load1:
900*10^3=1.732*11*10^3*I1
I1=[900*10^3]/[1.732*11*10^3]=47.23<-inverse cosine of 0.7=47.23<-45.57 amps.
load 2:
500*10^3=1.732*11*10^3*I2*0.6
I2=[500*10^3]/[1.732*11*10^3*0.6]=43.73<-inverse cosine of 0.6=43.73<-53.13 amps.
load 3:
350*10^3= 1.732*11*10^3*I3*SINE(INVERSE COSINE OF 0.7)
I3=[350*10^3]/[1.732*11*10^3*0.714]=25.72<-INVERSE COSINE OF 0.7=25.72<-45.57 amps.
load 4:
600*10^3=1.732*11*10^3*I4*0.6
I4=[600*10^3]/[1.732*11*10^3*0.6]=52.48<-INVERSE COSINE OF 0.6=52.48<-53.13 amps.
b) total load current=I1+I2+I3+I4=168.79<-49.87 amps.
a) s=3*[11000/1.732]*168.79<-49.87=3215881.414<-49.87 VA
c)per phase load voltage=11000/1.732=6350.85 volts.
Van=6350.85<0+(0.5+j0.25)*168.79<-49.87=6437.616<-0.3322 volts
Vbn=6437.616<(-0.3322-120)=6437.616<-120.3322 volts.
Vcn=6437.616<(-0.3322+120)=6437.616<119.6678 volts.
of shagah Name 120 96] Proble m 1: A balanced Y. connected, 50-Hz abc source is...
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