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Electromagnetic Induction question

2020 · 9 Jan · Shift 1 · Q43
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Electromagnetic Induction question

2020 · 9 Jan · Shift 1 · Q43

JEE MainPhysicsElectromagnetic InductionNumerical+4 / −1
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 ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 10

  1. Use the inductor emf relation

    For a choke (inductor), the induced emf magnitude is ∣V∣=L∣dIdt∣|V| = L\left|\frac{dI}{dt}\right|∣V∣=L​dtdI​​

  2. Given data

    • Reverse voltage: V=100 VV = 100\ \text{V}V=100 V
    • Current changes from 0.25 A0.25\ \text{A}0.25 A to 000 so ΔI=0.25 A\Delta I = 0.25\ \text{A}ΔI=0.25 A
    • Time duration: 0.025 ms=0.025×10−3 s=2.5×10−5 s0.025\ \text{ms} = 0.025 \times 10^{-3}\ \text{s} = 2.5 \times 10^{-5}\ \text{s}0.025 ms=0.025×10−3 s=2.5×10−5 s

  3. Compute rate of change of current

    Since the change is uniform, ∣dIdt∣=ΔIΔt=0.252.5×10−5=104 A/s\left|\frac{dI}{dt}\right| = \frac{\Delta I}{\Delta t} = \frac{0.25}{2.5\times 10^{-5}} = 10^4\ \text{A/s}​dtdI​​=ΔtΔI​=2.5×10−50.25​=104 A/s

  4. Find self-inductance

    L=VdI/dt=100104=10−2 HL = \frac{V}{dI/dt} = \frac{100}{10^4} = 10^{-2}\ \text{H}L=dI/dtV​=104100​=10−2 H

    10−2 H=0.01 H=10 mH10^{-2}\ \text{H} = 0.01\ \text{H} = 10\ \text{mH}10−2 H=0.01 H=10 mH

  5. Final answer

    10\boxed{10}10​

  6. Comparison with stored correct answer

    Stored correct answer = 10

    Our derived answer = 10

    Hence, the answer agrees with the stored correct answer.

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