Q 8. 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.


 

Current at the initial point, null

Current at the final pint, null

Therefore, change in current is, dI=I1I2=5A

Total time taken, t = 0.1 s

Average emf, e = 200 V

We have the relation, for self – inductance (L) and average emf of the coil : e=didt

L=e(dtdt)=20050.1=4H

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