JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Electromagnetic Induction

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}

is the time constant of the circuit) is

Answer: (A)

Solution

While the current builds up,

i (t) = \frac{ε}{R} (1 - e^{- t / τ})

with time constant

τ = \frac{L}{R}

. The charge delivered in one time constant is

q = \int_{0}^{τ} i d t = \frac{ε}{R} [t + τ e^{- t / τ}]_{0}^{τ} = \frac{ε}{R} (τ e^{- 1}) = \frac{ε}{R} \cdot \frac{τ}{e}

. Substituting

τ = \frac{L}{R}

gives

q = \frac{ε L}{e R ^{2}}

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Single correct option