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+//CHAPTER 7- SINGLE PHASE TRANSFORMER
+//Example 14
+
+disp("CHAPTER 7");
+disp("EXAMPLE 14");
+
+//VARIABLE INITIALIZATION
+va=22000;                       //apparent power
+v1=1100;                        //primary voltage in Volts
+v2=110;                         //secondary voltage in Volts
+R1=2;                           //in Ohms
+R2=0.02;                        //in Ohms
+X1=5;                           //in Ohms
+X2=0.045;                       //in Ohms 
+
+//SOLUTION
+//N1byN2=v1/v2;
+
+R_dash_2=R2*((v1/v2)^2);
+X_dash_2=X2*((v1/v2)^2);
+disp("SOLUTION (a)");
+disp(sprintf("The equivalent resistance of secondary referred to primary is %fΩ",R_dash_2));
+disp(sprintf("The equivalent reactance of secondary referred to primary is %f Ω",X_dash_2));
+//
+R_e1=R_dash_2+R1;
+X_e1=X_dash_2+X1;
+disp("SOLUTION (b)");
+disp(sprintf("The total resistance referred to primary is %f Ω",R_e1));
+disp(sprintf("The total reactance referred to primary is %f Ω",X_e1));
+//
+R_dash_1=R1*((v2/v1)^2);
+X_dash_1=X1*((v2/v1)^2);
+disp("SOLUTION (c)");
+disp(sprintf("The equivalent resistance of secondary referred to secondary is %f Ω",R_dash_1));
+disp(sprintf("The equivalent reactance of secondary referred to secondary is %f Ω",X_dash_1));
+//
+R_e2=R_dash_1+R2;
+X_e2=X_dash_1+X2;
+disp("SOLUTION (d)");
+disp(sprintf("The total resistance referred to secondary is %f Ω",R_e2));
+disp(sprintf("The total reactance referred to secondary is %f Ω",X_e2));
+//
+I1=va/v1;
+I2=va/v2;
+copperLoss=R1*I1^2+R2*I2^2;
+disp("SOLUTION (e)");
+disp(sprintf("The total copper loss is %f W",copperLoss));
+disp(" "); 
+// 
+//END