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(A) The mean value of current $i_{	ext{mean}}$ over a time interval $t_1$ to $t_2$ is given by $i_{	ext{mean}} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2

Vedclass · Accepted Answer

$\frac{i_0}{2}$

Vedclass · Answer

(A) The mean value of current $i_{	ext{mean}}$ over a time interval $t_1$ to $t_2$ is given by $i_{	ext{mean}} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} i(t) dt$.For the first half cycle ($0$ to $T/2$),the current $i(t)$ represents a straight line passing through the origin with slope $m = \frac{i_0}{T/2} = \frac{2i_0}{T}$.Thus,$i(t) = \frac{2i_0}{T} t$.Now,calculating the mean current for the interval $0$ to $T/2$:$i_{	ext{mean}} = \frac{1}{T/2 - 0} \int_{0}^{T/2} \frac{2i_0}{T} t dt$$i_{	ext{mean}} = \frac{2}{T} \cdot \frac{2i_0}{T} \int_{0}^{T/2} t dt$$i_{	ext{mean}} = \frac{4i_0}{T^2} \left[ \frac{t^2}{2} ight]_{0}^{T/2}$$i_{	ext{mean}} = \frac{4i_0}{T^2} \cdot \frac{1}{2} \left( \frac{T^2}{4} - 0 ight)$$i_{	ext{mean}} = \frac{2i_0}{T^2} \cdot \frac{T^2}{4} = \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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