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MHT CET 2024 (11 May, Shift 1) · Mutual Inductance

Mutual Inductance question from MHT CET 2024 (11 May, Shift 1)

Q24MHT CET 2024 · 11 MAY · SHIFT 1Easy

Mutual Inductance

The coefficient of mutual induction is 2 H and induced e.m.f. across secondary is 2 kV. Current in the primary is reduced from 6 A to 3 A. The time required for the change of current is

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Answer: (a)

  1. Induced emf in secondary: e=MΔIΔt⇒Δt=M ΔIee = M\dfrac{\Delta I}{\Delta t} \Rightarrow \Delta t = \dfrac{M\,\Delta I}{e}
  2. Δt=2×(6−3)2000=3×10−3 s\Delta t = \dfrac{2 \times (6-3)}{2000} = 3 \times 10^{-3}\text{ s}
  3. So the answer is (a) 3×10−3 s3 \times 10^{-3}\text{ s}.
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