240

AC POWER

[CHAP. 10

10.14 Determine the total power information for three parallel-connected loads: load #1, 250 VA, pf ¼ 0:50 lagging; load #2, 180 W, pf ¼ 0:80 leading; load #3, 300 VA, 100 var (inductive). Calculate the average power P and the reactive power Q for each load. Load #1 Given s ¼ 250 VA, cos  ¼ 0:50 lagging. Then, qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ P ¼ 250ð0:50Þ ¼ 125 W Q ¼ ð250Þ2  ð125Þ2 ¼ 216:5 var (inductive) Load #2

Given P ¼ 180 W, cos  ¼ 0:80 leading. Then,  ¼ cos1 0:80 ¼ 36:878 and Q ¼ 180 tanð36:878Þ ¼ 135 var (capacitive)

Given S ¼ 300 VA, Q ¼ 100 var (inductive). Then, qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ P ¼ ð300Þ2  ð100Þ2 ¼ 282:8 W

Combining componentwise: PT ¼ 125 þ 180 þ 282:8 ¼ 587:8 W QT ¼ 216:5  135 þ 100 ¼ 181:5 var (inductive) ST ¼ 587:8 þ j181:5 ¼ 615:2 17:168 Therefore, ST ¼ 615:2 VA and pf ¼ cos 17:168 ¼ 0:955 lagging.

10.15 Obtain the complete power triangle and the total current for the parallel circuit shown in Fig. 1019, if for branch 2, S2 ¼ 1490 VA.

Fig. 10-19 2 From S2 ¼ I2;eff Z2 ,

1490 2 I2;eff ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ 222 A2 32 þ 62 and, by current division, I1 3 þ j6 ¼ I2 2 þ j3

whence

2 I1;eff ¼

32 þ 62 2 45 ð222Þ ¼ 768 A2 I2;eff ¼ 13 22 þ 32

2 S1 ¼ I1;eff Z1 ¼ 768ð2 þ j3Þ ¼ 1536 þ j2304

Then,

2 Z2 ¼ 222ð3 þ j6Þ ¼ 666 þ j1332 S2 ¼ I2;eff ST ¼ S1 þ S2 ¼ 2202 þ j3636

that is, PT ¼ 2202 W, QT ¼ 3636 var (inductive), ST ¼

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð2202Þ2 þ ð3636Þ2 ¼ 4251 VA

and

pf ¼

2202 ¼ 0:518 lagging 4251

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