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