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+//CHAPTER 7- SINGLE PHASE TRANSFORMER
+//Example 32
+
+disp("CHAPTER 7");
+disp("EXAMPLE 32");
+
+//VARIABLE INITIALIZATION 
+va=50000;
+v1=4400;                             //primary voltage in Volts
+v2=220;                              //secondary voltage in Volts
+f=50;
+R1=3.45;
+X1=5.2;
+R2=0.0009;
+X2=0.015; 
+
+//SOLUTION 
+// 
+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 (i)");
+disp(sprintf("The equivalent resistance referred to primary %f Ω",R_e1));
+disp("SOLUTION (ii)");
+disp(sprintf("The equivalent resistance referred to secondaryy %f Ω",R_e2));
+disp("SOLUTION (iii)");
+disp(sprintf("The equivalent leakage reactance referred to primary %f Ω",X_e1)); 
+disp(sprintf("The equivalent leakage reactance referred to secondary %f Ω",X_e2));
+disp("SOLUTION (iv)");
+disp(sprintf("The equivalent impedance referred to primary %f Ω",magZ_e1)); 
+disp(sprintf("The equivalent impedance referred to secondary %f Ω",magZ_e2));
+//
+I1=va/v1;
+I2=va/v2;
+Pcu=I2^2*R_e2;
+disp("SOLUTION (d)");
+disp(sprintf("The copper loss at full load %f W",Pcu)); 
+disp(" "); 
+//The answers in the book on page 7.77 are wrong for all but Xe1 and Xe2 values.
+//END