Thank you for reporting, we will resolve it shortly
Q.
The mean value of current for half cycle for a current variation shown by the graph is
Alternating Current
Solution:
$I_{\text {mean }}=\frac{\int\limits_{0}^{T / 2} I d t}{T / 2}$
From 0 to $\frac{T}{2}$ graph is straight line so the function (I) will be $=\frac{i_{0}}{(T / 2)} t=\frac{2 i_{0}}{T} t$
So $I_{\text {mean }}=\frac{2}{T} \int\limits_{0}^{T / 2} \frac{2 i_{0}}{T} t d t=\frac{2}{T}\left(\frac{2}{T}\right) i_{0}\left(\frac{t^{2}}{2}\right)_{0}^{T / 2}$
$=\frac{4}{T^{2}} i_{0}\left(\frac{1}{2}\right)\left(\frac{T^{2}}{4}-0\right)^{\circ}=\frac{i_{0}}{2}$
