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=I1−I2=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.
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