Question

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

• A. $\frac{\textit{eR}^{2} \textit{E}}{\textit{L}}$
• B. $\frac{\textit{EL}}{\textit{R}}$
• C. $\frac{\textit{EL}}{\textit{eR}^{2}}$
• D. $\frac{\textit{eL}}{\textit{ER}}$

Answer 1

Answer: (C)

The current at time $t$ is given by
$i=i_0\left(1-e^{-t / \tau}\right)$
Here, $i_0=\frac{E}{R}$ and $\tau=\frac{L}{R}$
$\therefore q=\int\limits_0^\tau i d t=\int\limits_0^\tau(i)_0\left(1-( e )^{-t / \tau}\right) d t $
$=\frac{(i)_0 \tau}{e}=\frac{\left(\frac{E}{R}\right)\left(\frac{L}{R}\right)}{e}$
$=\frac{E L}{e R^2}$