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, $$P=3\frac{V_{ph}^{2}}{R}=\frac{3\times \frac{2}{3}\times 600^{2}}{30}$$

$$P = 24,000W=24kW$$