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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+// to calculate line currents of 3 ph side
+
+clc;
+N1=6600;
+N2=100;
+a=N1/N2;
+b=(sqrt(3)/2)*a;
+P=400; //kW
+pfa=.707;
+pfb=1;
+V=100;
+Ia=P*1000/(V*pfa);
+Ib=P*2*1000/(V*pfb);
+I_A=Ia/b;disp(I_A,'I_A(A)');
+I_BC=Ib/a;
+I_B=I_BC-49.5*complex(pfa,pfa); disp(abs(I_B),'I_B(A)');
+I_C=I_BC+49.5*complex(pfa,-1*pfa);disp(abs(I_C),'I_C(A)');
+