Solveeit Logo
Login

Question

Physics Question on Electromagnetic induction

The energy stored in an inductance of self inductance LL heavy carrying a current of II ampere is

A

12L2I\frac{1}{2} L^{2} I

B

12LI2\frac{1}{2} L I^{2}

C

LI2L I^{2}

D

L2IL^{2} I

Answer

12LI2\frac{1}{2} L I^{2}

Explanation

Solution

The energy stored in an inductor U=12LI2U=\frac{1}{2} L I^{2}