Question

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.

Answer 1

Initial current, $I_1$ = 5.0
A Final current, $I_2$ = 0.0
A Change in current, $dI$= $I_1-I_2$ = 5A
Time taken for the change, t = 0.1 s
Average emf, e = 200 V
For self-inductance (L) of the coil, we have the relation for average emf as:
e = $L^{\frac{di}{dt}}$
L = $\frac{e}{\bigg(\frac{di}{dt}\bigg)}$
= $\frac{200}{\frac{5}{0.1}}$ = 4H
Hence, the self induction of the coil is 4 H.