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CHAP. 11]

265

POLYPHASE CIRCUITS

Fig. 11-29 144:3 08 144:3 08 þ ¼ 42:0 9:98 A 5 08 10 308 pffiffiffi pffiffiffi P ¼ 3VL eff IL eff cos  ¼ 3ð176:8Þð29:7Þ cos 9:98 ¼ 8959 IL ¼

Then

11.9

W

Obtain the readings when the two-wattmeter method is applied to the circuit of Problem 11.8. The angle on IL , 9:98, is the negative of the angle on the equivalent impedance of the parallel combination of 5 08  and 10 308 . Therefore,  ¼ 9:98 in the formulas of Section 11.13. W1 ¼ VL eff IL eff cos ð þ 308Þ ¼ ð176:8Þð29:7Þ cos 39:98 ¼ 4028 W W2 ¼ VL eff IL eff cos ð  308Þ ¼ ð176:8Þð29:7Þ cos ð20:18Þ ¼ 4931 W As a check, W1 þ W2 ¼ 8959 W, in agreement with Problem 11.8.

11.10 A three-phase supply, with an effective line voltage 240 V, has an unbalanced -connected load shown in Fig. 11-30. Obtain the line currents and the total power.

Fig. 11-30 The power calculations can be performed without knowledge of the sequence of the system. effective values of the phase currents are IAB eff ¼

240 ¼ 9:6 A 25

IBC eff ¼

240 ¼ 16 A 15

ICA eff ¼

240 ¼ 12 A 20

Hence, the complex powers in the three phases are SAB ¼ ð9:6Þ2 ð25 908Þ ¼ 2304 908 ¼ 0 þ j2304 SBC ¼ ð16Þ2 ð15 308Þ ¼ 3840 308 ¼ 3325 þ j1920 SCA ¼ ð12Þ2 ð20 08Þ ¼ 2880 08 ¼ 2880 þ j0

The

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