The current in a coil falls from 5 A to 0 A in 0.1 s. If an average emf of 100 V is induced, then the self -inductance of the coil would be-
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2 H
Given:
Initial current: $I_i = 5\ \text{A}$
Final current: $I_f = 0\ \text{A}$
Time interval: $\Delta t = 0.1\ \text{s}$
Average induced emf: $\mathcal{E} = 100\ \text{V}$
Induced emf due to self-inductance: $\mathcal{E} = L \frac{\Delta I}{\Delta t}$
Change in current: $\Delta I = I_f - I_i = 0 - 5 = -5\ \text{A}$
Magnitude: $\mathcal{E} = L \frac{|\Delta I|}{\Delta t} \Rightarrow L = \frac{\mathcal{E} \Delta t}{|\Delta I|} = \frac{100 \cdot 0.1}{5} = 2\ \text{H}$
∴ Self-inductance of the coil = 2 H