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+//CHAPTER 7- SINGLE PHASE TRANSFORMER
+//Example 23
+
+disp("CHAPTER 7");
+disp("EXAMPLE 23");
+
+//VARIABLE INITIALIZATION 
+va=33000;
+v1=2200;                             //primary voltage in Volts
+v2=220;                              //secondary voltage in Volts
+f=50;
+R1=2.4;
+X1=6;
+R2=0.03;
+X2=0.07; 
+
+//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;
+
+disp("SOLUTION (a)");
+disp(sprintf("The primary resistance referred to secondary %f Ω",R_dash_1));
+disp(sprintf("The primary leakage reactance referred to secondary %f Ω",X_dash_1));
+//
+disp("SOLUTION (b)");
+disp(sprintf("The secondary resistance referred to secondary %f Ω",R_dash_2));
+disp(sprintf("The secondary leakage reactance referred to secondary %f Ω",X_dash_2));
+//
+disp("SOLUTION (C(i))");
+disp(sprintf("The equivalent resistance referred to primary %f Ω",R_e1));
+disp(sprintf("The equivalent leakage reactance referred to primary %f Ω",X_e1));
+//
+disp("SOLUTION (C(ii))");
+disp(sprintf("The equivalent resistance referred to secondaryy %f Ω",R_e2));
+disp(sprintf("The equivalent leakage reactance referred to secondary %f Ω",X_e2));
+//
+I1=va/v1;
+I2=va/v2;
+oLoss=I2^2*R_e2;
+disp("SOLUTION (d)");
+disp(sprintf("The ohmic loss at full load %f W",oLoss));
+//
+Z_e1=sqrt(R_e1^2+X_e1^2);
+//voltage to be applied on HV side
+V=160*(v2/v1)*Z_e1;
+P=(160*(v2/v1))^2*R_e1;
+disp("SOLUTION (e)");
+disp(sprintf("The voltage to be applied on HV side is %f V",V));
+disp(sprintf("The power input is %f W",P));
+disp(" "); 
+// 
+//END