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+//CHAPTER 7- SINGLE PHASE TRANSFORMER
+//Example 25
+
+disp("CHAPTER 7");
+disp("EXAMPLE 25");
+
+//15kVA 2200/110 V transformer
+//VARIABLE INITIALIZATION 
+va=25000;                             //power rating
+v1=2200;                             //primary voltage in Volts
+v2=110;                              //secondary voltage in Volts
+f=50;
+R1=1.75;
+X1=2.6;
+R2=0.0045;
+X2=0.0075; 
+
+//SOLUTION 
+// 
+//Primary resistance and leakage reactance referred to secondary
+//R'1 & X'1
+//Secondary resistance and leakage reactance referred to primary
+//R'2 & X'2
+//Equivalent resistance & leakage reactance referred to primary
+//Re1 & Xe1
+//Equivalent resistance & leakage reactance referred to secondary
+//Re2 & Xe2
+//
+R_dash_2=R2*(v1/v2)^2;
+R_e1=R1+R_dash_2;
+X_dash_2=X2*(v1/v2)^2;
+X_e1=X1+X_dash_2;
+//
+R_dash_1=R1*(v2/v1)^2;
+R_e2=R2+R_dash_1;
+X_dash_1=X1*(v2/v1)^2;
+X_e2=X2+X_dash_1;
+//
+Z_e1=R_e1+X_e1*%i;
+Z_e2=R_e2+X_e2*%i;
+magZ_e1=sqrt(real(Z_e1)^2+imag(Z_e1)^2);
+magZ_e2=sqrt(real(Z_e2)^2+imag(Z_e2)^2);
+//
+//
+disp("SOLUTION (a)");
+disp(sprintf("The equivalent resistance referred to primary %.2f Ω",R_e1));
+disp("SOLUTION (b)");
+disp(sprintf("The equivalent resistance referred to secondaryy %.5f Ω",R_e2));
+disp("SOLUTION (c)");
+disp(sprintf("The equivalent leakage reactance referred to primary %.1f Ω",X_e1));
+disp("SOLUTION (d)");
+disp(sprintf("The equivalent leakage reactance referred to secondary %.3f Ω",X_e2));
+disp("SOLUTION (e)");
+disp(sprintf("The equivalent impedance referred to primary %.5f Ω",magZ_e1));
+disp("SOLUTION (f)");
+disp(sprintf("The equivalent impedance referred to secondary %.5f Ω",magZ_e2));
+//
+//primary and secondary full load current and voltage relationship with power rating
+I1=va/v1;           //primary current
+I2=va/v2;           //secondary current
+cuLoss=I2^2*R_e2;   //copper loss     or also as I1^2.R1 + I2^2.R2
+disp("SOLUTION (d)");
+disp(sprintf("The copper loss at full load %f W",cuLoss)); 
+disp(" "); 
+// 
+//END