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[CHAP. 11

load; likewise, the second term in WC is PCB . Adding the two equations and recombining the middle terms then yields WA þ WC ¼ PAB þ Re ½ðVAB eff  VCB eff ÞIAC eff  þ PCB ¼ PAB þ PAC þ PCB since, by KVL, VAB  VCB ¼ VAC . The same reasoning establishes the analogous result for a Y-connected load. Balanced Loads When three equal impedances Z  are connected in delta, the phase currents make 308 angles with their resultant line currents. Figure 11-21 corresponds to Fig. 11-20 under the assumption of ABC sequencing. It is seen that VAB leads IA by  þ 308, while VCB leads IC by   308. Consequently, the two wattmeters will read WA ¼ VAB eff IA eff cos ð þ 308Þ

WC ¼ VCB eff IC eff cos ð  308Þ

or, since in general we do not know the relative order in the voltage sequence of the two lines chosen for the wattmeters, W1 ¼ VL eff IL eff cos ð þ 308Þ W2 ¼ VL eff IL eff cos ð  308Þ These expressions also hold for a balanced Y-connection.

Fig. 11-21

Elimination of VL eff IL eff between the two readings leads to   pffiffiffi W2  W1 tan  ¼ 3 W2 þ W1 Thus, from the two wattmeter readings, the magnitude of the impedance angle  can be inferred. The sign of tan  suggested by the preceding formula is meaningless, since the arbitrary subscripts 1 and 2 might just as well be interchanged. However, in the practical case, the balanced load is usually known to be inductive ð > 0Þ.

Solved Problems 11.1

The two-phase balanced ac generator of Fig. 11-22 feeds two identical loads. The two voltage sources are 1808 out of phase. Find (a) the line currents, voltages, and their phase angles, and (b) the instantaneous and average powers delivered by the generator.