Consider the LR circuit shown in the figure. If the switch S is closed at $\mathrm{t}=0$ then the amount of charge that passes through the battery between $\mathrm{t}=0$ and $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{R}}$ is:

The current in the LR circuit at any time $t$ is
$\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}\left(1-{\mathrm{e}}^{-\frac{\mathrm{R}\mathrm{t}}{\mathrm{L}}}\right)$

Where $E, R, L, I$ are the EMF, resistance, inductance and the current respectively.
$\therefore$ Charge flown through the circuit from time 0 to $\frac{L}{R}$ is 
(charge) $∆\mathrm{q}=\underset{0}{\overset{\frac{\mathrm{L}}{\mathrm{R}}}{\int }}\mathrm{I}.\mathrm{d}\mathrm{t}=\underset{0}{\overset{\frac{\mathrm{L}}{\mathrm{R}}}{\int }}\frac{\mathrm{E}}{\mathrm{R}}\left(1-{\mathrm{e}}^{-\frac{\mathrm{R}\mathrm{t}}{\mathrm{L}}}\right)\mathrm{d}\mathrm{t}$
$\Rightarrow ∆\mathrm{q}=\frac{\mathrm{E}}{\mathrm{R}}{\left[\mathrm{t}+\frac{\mathrm{L}}{\mathrm{R}}\mathrm{ }{\mathrm{e}}^{-\frac{\mathrm{R}\mathrm{t}}{\mathrm{L}}}\right]}_0^{\frac{\mathrm{L}}{\mathrm{R}}}$
$\Rightarrow ∆\mathrm{q}=\frac{\mathrm{E}}{\mathrm{R}}\left[\frac{\mathrm{L}}{\mathrm{R}}+\frac{\mathrm{L}}{\mathrm{R}}{\mathrm{e}}^{-1}-\frac{\mathrm{L}}{\mathrm{R}}\right]$
$\Rightarrow ∆\mathrm{q}=\frac{\mathrm{E}}{\mathrm{R}}\times \frac{\mathrm{L}}{\mathrm{e}\mathrm{R}}$
$\Rightarrow ∆\mathrm{q}=\frac{\mathrm{E}\mathrm{L}}{\mathrm{e}{\mathrm{R}}^2}=\frac{\mathrm{E}\mathrm{L}}{2.7{\mathrm{R}}^2}$