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An inductor of 5 H inductance carries a ...

An inductor of 5 H inductance carries a steady current of 2A. How can a 50 V self-induced emf be made to appear in the inductor ?

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L =5H, |e| = 50V
Let us produce the required emf by reducing current to zero.
Now , `|e| = L (dI)/(dt) " or " dt = (LdI)/(|e|) = (5 xx 2)/(50) s `
` = 10/50 s = 1/5 = 0.2 s` ?
So, the desired emf can be produced by reducing the given current to zero in 0.2 second.
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