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//Chapter 21, Problem 18
clc;
vi=200e3; //rated transformer
pf=0.85; //power factor
lcu=(1/2)^2*1.5e3; //copper loss
lfe=1e3; //iron loss
p0=(1/2)*vi*pf; //full-load output power
lt=lcu+lfe; //total losses
pi=p0+lt; //input power
Ef=(1-(lt/pi)); //efficiency
printf("Transformer efficiency at half load = %.3f percent",Ef*100);
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