Q. 84.4( 20 Votes )

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 :

Given: Initial current I1 = 5.0 A

Final current I2 = 0.0 A

Time = 0.1 seconds

Average e.m.f = 200 V

Change in the current can be calculated as:

dI = I1 - I2 = 5.0A – 0.0A

dI = 5 A

The self-conductance of the circuit is calculated using the formula:


substituting the values, we get

L = 4 Henry

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