The current in a coil falls from 5.0 to 0.0 A in 0.1 s. If average emf of 200 V is induced, the value of self inductance of coil is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 4\,\mathrm{H} \)
A changing current in a coil induces an emf in the same coil by self-induction; its magnitude is the self-inductance times the rate of change of current.
\( \displaystyle |\varepsilon| = L\,\frac{|\Delta I|}{\Delta t} \)
\( |\Delta I| = 5.0 - 0.0 = 5.0\,\mathrm{A} \)
\( \Delta t = 0.1\,\mathrm{s} \)
\( \displaystyle \frac{|\Delta I|}{\Delta t} = \frac{5.0}{0.1} = 50\,\mathrm{A\,s^{-1}} \)
\( \displaystyle L = \frac{|\varepsilon|}{|\Delta I|/\Delta t} = \frac{200}{50} = 4\,\mathrm{H} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
