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Q6.8  Current in a circuit falls from 5.0A to 0.0A  in 0.1s . If an average emf  of 200V  induced, give an estimate of the self-inductance of the circuit.

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Given

Initial current I_{initial}=5A

Final current I_{final}=0A

Change in time I_{final}=\Delta t=0.1s

Average emf e=200V

Now,

As we know, in an inductor

e=L\frac{di}{dt}=L\frac{\Delta I}{\Delta t}=L\frac{I_{final}-I_{initial}}{\Delta t}

L= \frac{e\Delta t}{I_{final}-I_{initial}}=\frac{200*0.1}{5-0}=4H

Hence self-inductance of the circuit is 4H.

Posted by

Pankaj Sanodiya

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