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

Initial current, I₁= 5.0 A

Final current, I₂ = 0.0 A

Change in current, dl = I₁ — I₂ = 5 A

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 =di/dt

L = e/(di/dt)

=200/(5/0.1) = 4 H

Hence, the self-induction of the coil is 4 H.