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//CHAPTER 7- SINGLE PHASE TRANSFORMER
//Example 25
disp("CHAPTER 7");
disp("EXAMPLE 25");
//VARIABLE INITIALIZATION
va=25000;
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
//
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 (C(i))");
disp("SOLUTION (a)");
disp(sprintf("The equivalent resistance referred to primary %f Ω",R_e1));
disp("SOLUTION (b)");
disp(sprintf("The equivalent resistance referred to secondaryy %f Ω",R_e2));
disp("SOLUTION (c)");
disp(sprintf("The equivalent leakage reactance referred to primary %f Ω",X_e1));
disp("SOLUTION (d)");
disp(sprintf("The equivalent leakage reactance referred to secondary %f Ω",X_e2));
disp("SOLUTION (e)");
disp(sprintf("The equivalent impedance referred to primary %f Ω",magZ_e1));
disp("SOLUTION (f)");
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(" ");
//
//END
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