Question

As shown in the figure, a battery of emf$ \epsilon $ 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}$ is the time constant of the circuit) is:

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

Answer 1

The correct option is D

$\frac{\epsilon L}{e{R}^2}$

Step 1. Given Data,

Electrmagnetic Force of bettery $\epsilon$,

Inductor$L$,

Resistance $R$ ,

The switch is closed at Time$t=0$,

$t_c$ is the time constant of the circuit,

Let if $i$ be the currecnt and $q$ be the charge then we have to find the total charge that flows from the battery, between $t=0$ and$t=t_c$ ($t_c$ is the time constant of the circuit).

Step 2. Find the total charge between $t=0$ and$t=t_c$ ,

As we all know,

Current, $i=i_0(1–e^{–Rt/L})$ (where $i_0$ is a mean current)

Since we know that time constant $t_c=\frac{L}{R}$Then,

$\Rightarrow i=i_0(1–e^{–t/t_c})$

Charge, $q={\int }_0^{t_c}idt$

$$\begin{gathered} ={\int }_0^{t_c}{i}_0(1–e^{–t/t_c})dt \\ ={\int }_0^{t_c}\frac{\epsilon }{R}(1–e^{–t/t_c})dt[As,i_0=\frac{\epsilon }{R}] \\ =\frac{\epsilon L}{e{R}^2} \end{gathered}$$

Hence, option (D) is correct.