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If a current of 3.0 amperes flowing in ...

If a current of `3.0` amperes flowing in the primary coil is reduced to zero in `0.001` second, then the induced e.m.f. in the secondary coil is `15000 "volts"`. The mutual inductance between the two coils is

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Here, `(dI)/(dt) = (0 - 3.0)/(0.001) = - 3000 A//s`
`e = 15000 V, M = ?`
From `e = - M (dI)/(dt)`
`M = - e (dt)/(dI) = - 15000 ((-1)/(3000)) = 5 H`
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