The coefficient of mutual induction is 2 H and induced e.m.f. across secondary is 2 kV . Current in the…

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
  1. $4 \times 10^{-3} s$
  2. $6 \times 10^{-3} \mathrm{~s}$
  3. $2 \times 10^{-3} \mathrm{~s}$
  4. $3 \times 10^{-3} s$

Solution

Using the formula for induced electromotive force, $\varepsilon = M \frac{\Delta I}{\Delta t}$, with mutual inductance $M = 2\text{ H}$, e.m.f. $\varepsilon = 2000\text{ V}$, and current change $\Delta I = 3\text{ A}$, we solve for the time interval:

$\Delta t = \frac{M \Delta I}{\varepsilon} = \frac{2 \times 3}{2000}$

Simplifying gives $\Delta t = 0.003\text{ s}$ or $3 \times 10^{-3}\text{ s}$, which corresponds to option D.

Asked in: MHT CET 2025 (20 April Shift 1)

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