JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Electromagnetic Induction

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A 20 Henry inductor coil is connected to a 10 ohm resistance in series as shown in figure. The time at which rate of dissipation of energy (Joule's heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor, is

(A) $\frac{2}{l n 2}$
(B) $ln 2$
(C) $\frac{1}{2} ln 2$
(D) $2 ln 2$ ✓ Correct

Answer: (D)

Solution

In the growing

L R

circuit the current follows

I = I_{0} (1 - e^{- t / τ})

with

τ = L / R

. The resistor dissipates power at rate

I^{2} R

while the inductor stores magnetic energy at rate

L I \frac{d I}{d t}

. Setting the two equal,

I^{2} R = L I \frac{d I}{d t}

, and substituting

I = \frac{E}{R} (1 - e^{- t / τ})

with

\frac{d I}{d t} = \frac{E}{L} e^{- t / τ}

, the common factors cancel to leave

1 - e^{- t / τ} = e^{- t / τ}

. Hence

e^{- t / τ} = \frac{1}{2}

and

t = τ ln 2 = \frac{L}{R} ln 2 = \frac{20}{10} ln 2 = 2 ln 2

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Electromagnetic Induction

Solution

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