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  4. The energy stored in a 50 mH inductor carrying a current of 4 A will be

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Q. The energy stored in a $50\, mH$ inductor carrying a current of $4\, A$ will be

Electromagnetic Induction

Solution:

$U=\frac{1}{2} L i^{2}=\frac{1}{2} \times\left(50 \times 10^{-3}\right) \times(4)^{2}$
$=400 \times 10^{-3}=0.4\, J$