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.

Question

Philoid · Accepted 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:

or

substituting the values, we get

∴ L = 4 Henry