The rate of change of current which produces an induced emf of 100 V in an inductor of inductance 10 H is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 10 A/s
The emf induced in an inductor by a changing current is given by self-induction: its magnitude is the inductance times the rate of change of current.
\( \displaystyle |\varepsilon| = L\,\frac{dI}{dt} \)
\( \displaystyle \frac{dI}{dt} = \frac{|\varepsilon|}{L} = \frac{100\,\mathrm{V}}{10\,\mathrm{H}} = 10\,\mathrm{A/s} \)