JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Electromagnetic Induction

A series L-R circuit is connected to a battery of emf V. If the circuit is switched on at

t = 0

, then the time at which the energy stored in the inductor reaches

(1/ n)

times of its maximum value, is:

Answer: (D)

Solution

In an

L

R

circuit switched on at

t = 0

the current grows as

i = \frac{V}{R} (1 - e^{- R t / L})

, and the energy stored in the inductor is

U = \frac{1}{2} L i^{2}

, with maximum value

U_{m a x} = \frac{1}{2} L \frac{V ^{2}}{R ^{2}}

when the current is steady. Requiring

U = \frac{U _{m a x}}{n}

gives

(1 - e^{- R t / L})^{2} = \frac{1}{n}

, so

1 - e^{- R t / L} = \frac{1}{n}

and

e^{- R t / L} = \frac{n - 1}{n}

. Inverting,

t = \frac{L}{R} ln (\frac{n}{n - 1})

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option