The output sinusoidal current versus time cur | vedclass

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Question

Average value of output current is given by

$ \mathrm{I}_{\mathrm{av}} =\frac{\int_{0}^{\mathrm{T} / 2} \mathrm{I} 	ext { dt }}{\int_{0}^{\mathrm{

Vedclass · Accepted Answer

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

Vedclass · Answer

Average value of output current is given by

$ \mathrm{I}_{\mathrm{av}} =\frac{\int_{0}^{\mathrm{T} / 2} \mathrm{I} 	ext { dt }}{\int_{0}^{\mathrm{T} / 2} \mathrm{dt}}=\frac{\int_{0}^{\mathrm{T} / 2} \mathrm{I}_{0} \sin \omega \mathrm{t} \mathrm{dt}}{\mathrm{T} / 2}$

$=\frac{2 \mathrm{I}_{0}}{\mathrm{T}}\left[-\frac{\cos \omega \mathrm{t}}{\omega}ight]_{0}^{\mathrm{T} / 2} $

$=\frac{2 \mathrm{I}_{0}}{\mathrm{T}}\left[-\frac{\cos (\omega \mathrm{T} / 2)}{\omega}+\frac{\cos 0^{\circ}}{\omega}ight] $

$=\frac{2 \mathrm{I}_{0}}{\omega \mathrm{T}}\left[-\cos \pi+\cos 0^{\circ}ight]$

$=\frac{2 \mathrm{I}_{0}}{2 \pi}[1+1]=\frac{2 \mathrm{I}_{0}}{\pi}$

The output sinusoidal 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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