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Q.

The output current versus time curve of a rectifier is shown in the figure. The average value of output current in this case is

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a

${I_0}$

b

$\frac{{2{I_0}}}{\pi }$

c

0

d

$\frac{{{I_0}}}{2}$

answer is C.

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Detailed Solution

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${I_{av}} = \frac{{\int_0^{T/2} {\,\,\,\,\,i\,dt} }}{{\int_0^{T/2} {\,\,\,\,\,dt} }}$ $= \frac{{\int_0^{T/2} {\,\,\,\,\,{I_0}\sin (\omega \,t)dt} }}{{T/2}}$

$= \frac{{2{I_0}}}{T}\left[ {\frac{{ - \cos \omega \,t}}{\omega }} \right]_0^{T/2}$ $=\frac{2I_0}{T}$ $\left[ {-\frac{{ \cos \frac{\omega \,T}{2}}}{\omega }} +\frac{\cos\,0^o}{\omega}\right]$

$= \frac{{2{I_0}}}{{\omega \,T}}[ - \cos \pi + \cos {0^o}]$ $= \frac{{2{I_0}}}{{2\pi }}[1 + 1] = \frac{{2{I_0}}}{\pi }$

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