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, be tween $t=0$ and $t=t_c$ ($t_c$ is the time constant of the circuit) is :

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

Answer 1

Answer: (D)

$i=i_0\left(1-e^{-Rt/L}\right)=i_0\left(1-e^{-t/T_C}\right)<br/>q=$$\underset{0}{\overset{T_C}{\int }}idt$
$=\underset{0}{\overset{T_C}{\int }}$$\frac{\epsilon }{R}\left(1-e^{-t/T_C}\right)$
$=\frac{\epsilon }{R}\left(t-\frac{e^{-t/T_C}}{-1/T_C}\right){|}_0^{T_C}$
$=\frac{e}{R}\left(T_C-T_C{e}^{-1}\right)-\frac{e}{R}\left(0+T_C\right)$
$q=\frac{e}{R}\times T_C{e}^{-1}$
$=\frac{\epsilon }{R}\times \frac{L}{R}\frac{1}{e}$
$=\frac{\epsilon L}{e{R}^2}$