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A three-phase, three-wire system, with an e ective line voltage 176.8 V, supplies two balanced loads, one in delta with Z 15 08  and the other in wye with ZY 10 308 . Obtain the total power.
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First convert the -load to Y, and then use the single-line equivalent circuit, Fig. 11-29, to obtain the line current.
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Fig. 11-29 144:3 08 144:3 08 42:0 9:98 A 5 08 10 308 p p P 3VL eff IL eff cos  3 176:8 29:7 cos 9:98 8959 IL
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Obtain the readings when the two-wattmeter method is applied to the circuit of Problem 11.8.
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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,  9:98 in the formulas of Section 11.13. W1 VL eff IL eff cos  308 176:8 29:7 cos 39:98 4028 W W2 VL eff IL eff cos  308 176:8 29:7 cos 20:18 4931 W As a check, W1 W2 8959 W, in agreement with Problem 11.8.
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11.10 A three-phase supply, with an e ective line voltage 240 V, has an unbalanced -connected load shown in Fig. 11-30. Obtain the line currents and the total power.
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Fig. 11-30 The power calculations can be performed without knowledge of the sequence of the system. e ective values of the phase currents are IAB eff 240 9:6 A 25 IBC eff 240 16 A 15 ICA eff 240 12 A 20 The
Hence, the complex powers in the three phases are SAB 9:6 2 25 908 2304 908 0 j2304 SBC 16 2 15 308 3840 308 3325 j1920 SCA 12 2 20 08 2880 08 2880 j0
POLYPHASE CIRCUITS
[CHAP. 11
and the total complex power is their sum, ST 6205 j4224 that is, PT 6205 W and QT 4224 var (inductive). To obtain the currents a sequence must be assumed; let it be ABC. IAB IBC ICA 339:4 1208 13:6 308 A 25 908 339:4 08 22:6 308 A 15 308 339:4 2408 17:0 2408 A 20 08
Then, using Fig. 11-7(a),
The line currents are obtained by applying KCL at the junctions. IA IAB IAC 13:6 308 17:0 2408 29:6 46:78 A
IB IBC IBA 22:6 308 13:6 308 19:7 66:78 A IC ICA ICB 17:0 2408 22:6 308 28:3 173:18 A
11.11 Obtain the readings of wattmeters placed in lines A and B of the circuit of Problem 11.10 (Line C is the potential reference for both meters.)
   29:6 WA Re VAC eff I eff Re 240 608 p 46:78 A 2 Re 5023 13:38 4888 W    19:7 WB Re VBC eff I eff Re 240 08 p 66:78 B 2 Re 3343 66:78 1322 W Note that WA WB 6210 W, which agrees with PT as found in Problem 11.10.
11.12 A three-phase, four-wire, ABC system, with line voltage VBC 294:2 08 V, has a Y-connected load of ZA 10 08 , ZB 15 308 , and ZC 10 308  (Fig. 11-31). Obtain the line and neutral currents.
Fig. 11-31
CHAP. 11]
POLYPHASE CIRCUITS
169:9 908 16:99 908 A 10 08 169:9 308 11:33 608 A IB 15 308 169:9 1508 16:99 1208 A IC 10 308 IN IA IB IC 8:04 69:58 A IA
11.13 The Y-connected load impedances ZA 10 08 , Z 15 308 , and ZC 10 308 , in Fig. 11-32, are supplied by a three-phase, three-wire, ABC system in which VBC 208 08 V. Obtain the voltages across the impedances and the displacement neutral voltage VON .
Fig. 11-32
The method of Example 11.7 could be applied here and one node-voltage equation solved. However, the mesh currents I1 and I2 suggested in Fig. 11-32 provide another approach.      I1 10 08 15 308 208 1208 15 308 15 308 10 308 I2 15 308 208 08 Solving, I1 14:16 86:098 A and I2 10:21 52:418 A. IA I1 14:16 86:098 A The line currents are then IC I2 10:21 127:598 A
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