A) \[\frac{{{V}_{0}}}{\sqrt{2}}\]
B) \[\frac{{{V}_{0}}}{2}\]
C) zero
D) \[\frac{2{{V}_{0}}}{\pi }\]
Correct Answer: D
Solution :
(d) \[\frac{2{{V}_{0}}}{\pi }\] The average value of the voltage is \[{{V}_{av}}=\frac{\int{_{0}^{2\pi /\omega }\,V\,dt}}{\int{_{0}^{2\pi /\omega }dt}}\] \[=\frac{\int{_{0}^{\pi /\omega }{{V}_{0}}\sin \omega t\,\,dt+\int{_{\pi /\omega }^{2\pi /\omega }(-{{V}_{0}}sin\omega t)dt}}}{\frac{2\pi }{\omega }}\] \[=\frac{\omega }{2\pi }\left[ \left| \frac{-{{V}_{0}}\cos \,\omega t}{\omega } \right|_{0}^{\pi /\omega }+\left| \frac{{{V}_{0}}\cos \,\omega t}{\omega } \right|_{\pi /\omega }^{2\pi /\omega } \right]=\frac{2{{V}_{0}}}{\pi }\]
You need to login to perform this action.
You will be redirected in
3 sec
