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+//CHAPTER 7- SINGLE PHASE TRANSFORMER
+//Example 20
+
+disp("CHAPTER 7");
+disp("EXAMPLE 20");
+
+//VARIABLE INITIALIZATION
+va=4000;                            //apparent power
+v1=200;                             //primary voltage in Volts
+v2=400;                             //secondary voltage in Volts
+f=50;
+R_e1=0.15;
+Pi=60;                              //core losses iron core
+pf1=0.9;
+pf2=0.8;
+
+//SOLUTION  
+R_e2=(v2/v1)^2*R_e1;
+I1=va/v1;
+I2=va/v2;
+Pcu=I2^2*R_e2;                      //cu losses
+disp("SOLUTION (i)");
+disp(sprintf("The value of Copper Losses at full load is %f W",Pcu)); 
+//
+Pout=va*pf1;
+Pin=Pout+Pi+Pcu;
+eff=Pout*100/Pin;
+disp("SOLUTION (ii)");
+disp(sprintf("The percent efficiency at full load %f PF is %f",pf1,eff)); 
+//
+//at half load
+Pout=va*pf2/2;
+Pin=Pout+Pi+Pcu*(1/2)^2;
+eff=Pout*100/Pin;
+disp("SOLUTION (ii)");
+disp(sprintf("The percent efficiency at full load %f PF is %f",pf2,eff)); 
+  
+disp(" "); 
+// 
+//END