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=5.0A Final current, I2=0A Change in current, dl=I1−I2=5A Time taken for the charge, t=0.1s Average emf, e=200V For self inductance(L) of the coil, we have the relation for average emf as: e=Ldtdi L=dtdie =0.15=4H200 Hence the self inductance of the coil is 4H.
