First slide

Inductance

Question

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

Moderate

A

0.4 J
B

4.0 J
C

0.8 J
D

0.04 J

Solution

$U = \frac{1}{2} {Li}^{2} = \frac{1}{2} \times (50 \times 10^{- 3}) \times (4)^{2} = 400 \times 10^{- 3} = 0 .4 J$

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