JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Electromagnetic Induction

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An inductor coil stores 64 J of magnetic field energy and dissipates energy at the rate of 640 W when a current of 8 A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit in seconds :

(A)0.8
(B)0.2✓ Correct
(C)0.125
(D)0.4

Answer: (B)

Solution

The magnetic energy stored in the coil fixes the inductance:

\frac{1}{2} L I^{2} = 64 J \Rightarrow L = \frac{2 \times 64}{8 ^{2}} = 2 H

The rate of dissipation fixes the resistance:

I^{2} R = 640 W \Rightarrow R = \frac{640}{64} = 10 Ω

When the coil alone is joined across an ideal battery the current grows with time constant

τ = \frac{L}{R} = \frac{2}{10} = 0.2 s

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option

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