# GATE 2015 EE - SET 1 Question 51

Solution :

Transformer rating = 500 kVA

$Z_{1}$= 1 +j6 ,$Z_{2}$ = 0.8 + j 4.8 , S = 1000 kVa , $S_{T2}$ = ?

$Cos\phi$ = 0.8 lagging

$S_{T2}$ =$\frac{s}{\frac{Z_{1}}{Z_{1}+Z_{2}}}$

$S_{T2}$ = $\frac{1000*(i+j6)}{(i+j6)+(0.8+ j4.8)}$

$S_{T2}$ = $\frac{1000(1+j6)}{1.8+j10.8}$

$S_{T2}$ = $\frac{1000*6.08\angle 80.54}{10.94\angle 80.54}$ = 555.7KVA