The rate of change of current which produces an induced emf of 100 V in an inductor of inductance 10 H is

10 A/s

The correct answer is Option (2) → 10 A/s

Given:

Induced emf, $E = 100\ \text{V}$

Inductance, $L = 10\ \text{H}$

For an inductor, induced emf is given by:

$E = L \frac{dI}{dt}$

Rearranging:

$\frac{dI}{dt} = \frac{E}{L} = \frac{100}{10} = 10\ \text{A/s}$

∴ Rate of change of current = 10 A/s