GATE 2015 EE – SET 2 – Question 43

Solution :

Per Phase current on the secondary side ,

$$\left ( I_{ph} \right )_{s}$$ = $$\frac{100}{\sqrt{3}}\angle 30^{0}$$

Turns ratio of primary to secondary ,

$$\frac{\left ( V_{ph} \right )_{p}}{\left ( V_{ph} \right )_{s}} $$= $$\frac{(\frac{230}{\sqrt{3})}}{115}$$=$$\frac{2}{\sqrt{3}}$$

$$\frac{\left ( V_{ph} \right )_{p}}{\left ( V_{ph} \right )_{s}}$$  = $$\frac{\left ( I_{ph} \right )_{s}}{\left ( I_{ph} \right )_{p}}$$

$$\left ( I_{ph} \right )_{p}$$ =$$\left ( I_{ph} \right )_{s}*\frac{(V_{ph})_{s}}{(V_{ph})_{p}}$$

$$\left ( I_{ph} \right )_{p}$$ =$$ \frac{\sqrt{3}}{2}*\frac{100}{\sqrt{3}}\angle 30^{0}$$

$$\Rightarrow $$    50 $$\angle 30^{0}$$