Solution:
Formula for rate of change of energy in an inductor is \(\frac{dU}{dt} = LI\frac{dI}{dt}\). Given \(L = 2\text{ H}\), \(I = 2\text{ A}\), and \(\frac{dI}{dt} = 4\text{ A/s}\), we get \(\frac{dU}{dt} = 2 \times 2 \times 4 = 16\text{ J/s}\).
A current of \(2\text{ A}\) is increasing at a rate of \(4\text{ A/s}\) through a coil of inductance \(2\text{ H}\). The energy stored in the inductor per unit time in given instant is:
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