Gate 2016 EE -Set 2 – Question 45

$$V_{m}(t)=200\pi \sin \left ( 100\pi t \right )$$

$$R=20\Omega ; E=800V$$

Using the formula, $$V_{0}=\frac{2V_{m}}{\pi }\cos \alpha \Rightarrow \frac{2\times 200\pi }{\pi }\cos \left ( 120^{\circ} \right )$$

$$= 400\cos \left ( 120^{\circ} \right )=-200V$$

$$\left | V_{0} \right |=200$$

Power balance equation $$\Rightarrow EI_{0}=I_{0}^{2}R+V_{0}I_{0}$$

$$800I_{0}=I_{0}^{2}(20)+200I_{0}$$

$$I_{0}=30A$$

Real power fed back to source, $$P_{0}=V_{0}I_{0}=200\times 30=6000=6kW$$