Question

Solve this:
Q. In a R-L circuit discharging current is given by $I=I_0{e}^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$\tau $}\right.}$, where $\tau$ is the time constant of the circuit. The average current for the period t = 0 to t = $\tau$ is

Answer 1

Dear Student,
$Averagecurrent=\frac{{\int }_0^{\tau }Id\tau }{{\int }_0^{\tau }d\tau }=\frac{{\int }_0^{\tau }{I}_0{e}^{-t/\tau }d\tau }{\tau }=\frac{I_0(1-1/e)}{{\tau }^2}=\frac{I_0}{{\tau }^2}\times \frac{(e-1)}{e}$
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