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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+//Caption:Calculate the secondary current at which max efficiency will occur and also calculate the max efficiency at 0.8pf lagging
+//Exa:3.18
+clc;
+clear;
+close;
+a=1000/200;
+R_1=0.25;//in ohms
+R_2=0.018;//in ohms
+R_O2=R_2+R_1/a^2;
+P_i=240;//in watts
+I_2=sqrt(P_i/R_2);
+disp(I_2,'The secondary current at which max efficiency will occur (in Amperes)=');
+P_o=200*I_2*0.8;//in watts
+P_t=2*P_i;//in watts
+Eff=P_o/(P_o+P_t);
+disp(Eff*100,'Max efficiency at 0.8pf lagging(in %)=') \ No newline at end of file