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+// Example 4.5;efficiency and load for maximum efficiency
+clc;
+close;
+clear;
+format('v',5)
+// given 
+cl=125;//copper losses
+fcl=2^2*cl;//full load copper losses
+il=457;//iron losses
+pf=0.8;//power factor
+kba=30;//loss
+disp("part (a)")
+fle=((kba*pf)/((kba*pf)+(fcl+il)*10^-3))*100;//full load efficiency  in %
+disp(fle,"full load efficiency at 0.8 pf is,(%)=")
+lme=kba*sqrt(il/fcl);//variable
+pfl=(lme/kba)*100;//percentage of full load on which efficiency will be maximum 
+disp("part (b)")
+disp(pfl,"percentage of full load on which efficiency will be maximum is,(%)=")