In a coil, an increase in current from 5 A to 10 A in 100 ms induces an induced emf of 100 V. The self inductance of the coil is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2 H
The emf induced in a coil by its own changing current is given by self-induction:
\( \displaystyle |\varepsilon| = L\,\frac{\Delta I}{\Delta t} \;\Rightarrow\; L = \frac{|\varepsilon|\,\Delta t}{\Delta I} \)
\( \Delta I = 10 - 5 = 5\,\mathrm{A} \)
\( \Delta t = 100\,\mathrm{ms} = 0.1\,\mathrm{s} \)
\( \displaystyle L = \frac{100 \times 0.1}{5} = 2\,\mathrm{H} \)
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 – …
