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+//CHAPTER 7- SINGLE PHASE TRANSFORMER
+//Example 37
+
+disp("CHAPTER 7");
+disp("EXAMPLE 37");
+
+//VARIABLE INITIALIZATION
+va=500000;                                  //apparent power
+v1=3300;                                    //primary voltage in Volts
+v2=500;                                     //secondary voltage in Volts
+f=50;
+//loads
+pf=1;
+eff=0.97;                                   // at 3/4 full load at unity pf
+pf2=0.8;                                        
+//
+//SOLUTION  
+I1=va/v1;
+loss=(1-eff)*va*(3/4)*pf/eff;              //=Pc+Pcu losses at 3/4 load
+//since the eff value is maximum, Pcu=Pc; therefore, 2*Pc=loss
+Pc=loss/2;
+//(3/4)^2*Pcu=Pc;  
+f=(3/4)^2;                                 //3/4 load
+//Pcu=Pc/f
+Pcu=Pc/f;
+//disp(sprintf("The Pc is %f W",Pc));
+//disp(sprintf("The Pcu is %f W",Pcu));
+//
+R_e1=Pcu/I1^2;
+disp(sprintf("The value of Re1 is %f W",R_e1));
+//10% impedance
+Z_e1=v1*0.1/I1;
+X_e1=sqrt(Z_e1^2-R_e1^2);
+phi=acos(0.8);
+%reg=(I1*R_e1*cos(phi)+I1*X_e1*sin(phi))*100/v1;
+disp(sprintf("The percent regulation at full load 0.8 pf is %f W",%reg));
+disp(" "); 
+// 
+//END