A shown in the figure, a battery of emf \varepsilon 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 = tc (tc is the time constant of the circuit) is :

\varepsilon L \over R^2( 1 - 1 \over e )

Electromagnetic Induction PYQ — JEE Main 2020 · 8 Jan · Shift 2 · Q47 with Solution

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Correct answer: A

When a battery of emf $\varepsilon$ is connected in series with resistance $R$ and inductance $L$ , the current as a function of time is

i(t)=\frac{\varepsilon}{R}\left(1-e^{-t/\tau}\right)

where the time constant is

\tau=\frac{L}{R}

Given $t_c=\tau=\dfrac{L}{R}$ .

Charge is

q=\int_0^{t_c} i(t)\,dt

Substitute $i(t)$ :

q=\int_0^{L/R} \frac{\varepsilon}{R}\left(1-e^{-t/\tau}\right)dt

Since $\tau=L/R$ ,

q=\frac{\varepsilon}{R}\int_0^{\tau} \left(1-e^{-t/\tau}\right)dt

\int \left(1-e^{-t/\tau}\right)dt = t + \tau e^{-t/\tau}

So,

q=\frac{\varepsilon}{R}\left[ t+\tau e^{-t/\tau} \right]_0^{\tau}

At $t=\tau$ :

\tau+\tau e^{-1}

At $t=0$ :

0+\tau

Therefore,

q=\frac{\varepsilon}{R}\left(\tau+\frac{\tau}{e}-\tau\right)

q=\frac{\varepsilon}{R}\cdot \frac{\tau}{e}

Now use $\tau=\dfrac{L}{R}$ :

q=\frac{\varepsilon}{R}\cdot \frac{L}{eR}

q=\frac{\varepsilon L}{eR^2}

q=\frac{\varepsilon L}{eR^2}

This is Option A.

Stored correct answer: A

Derived answer: A

They match.