CHAP. 11]

261

POLYPHASE CIRCUITS

Fig. 11-22 Let Z ¼ jZj  and Ip ¼ Vp =jZj. (a) The voltages and currents in phasor domain are VAN ¼ Vp 0

VBN ¼ Vp 1808 ¼ Vp 0

VAB ¼ VAN  VBN ¼ 2Vp 0

Now, from Ip and Z given above, we have IA ¼ Ip 

IB ¼ Ip 1808   ¼ Ip 

IN ¼ IA þ IB ¼ 0

(b) The instantaneous powers delivered are pa ðtÞ ¼ va ðtÞia ðtÞ ¼ Vp Ip cos  þ Vp Ip cos ð2!t  Þ pb ðtÞ ¼ vb ðtÞib ðtÞ ¼ Vp Ip cos  þ Vp Ip cos ð2!t  Þ The total instantaneous power pT ðtÞ is pT ðtÞ ¼ pa ðtÞ þ pb ðtÞ ¼ 2Vp Ip cos  þ 2Vp Ip cos ð2!t  Þ The average power is Pavg ¼ 2VP Ip cos .

11.2

Solve Problem 11.1 given Vp ¼ 110 Vrms and Z ¼ 4 þ j3 . (a) In phasor form, Z ¼ 4 þ j3 ¼ 5 36:98 . VAN ¼ 110 0 V

Then, VBN ¼ 110 1808 V

VAB ¼ VAN  VBN ¼ 110 0  110 1808 ¼ 220 0 and

IA ¼ VAN =Z ¼ 22 36:98 A

V

IB ¼ VBN =Z ¼ 22 216:98 ¼ 22 36:98 A IN ¼ IA þ IB ¼ 0

ðbÞ

pa ðtÞ ¼ 110ð22Þ½cos 36:98 þ cos ð2!t  36:98Þ ¼ 1936 þ 2420 cos ð2!t  36:98Þ

ðWÞ

pb ðtÞ ¼ 110ð22Þ½cos 36:98 þ cos ð2!t  36:98  3608Þ ¼ 1936 þ 2420 cos ð2!t  36:98Þ pðtÞ ¼ pa ðtÞ þ pb ðtÞ ¼ 3872 þ 4840 cos ð2!t  36:98Þ ðWÞ Pavg ¼ 3872 W

ðWÞ

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An