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A current of 2 A is increasing at a rate...

A current of `2 A` is increasing at a rate of `4 A s^(-1)` through a coil of inductance `1 H`. Find the energy stored in the inductor per unit time in the units of `J s^(-1)`.

Text Solution

Verified by Experts

The correct Answer is:
`(8)`
Potential difference across the coil is `V = L(di)/(dt)`
or `V = (1)(4) = 4V`
Now energy stored per unit time = power
`= Vi = (4)(2) = 8 J//s`
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