First slide

Inductance

Question

A 50 mH coil carries a current of 2 A. The energy stored, in joules, is

Easy

A

1
B

0.1
C

0.05
D

0.5

Solution

Energy stored, $E = \frac{1}{2} {Li}^{2} = \frac{1}{2} \times 50 \times 10^{- 3} \times 4 = 0.1 J$

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