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Q.
In the circuit shown, switch $S$ is closed at time $t=0$. The charge which passes through the battery in one time constant is
Electromagnetic Induction
Solution:
The current at time $t$ is given by $i=i_{0}\left(1-e^{-t / \tau}\right)$
Where $i_{0}=\frac{E}{R}$ and $\tau=\frac{L}{R}$
So $d Q=i d t=i_{0}\left(1-e^{-t / \tau}\right) d t$
$Q=\int d Q=\int\limits_{0}^{\tau} i_{0}\left(1-e^{t / \tau}\right) d t$
$=i_{0}\left\{\left[\int\limits_{0}^{\tau} d t-\int\limits_{0}^{\tau} e^{-t / \tau} d t\right]\right\}$
$=\frac{i_{0} \tau}{e} \Rightarrow Q=\frac{\frac{E}{R} \cdot \frac{L}{R}}{e}=\frac{E L}{e R^{2}}$
$\Rightarrow Q=\frac{E L}{e R^{2}}$
