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

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 $\tau =\frac{L}{R}$
$\therefore q=\underset{0}{\overset{\tau }{\int }}idt=\underset{0}{\overset{\tau }{\int }}(i{)}_0\left(1-(e{)}^{-t/\tau }\right)dt$
$=\frac{(i{)}_0\tau }{e}=\frac{\left(\frac{E}{R}\right)\left(\frac{L}{R}\right)}{e}$
$=\frac{EL}{e{R}^2}$