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+//CHAPTER 7- SINGLE PHASE TRANSFORMER
+//Example 24
+
+disp("CHAPTER 7");
+disp("EXAMPLE 24");
+
+//VARIABLE INITIALIZATION
+va=10000;
+v1=2500; //primary voltage in Volts
+v2=250; //secondary voltage in Volts
+R1=4.8;
+X1=11.2;
+R2=0.048;
+X2=0.112;
+
+//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;
+//leakage impedence
+z0=R_e2+X_e2*%i;
+//applied load
+Zl=5+3.5*%i;
+//total impedence in series
+Z=z0+Zl;
+magZ=sqrt(real(Z)^2+imag(Z)^2);
+magZl=sqrt(real(Zl)^2+imag(Zl)^2);
+I2=v2/magZ;
+V2=I2*magZl
+disp("SOLUTION (a)");
+disp(sprintf("The secondary terminal voltage is %f V",V2));
+//
+//part (b) of the problem cannot be solved mathematically alone.
+disp(" ");
+//
+//END