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