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+//CHAPTER 7- SINGLE PHASE TRANSFORMER
+//Example 39
+
+disp("CHAPTER 7");
+disp("EXAMPLE 39");
+
+//VARIABLE INITIALIZATION
+v1=220; //primary voltage in Volts
+v2=440; //secondary voltage in Volts
+f1=50;
+f2=25;
+//loads
+V=110
+//say, else computation may not be possible using computer
+Pout1=100; //in watt, just assumed for computational purposes for the 220V supply
+We1=0.01*Pout1; //in Watts at 220 V, eddy losses which are 1% of the output at 220V
+Wh1=0.01*Pout1; //in Watts at 220 V, hysteresis losses which are 1% of the output at 220V
+//Pc1=We1+Wh1; //Total iron losses which equals We+Wh due to eddy and hysteresis
+Pcu1=0.01*Pout1; //copper losses
+//
+//SOLUTION
+//since on connecting to half the power ie 110V, the output would get halved
+Pout2=Pout1/2;
+xPcu=Pcu1/Pout2;
+disp(sprintf("The copper losses at 110 V would be %f percent of the output",xPcu*100));
+//now coming to frequency dependant losses ie eddy and hysteresis
+//since we know that We=kh.f.B^1.6 and Wh=Ke.Kf^2.f^2.B^2
+//since all being constant exept frequency, we may take We2/We1=f2^2/f1^2
+//and Wh2/Wh1=f2/f1
+//find values for We2 and Wh2, whence Pc2=We2+Wh2
+We2=f2^2*We1/f1^2;
+Wh2=f2*Wh1/f1;
+xWe=We2/Pout2;
+xWh=Wh2/Pout2;
+disp(sprintf("The eddy losses at 110 V would be %f percent of the output",xWe*100));
+disp(sprintf("The hysteresis losses at 110 V would be %f percent of the output",xWh*100));
+disp(" ");
+//
+//END