39. For a inductor coil $L = 0.04 H$, then work done by source to establish a current of $5 A$ in it is: (1999)
Solution:
Work done to establish a current in an inductor is stored as magnetic energy: $W = \frac{1}{2} L I^2$.
Given $L = 0.04 H$ and $I = 5 A$.
$W = \frac{1}{2} \times 0.04 \times 5^2 = 0.02 \times 25 = 0.5 J$.
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