Question

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 1

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%.