Question

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

• A. 0.8 J
• B. 8 J
• C. 0.08 J
• D. 80 J

Answer 1

Answer: (C)

0.08 J

Answer 2

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