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If the pu impedance of a single phase transformer is 0.01+j0.05, then its regulation at p.f. of 0.8 leading will be?(a) 3.8%(b) 2.2%(c) -3.8%(d) -2.2%I have been asked this question by my college director while I was bunking the class.This intriguing question originated from Voltage Regulation of Transformer in section Transformers of Electrical Machines

Answer»

The correct option is (d) -2.2%

Easiest EXPLANATION: V.R. = (r(pu)*cosθ-x(pu)*sinθ)*100 % = (0.01*0.8 – 0.05*0.6)*100 = -2.2%.



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