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+// Ex 45 Page 389
+
+clc;clear;close;
+// Given
+Edc=500;//V
+fim=.085;//Wb
+f=50;//Hz
+E1=11000;//V
+P=1500;//kW
+p=8;//pole
+pf=0.9
+V=500;//V
+J=3;//A/mm^2
+
+E2=Edc/sqrt(2)//V
+N2=E2/4.44/f/fim;//no. of turns
+N1=E1/E2*N2;//no. of turns
+printf("no. of turns in primary = %d",N1)
+printf("\n no. of turns in secondary = %d",N2)
+Idc=P*10**3/V;//A
+eta=1;//because of no loss
+ISR=0.472*Idc/(eta*pf)
+A1=ISR/J*10**-6;//m^2 (cross section area)
+I1=N2/N1*ISR;//A
+A2=I1/J*10**-6;//m**2 (cross section area of primary winding)
+printf("\n\n cross section of primary winding=%.2e m^2",A1)
+printf("\n cross section of secondary winding=%.1e m^2",A2)
+//ans in the book are not accurate.