Question

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 1

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%.