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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.
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CHAP. 11]
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Fig. 11-22 Let Z jZj  and Ip Vp =jZj. (a) The voltages and currents in phasor domain are VAN Vp 0 VBN Vp 1808 Vp 0 VAB VAN VBN 2Vp 0
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Now, from Ip and Z given above, we have IA Ip  IB Ip 1808  Ip  IN IA IB 0
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(b) The instantaneous powers delivered are pa t va t ia t Vp Ip cos  Vp Ip cos 2!t  pb t vb t ib t Vp Ip cos  Vp Ip cos 2!t  The total instantaneous power pT t is pT t pa t pb t 2Vp Ip cos  2Vp Ip cos 2!t  The average power is Pavg 2VP Ip cos .
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Solve Problem 11.1 given Vp 110 Vrms and Z 4 j3 .
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(a) In phasor form, Z 4 j3 5 36:98 . VAN 110 0 V Then, VBN 110 1808 V V
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VAB VAN VBN 110 0 110 1808 220 0 and IA VAN =Z 22 36:98 A
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IB VBN =Z 22 216:98 22 36:98 A IN IA IB 0
pa t 110 22 cos 36:98 cos 2!t 36:98 1936 2420 cos 2!t 36:98
W W
pb t 110 22 cos 36:98 cos 2!t 36:98 3608 1936 2420 cos 2!t 36:98 p t pa t pb t 3872 4840 cos 2!t 36:98 W Pavg 3872 W
POLYPHASE CIRCUITS
[CHAP. 11
Repeat Problem 11.2 but with the two voltage sources of Problem 11.1 908 out of phase.
(a) Again, Z 5 36:98. Then, VAN 110 0 V VBN 110 908 V
p VAB VAN VBN 110 0 110 908 110 2 458 155:6 458 V and IA VAN =Z 22 36:98 A IB VBN =Z 22 126:98 A
p IN IA IB 22 36:98 22 126:98 22 2 81:98 31:1 81:98 A b pa t 110 22 cos 36:98 cos 2!t 36:98 1936 2420 cos 2!t 36:98 W pb t 110 22 cos 36:98 cos 2!t 36:98 1808 1936 2420 cos 2!t 36:98 p t Pa Pb 2 1936 3872 W Pavg 3872 W W
Show that the line-to-line voltage VL in a three-phase system is voltage VPh .
See the voltage phasor diagram (for the ABC sequence), Fig. 11-23.
p 3 times the line-to-neutral
Fig. 11-23
A three-phase, ABC system, with an e ective voltage 70.7 V, has a balanced -connected load with impedances 20 458 . Obtain the line currents and draw the voltage-current phasor diagram.
The circuit is shown in Fig. 11-24. The phasor voltages have magnitudes Vmax Phase angles are obtained from Fig. 11-7(a). Then, IAB VAB 100 1208 5:0 758 A Z 20 458 p 2Veff 100 V.
Similarly, IBC 5:0 458 A and ICA 5:0 1958 A. The line currents are: IA IAB IAC 5 758 5 1958 8:65 458 A Similarly, IB 8:65 758 A, IC 8:65 1658 A. The voltage-current phasor diagram is shown in Fig. 11-25.
CHAP. 11]
POLYPHASE CIRCUITS
Fig. 11-24
Fig. 11-25
A three-phase, three-wire CBA system, with an e ective line voltage 106.1 V, has a balanced Yconnected load with impedances 5 308  (Fig. 11-26). Obtain the currents and draw the voltage-current phasor diagram.
With balanced Y-loads, the neutral conductor carries no current. Even though this system is threewire, the neutralp be added to simplify computation of the line currents. The magnitude of the line may p voltage is VL 2 106:1 150 V. Then the line-to-neutral magnitude is VLN 150= 3 86:6 V. IA VAN 86:6 908 17:32 608 A Z 5 308
Fig. 11-26
POLYPHASE CIRCUITS
[CHAP. 11
Similarly, IB 17:32 608 A, IC 17:32 1808 A. See the phasor diagram, Fig. 11-27, in which the balanced set of line currents leads the set of line-to-neutral voltages by 308, the negative of the angle of the impedances.
Fig. 11-27
A three-phase, three-wire CBA system, with an e ective line voltage 106.1 V, has a balanced connected load with impedances Z 15 308 . Obtain the line and phase currents by the single-line equivalent method.
p p Referring to Fig. 11-28, VLN 141:4 2 = 3 115:5 V, and so 115:5 08 23:1 308 A 15=3 308
IL
Fig. 11-28 The line currents lag the ABC-sequence, line-to-neutral voltages by 308: IA 23:1 608 A IB 23:1 608 A IC 23:1 1808 A
p The phase currents, of magnitude IPh IL = 3 13:3 A, lag the corresponding line-to-line voltages by 308: IAB 13:3 908 A IBC 13:3 308 A ICA 13:3 2108 A
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