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Numerical

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.

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#### Solution

Initial current, I_{1} = 5.0 A

Final current, I_{2} = 0.0 A

Change in current, dI = I_{1} − I_{2} = 5 A

Time taken for the change, t = 0.1 s

Average emf, e = 200 V

For self-inductance (L) of the coil, we have the relation for average emf as:

e = `"L""di"/"dt"`

`"L" = "e"/((("di")/("dt")))`

= `200/(5/0.1)`

= 4 H

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

Concept: Inductance - Self Inductance

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