The current in a circuit drops from 2.0 A to 0.0 A in 50 ms. If the average induced emf in the circuit is 200 V, the self inductance of the circuit is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5.0 H
A changing current induces a self-induced emf \( |\varepsilon| = L\,|\Delta I/\Delta t| \).
\( \displaystyle \frac{\Delta I}{\Delta t} = \frac{2.0\,\mathrm{A}}{50\times10^{-3}\,\mathrm{s}} = 40\,\mathrm{A\,s^{-1}} \)
\( \displaystyle L = \frac{|\varepsilon|}{\Delta I/\Delta t} = \frac{200}{40} \)
\( \displaystyle L = 5.0\,\mathrm{H} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
