Question

In the circuit shown in Fig, switch S is closed at time $t=0$. The charge that 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

The correct option is C $\frac{EL}{e{R}^2}$

The current at one time second is given by:
$i=i_0(1-e^{-1/\tau })$
Here $i_0=\frac{E}{R}$ and $\tau =\frac{L}{R}$
$\therefore q={\int }_0^{\tau }idt={\int }_0^{\tau }{i}_0(1-e^{-1/\tau })dt$
$=\frac{i_0\tau }{e}=\frac{\left(\frac{E}{R}\right)\left(\frac{L}{R}\right)}{e}=\frac{EL}{e{R}^2}$