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:
Answer & explanation
Correct answer: option 2
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]
$1/L=\frac{E\Delta I}{\Delta t}=\frac{200×0.1}{5}$
$L = 4 H$