Question

Question 5 6+3j ohm 4 o | 110/0° Ꮩ 11 +9j Ohm 110-240 V 110-120° v B 11 +9j Ohm N 6+3j ohm 11 +9j Ohm 6+3j ohm Figure 1. For the circuit of the three-wire Y-Y system in Figure 1 above, a) Calculate the total impedance Z b) Calculate the current in phase a c) Calculate the load phase voltage VAN d) Calculate the current in phase b e) Calculate the load phase voltage Von f) Calculate the current in phase c g) Calculate the load phase voltage Von h) Calculate the magnitude of the line voltages Vab, Voc, and ca

Answer 1

Qusteions 6+3) ohm a IAN ZTL Van Inoco 110C-120 ZL - 11+ ja ohm B 17j9 Ohm 1102-2400 b von Von 6+3j Ohm + ZTLI 6+3j ohm IBN ZIL IN ist ja ohm we may replace it colth its The above three phase arut is balanced Single phase equivalent ciruit. ZTL IAN Van f JZL VAN Frg i 108 n ZT = ZTL+ ZL Senes) + 6+3j+11+ ja 17+12) i IAN= Van ZT llocoo : 5.2862 2-35.217° 17+12) Since balanced we can draw current vector as IAN 128 120° 120 IN IBN IBN = TANL-1200 and ICN = IAN <-2400 IBN = 5.2862<-35217-120° IBN = 5.28624155.217° ICN= 5.2862-35.217-240° Icn = 5.2862<-295.2170

B ZTL ZTL С. IBN Von 2 IN Ven 2 N Figa figz From FigI VAN=IAN*Z. -(5.26624-35-217)*(1+)q) = 75-13 10 24.022° From Fig 2 VBN = IBN* 2L -(5.2862<-I5S217) (11tj9) 75.1310C-15.9270 From Eg 3 VEN= ICN*2L = (5-2962_-245-217) CI+ 39) = 75.1310C 124.072° Vab= Van - Von = 11020 - 1104-120 190.5252 30° Voc= Von- ven - 1100-120 - 1100-240 = 190.525 <-90° vca = &ven- Vana 110-240 – 11oco = 190.525 < 150° VAN = 95.1310540720 Vab Van -Von -Von •IAN = 5.2862235.2120 1200 35.217° ICN 1200 • Voc -van Von VEN IBN VBN Vca Summary a) 25= 19 +12) b) TN = 5.286 24 -3S 215° C) VAN = 75. 1310 – 4.072° d) IBN = 5.2862 <-155.2170 e) VBN = 75.1310<-115.9270 f) ICN = 5.28624-275.2170 g) VCN = 75.1310 <124.092 b) Vab=190.525230 Voc=190.525 4.90 Vca = 190.5252-1500 In case of douats please comment. Ready to solve your douats on above Solubon