A 50 mH coil carries a current of 2 A. The energy stored in this coil is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) - 0.1 J
The energy stored in an inductor is given by the formula -
$E=\frac{1}{2}LI^2$
and,
E, Energy level
L, Inductance of coil = $50mH = 50 ×10^{-3}H$ [given]
I, Current = 2A [given]
$∴E=\frac{1}{2}×50×10^{-3}×(2)^2$
$=0.1J$