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

The correct option is C 0.08 J

$U=\frac{1}{2}L{i}^2$
$=\frac{1}{2}40\times {10}^{-3}\times (2{)}^2$
$=80\times {10}^{-3}$
$U=0.08J$