In the circuit shown in figure switch S is closed at time t = 0. The charge which passes through the battery in one-time constant is

In L-R circuit, the growing current at time t is given by
$i=i_0[1-e^{-t/\tau }]where,i_0=\frac{E}{R}and\tau =\frac{L}{R}$
Charge passed through the battery in one-time constant is
 $\begin{array}{l}q=\underset{0}{\overset{\tau }{\int }}idt=\underset{0}{\overset{\tau }{\int }}{i}_0(1-e^{-t/\tau })dt\\ q=i_0\tau -{\left[\frac{i_0{e}^{-t/\tau }}{-2/\tau }\right]}_0^{\tau }=i_0\tau +i_0\tau [e^{-1}-1]\\ =i_0\tau -i_0\tau +\frac{i_0\tau }{e}\\ q=\frac{i_0\tau }{e}=\frac{(E/R)(L/R)}{e}=\frac{EI}{e{R}^2}\end{array}$