#### Question

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.

#### Solution

Initial current, *I*_{1} = 5.0 A

Final current, *I*_{2} = 0.0 A

Change in current`dI=I_1-I_2=5A`,

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)=4H`

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

Is there an error in this question or solution?

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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. Concept: Inductance - Self-Inductance.

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