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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 lagging will be?(a) 3.8%(b) 2.2%(c) -3.8%(d) -2.2%The question was posed to me during an interview for a job.The question is from Voltage Regulation of Transformer in portion Transformers of Electrical Machines

Answer»

The correct CHOICE is (a) 3.8%

EXPLANATION: V.R. = (r(PU)*cosθ+x(pu)*sinθ)*100 % = (0.01*0.8 + 0.05*0.6)*100 = 3.8%.



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