The current in a coil falls from $5.0 \mathrm{~A}$ to $0.0 \mathrm{~A}$ in $0.1 \mathrm{~s}$. If average emf of $200 \mathrm{~V}$ is induced, the value of self inductance of coil is:

4 H

The correct answer is Option (2) → 4 H

Given,

ΔI, change in current = 5 - 0 = 5A

Δt, time taken = 0.1s

E = 200V

then,

$ E = L \frac{\Delta I}{\Delta t}$ [formula]

$1/L=\frac{E\Delta I}{\Delta t}=\frac{200×0.1}{5}$

$L = 4 H$