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//CHAPTER 7- SINGLE PHASE TRANSFORMER
//Example 39
disp("CHAPTER 7");
disp("EXAMPLE 39");
//VARIABLE INITIALIZATION
v1=220; //primary voltage in Volts
v2=440; //secondary voltage in Volts
f1=50;
f2=25;
//loads
V=110
//say, else computation may not be possible using computer
Pout1=100; //in watt, just assumed for computational purposes for the 220V supply
We1=0.01*Pout1; //in Watts at 220 V, eddy losses which are 1% of the output at 220V
Wh1=0.01*Pout1; //in Watts at 220 V, hysteresis losses which are 1% of the output at 220V
//Pc1=We1+Wh1; //Total iron losses which equals We+Wh due to eddy and hysteresis
Pcu1=0.01*Pout1; //copper losses
//
//SOLUTION
//since on connecting to half the power ie 110V, the output would get halved
Pout2=Pout1/2;
xPcu=Pcu1/Pout2;
disp(sprintf("The copper losses at 110 V would be %f percent of the output",xPcu*100));
//now coming to frequency dependant losses ie eddy and hysteresis
//since we know that We=kh.f.B^1.6 and Wh=Ke.Kf^2.f^2.B^2
//since all being constant exept frequency, we may take We2/We1=f2^2/f1^2
//and Wh2/Wh1=f2/f1
//find values for We2 and Wh2, whence Pc2=We2+Wh2
We2=f2^2*We1/f1^2;
Wh2=f2*Wh1/f1;
xWe=We2/Pout2;
xWh=Wh2/Pout2;
disp(sprintf("The eddy losses at 110 V would be %f percent of the output",xWe*100));
disp(sprintf("The hysteresis losses at 110 V would be %f percent of the output",xWh*100));
disp(" ");
//
//END
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