In an inductive circuit, the current falls from 7.5 A to 1.5 A in 0.15 s. It induces emf of 220 V in the inductive coil. The inductance of the circuit is

5.5 H

The correct answer is Option (2) → 5.5 H

Given:

Initial current, $I_1 = 7.5\,A$

Final current, $I_2 = 1.5\,A$

Change in time, $\Delta t = 0.15\,s$

Induced emf, $E = 220\,V$

Induced emf in an inductor:

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

Therefore,

$L = \frac{E \times \Delta t}{\Delta I}$

Substitute values:

$L = \frac{220 \times 0.15}{(7.5 - 1.5)}$

$L = \frac{33}{6} = 5.5\,H$

Final Answer: $L = 5.5\,H$