Question

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

• A. $\frac{e{R}^{2}E}{L}$
• B. $E\left(\frac{L}{R}\right)$
• C. $\frac{EL}{e{R}^2}$
• D. $\frac{eL}{ER}$

Answer 1

Answer: (C)

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 $dQ=idt=i_0\left(1-e^{-t/\tau }\right)dt$
$Q=\int dQ=\underset{0}{\overset{\tau }{\int }}{i}_0\left(1-e^{t/\tau }\right)dt$
$=i_0\left\{\left[\underset{0}{\overset{\tau }{\int }}dt-\underset{0}{\overset{\tau }{\int }}{e}^{-t/\tau }dt\right]\right\}$
$=\frac{i_0\tau }{e}\Rightarrow Q=\frac{\frac{E}{R}\cdot \frac{L}{R}}{e}=\frac{EL}{e{R}^2}$
$\Rightarrow Q=\frac{EL}{e{R}^2}$