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+//Problem 20.17: Determine the efficiency of the transformer in Problem 20.16 at half full-load and 0.85 power factor.
+
+//initializing the variables:
+S = 200000; // in VA
+Pc = 1500; // in Watt
+Pi = 1000; // in Watt
+pf = 0.85; // power factor
+
+//calculation:
+//Efficiency = output power/input power = (input power—losses)/input power
+//Efficiency = 1 - losses/input power
+//Half full-load power output = V*I*pf/2
+Po = S*pf/2
+//Copper loss (or I*I*R loss) is proportional to current squared
+//Hence the copper loss at half full-load is
+Pch = Pc/(2*2)
+//Iron loss = 1000 W (constant)
+//Total losses
+Pl = Pch + Pi
+//Input power at half full-load = output power at half full-load + losses
+PI = Po + Pl
+//efficiency
+eff = (1-(Pl/PI))*100
+
+printf("\n\n Result \n\n")
+printf("\n the transformer efficiency at half full load is %.2f percent", eff)
+\ No newline at end of file