First slide

Inductance

Question

Energy stored in a coil of self-inductance 40 mH carrying a steady current of 2 A is

Easy

A

0.8 J
B

8 J
C

0.08 J
D

80 J

Solution

$U = \frac{1}{2} {Li}^{2} \Rightarrow U = \frac{1}{2} \times 40 \times 10^{- 3} \times (2)^{2} = 0.08 J$

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