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Current in a coil falls from 2.5a to 0.0...

Current in a coil falls from 2.5a to 0.0a in 0.1 second inducing an emf of 200v.calculate the value of self inductance .

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given , e= 200v, dl = 2.5A, dt= 0.1s
we have
`e= L(dl)/(dt) Rightarrow 200= L (2.5)/(0.1) = 25L`
L= 8H
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