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GATE 2015 EE - SET 2 - Question 33

Solution :

V_{s} = 100 Sin(100\pi t)

(V_{0})_{avg} =\frac{1}{2\pi }\int_{0}^{2\pi }V_{s}d(wt)

(V_{0})_{avg}=\frac{1}{2\pi }\left [\int_{\frac{\pi }{6}}^{\pi } 100 Sin(100\pi t)dwt - \int_{\pi }^{2\pi } 100 Sin(100\pi t)dwt\right ]

(V_{0})_{avg} =\frac{1}{2\pi }\left [\int_{\frac{\pi }{6}}^{\pi } 100Sin wt dt - \int_{\pi }^{2\pi } 100 Sinwtdt\right ]

(V_{0})_{avg} =\left [ 100(-coswt)_{\frac{\pi }{6}}^{\pi }-(100) (-coswt)_{\pi }^{2\pi }\right ]

\frac{100}{2\pi } \left [ (1 + Cos30)+(1+1) \right ]  \Rightarrow \frac{100}{2\pi } \left [ 3 +0.866 \right ]

\Rightarrow \frac{100}{2\pi }\left [ 3.866 \right ] \Rightarrow  61.56

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