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//CHAPTER 7- SINGLE PHASE TRANSFORMER
//Example 34
disp("CHAPTER 7");
disp("EXAMPLE 34");
//VARIABLE INITIALIZATION
va=50000; //apparent power
I2=200; //secondary full load current
R1=0.55;
R2=0.023;
pf=0.8;
//turn ratio
K=1/5;
//SOLUTION
R_dash_1=K^2*R1;
R_e2=R2+R_dash_1;
Pcu=I2^2*R_e2;
//cu loss at 2/3 of the load
Pcu23=(2/3)^2*Pcu;
Pc=Pcu23; //at maximum efficiency Pc=Pcu
//full load output
Pout=va*pf;
loss=Pc+Pcu; //at full load
Pin=Pout+loss;
eff=Pout*100/Pin;
disp(sprintf("The percent efficiency at full load is %f",eff));
disp(" ");
//
//END
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