A) 0
B) \[\frac{{{I}_{0}}}{2}\]
C) \[\frac{2{{I}_{0}}}{\pi }\]
D) \[{{I}_{0}}\]
Correct Answer: C
Solution :
\[{{I}_{av}}=\frac{\int_{0}^{T/2}{\,\,\,\,\,i\,dt}}{\int_{0}^{T/2}{\,\,\,\,\,dt}}\]\[=\frac{\int_{0}^{T/2}{\,\,\,\,\,{{I}_{0}}\sin (\omega \,t)dt}}{T/2}\] \[=\frac{2{{I}_{0}}}{T}\left[ \frac{-\cos \omega \,t}{\omega } \right]_{0}^{T/2}\]\[=\frac{2{{I}_{0}}}{T}\left[ -\frac{\cos \,\left( \frac{\omega T}{2} \right)}{\omega }+\frac{\cos {{0}^{o}}}{\omega } \right]\] \[=\frac{2{{I}_{0}}}{\omega \,T}[-\cos \pi +\cos {{0}^{o}}]\]\[=\frac{2{{I}_{0}}}{2\pi }[1+1]=\frac{2{{I}_{0}}}{\pi }\]
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