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