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Gate 2016 EE -Set 2 Question 46

Load voltage V_{L}=V_{ph}\left ( delta \right )=\sqrt{\frac{2}{3}}V_{S}=\sqrt{\frac{2}{3}}\times 600

Power formula to be used is, P=3\frac{V_{ph}^{2}}{R}=\frac{3\times \frac{2}{3}\times 600^{2}}{30}

P = 24,000W=24kW

 

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