Question

The energy stored in an inductor of self inductance $L$ henry carrying a current of $IA$ is

• A. $L^{2}l$
• B. $-L{l}^2$
• C. $\frac{1}{2}L{l}^2$
• D. $\frac{1}{2}{L}^{2}l$

Answer 1

The correct option is C

$\frac{1}{2}L{l}^2$

Step 1: Given data:

1. Self-inductance of an inductor $=L$
2. Current $=IA$

Step 2: Formula used:

The voltage across the inductor,

$V=L\frac{dI}{dt}$

The rate at which energy is delivered to the inductor,

$P=VI$ (Where $P=$Power dissipated)

Step 3: Calculating the energy stored in an inductor:

An increase of energy within the inductor is -

$\begin{array}{rcl}dU& =& Pdt\\ & =& VIdt\\ & =& L\frac{dI}{dt}Idt\\ & =& LIdI\end{array}$

The energy stored in the inductor can be calculated by integrating the above equation

$U=\int dU=\int LIdI$

$\begin{array}{rcl}U& =& L{\int }_0^{I}IdI\\ & =& L{{\left[\frac{I^2}{2}\right]}_0}^I\\ & =& \frac{1}{2}L{l}^2\end{array}$

Hence, option C is the correct answer.