12 V battery connected to a coil of resistance 6 Ω through a switch, drives a constant current in the circuit. The switch is opened in 1 ms. The emf induced across the coil is 20 V. The inductance of the coil is: (a) 5 mH (b) 8 mH (c) 10 mH (d) 12 mH

Answer: (c). Steady current I = V/R = 12/6 = 2 A, falls to 0 in 1 ms. e = L (Δ I)/(Δ t) ⇒ L = (e Δ t)/(Δ I) = 20×10⁻³2 L = 10×10⁻³ H So the answer is (c) 10 mH.

12 V battery connected to a coil of resistance 6 Ω through a switch… | JEE Main 2023 (15 Apr, Shift 1)

Q58JEE Main 2023 · 15 APR · SHIFT 1Medium

Self-Inductance & Energy

$12\text{ V}$ battery connected to a coil of resistance $6\ \Omega$ through a switch, drives a constant current in the circuit. The switch is opened in $1\text{ ms}$ . The emf induced across the coil is $20\text{ V}$ . The inductance of the coil is:

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Answer: (c)

Steady current $I = \dfrac{V}{R} = \dfrac{12}{6} = 2\text{ A}$ , falls to 0 in 1 ms.
$e = L\,\dfrac{\Delta I}{\Delta t} \Rightarrow L = \dfrac{e\,\Delta t}{\Delta I} = \dfrac{20\times10^{-3}}{2}$
$L = 10\times10^{-3}\text{ H}$
So the answer is (c) $10\text{ mH}$ .

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