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+//CHAPTER 7- SINGLE PHASE TRANSFORMER
+//Example 38
+
+disp("CHAPTER 7");
+disp("EXAMPLE 38");
+
+//VARIABLE INITIALIZATION
+v1=220;                                    //primary voltage in Volts
+v2=115;                                     //secondary voltage in Volts
+f1=25;
+f2=50;
+//loads
+V=440
+We1=100;                                     //in Watts at 220 V, eddy losses
+Pc1=2*We1;                                    //Total iron losses which equals We+Wh due to eddy and hysteresis
+Wh1=Pc1-We1;
+//
+//SOLUTION  
+//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;
+Pc2=We2+Wh2;
+disp(sprintf("The total no load losses at 400 V is %f W",Pc2));
+disp(" "); 
+// 
+//END