistor. Find and plot (a) v across the series RC combination and (b) the instantaneous power p entering RC. (c) Compare the results with Examples 10.1 and 10.2. (a) Referring to Fig. 10-2(a), in one cycle of the current the voltages are  1V ð0 < t < 1 msÞ vR ¼ 1 V ð1 < t < 2 msÞ  ðt 2000t ðVÞ ð0 < t < 1 msÞ 1 i dt ¼ vC ¼ C 0 4  2000t ðVÞ ð1 < t < 2 msÞ  1 þ 2000t ðVÞ ð0 < t < 1 msÞ v ¼ vR þ vC ¼ [See Fig. 10-13ðaÞ 3  2000t ðVÞ ð1 < t < 2 msÞ (b) During one cycle, pR ¼ Ri2 ¼ 1 mW  2000t ðmWÞ ð0 < t < 1 msÞ pC ¼ vC i ¼ 2000t  4 ðmWÞ ð1 < t < 2 msÞ  1 þ 2000t ðmWÞ ð0 < t < 1 msÞ p ¼ vi ¼ pR þ pC ¼ 2000t  3 ðmWÞ ð1 < t < 2 msÞ (c)

10.2

The average power entering the circuit during one cycle is equal to the average power absorbed by the resistor. It is the same result obtained in Example 10.1. The power exchanged between the source and the circuit during one cycle also agrees with the result obtained in Example 10.2.

A 1-V ac voltage feeds (a) a 1- resistor, (b) a load Z ¼ 1 þ j, and (c) a load Z ¼ 1  j. Find P in each of the three cases. (a) P ¼ V 2 =R ¼ 1=1 ¼ 1 W pﬃﬃﬃ (b) and (c) jZj ¼ j1  jj ¼ 2.

10.3

[(See Fig. 10-13ðbÞ

pﬃﬃﬃ I ¼ V=jZj ¼ 1= 2.

P ¼ RI 2 ¼ 0:5 W

Obtain the complete power information for a passive circuit with an applied voltage v ¼ 150 cos ð!t þ 108Þ V and a resulting current i ¼ 5:0 cos ð!t  508Þ A. Using the complex power

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An