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+//Caption:Calculate the core loss and copper loss and also find the value of load current at which max efficiency will be attained
+//Exa:3.23
+clc;
+clear;
+close;
+KVA=100;
+V2=11000;//in volts
+x1=1;
+x2=0.5;
+pf1=0.8;
+pf2=1;
+Eff_1=0.985;
+Eff_2=0.99;
+P_cu=(KVA*1000*x1*pf1*((1/Eff_1)-1)-x2*KVA*1000*((1/Eff_2)-1))*(4/3);// in watts
+P_i=1218-P_cu;// in watts
+I_fl=KVA*1000/V2;
+I_2=I_fl*sqrt(P_i/P_cu);
+disp(P_cu,'Copper loss (in watts)=');
+disp(P_i,'Core loss (in watts)=');
+disp(I_2,'Load current (in amperes)=') \ No newline at end of file