JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Electromagnetic Induction

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In a fluorescent lamp choke (a small transformer) 100 V of reverse voltage is produced when the choke current changes uniformly from 0.25 A to 0 in a duration of 0.025 ms. The self-inductance of the choke (in mH) is estimated to be ____.

Answer: 10

Solution

The size of the self-induced emf is

∣ ε ∣ = L \frac{Δ I}{Δ t}

. The current falls by

0.25

A in

0.025

ms, that is

2.5 \times 1 0^{- 5}

s, so

100 = L \times \frac{0.25}{2.5 \times 1 0 ^{- 5}} = L \times 1 0^{4}

. This gives

L = 1 0^{- 2}

H, which is

10

mH.

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Integer answer

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