Question

As shown in the figure, a battery of emf $ϵ$ is connected to an inductor $L$ and resistance $R$ in series. The switch is closed at $t=0$. The total charge that flows from the battery, between $t=0$ and $t=t_c(t_c\text{is the time constant of the circuit) is}$

• A. $\frac{ϵL}{R^{2}e}$
• B. $\frac{ϵL}{R^2}(1-\frac{1}{e})$
• C. $\frac{ϵL}{R^2}$
• D. $\frac{ϵR}{e{L}^2}$

Answer 1

The correct option is A $\frac{ϵL}{R^{2}e}$

For series connection of a resistor and inductor, time variation of current is $I=I_0(1-e^{-t/t_c})=\frac{ϵ}{R}(1-e^{-t/t_c})$

Here,$t_C=\frac{L}{R}$

$q=\underset{0}{\overset{T_c}{\int }}Idt$

$\Rightarrow \int dq=\int \frac{ϵ}{R}(1-e^{-t/t_c})dt$

$\Rightarrow q=\frac{ϵ}{R}{[t+t_C{e}^{-t/t_c}]}_0^{t_c}$

$\Rightarrow q=\frac{ϵ}{R}[t_C+\frac{t_C}{e}-t_C]$

$\Rightarrow q=\frac{ϵ}{R}\frac{L}{Re}$

$\therefore q=\frac{ϵL}{R^{2}e}$