The mean value of current for half cycle for | vedclass

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Question

(a)

$l_{\max }=\frac{\int \limits_0^{T / 2} / d t}{T / 2}$

From 0 to $\frac{T}{2}$ graph is straight line so the function $(l)$ will be $=\frac{

Vedclass · Accepted Answer

$\frac{i_0}{2}$

Vedclass · Answer

(a)

$l_{\max }=\frac{\int \limits_0^{T / 2} / d t}{T / 2}$

From 0 to $\frac{T}{2}$ graph is straight line so the function $(l)$ will be $=\frac{i_0}{(T / 2)} t=\frac{2 i_0}{T} t$

$	ext { So }$

$l_{	ext {mean }}=\frac{2}{T} \int \limits_0^{T / 2} \frac{2 i_0}{T} t d t=\frac{2}{T}\left(\frac{2}{T}ight) i_0\left(\frac{t^2}{2}ight)_0^{T / 2}$

$=\frac{4}{T^2} i_0\left(\frac{1}{2}ight)\left(\frac{T^2}{4}-0ight)=\frac{i_0}{2}$

The mean value of current for half cycle for a current variation shown by the graph is

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