//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