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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}

 

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