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
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