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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

 

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