Question

A coil of inductance $1\text{H}$ and resistance $100\Omega$ is connected to a battery of emf $12\text{V}.$ Find the energy stored in the coil at an instant $t=10\text{ms}$ after the circuit is switched on.
[Use $e^{-1}=0.367$]

• A. $1\text{mJ}$
• B. $1.5\text{mJ}$
• C. $2.8\text{mJ}$
• D. $4.5\text{mJ}$

Answer 1

The correct option is C $2.8\text{mJ}$

Given:
$L=1\text{H};R=100\Omega E=12\text{V};t=10\text{ms}$

We know that energy stored in an inductor is given by,

$U=\frac{1}{2}L{i}^2$

Current through the $L-R$ circuit at any instant $t$ is,

$i=i_0(1-e^{\frac{-t}{\tau }})=\frac{E}{R}(1-e^{\frac{-t}{\tau }})$

$i=\frac{12}{100}(1-e^{\frac{-t}{\left(\frac{1}{100}\right)}})[\because \tau =\frac{L}{R}=\frac{1}{100}]$

$i=0.12(1-e^{-100t})$

$\therefore$ Current $i$ at $t=10\text{ms}$ is,

$i=0.12(1-e^{-100\times (10\times {10}^{-3})})$

$i=0.12(1-e^{-1})=0.075\text{A}$

$\therefore$ Energy stored in the inductor at $t=10\text{ms}$ is,

$E=\frac{1}{2}L{i}^2=\frac{1}{2}\times 1\times [0.075{]}^2$

$\therefore E=2.8\text{mJ}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{C}\right)$ is the correct answer.

Alternate solution: Energy stored in the coil is, $\int dU=\int Li\frac{di}{dt}dt$ $\int dU=\int L{i}_0(1-e^{-t/\tau })\times i_0(\frac{e^{-t/\tau }}{\tau })dt$ $U=L{i}_0^2\int (1-e^{-t/\tau })(\frac{e^{-t/\tau }}{\tau })dt$ $U=L(\frac{E}{R}{)}^2{\int }_0^t(1-e^{-t/\tau })(\frac{e^{-t/\tau }}{\tau })dt$ $U=L(\frac{E}{R}{)}^2\frac{{(1-e^{-t/\tau })}^2}{2}$ Put, $L=1\text{H};R=100\Omega E=12\text{V};t=10\text{ms}$ $\tau =\frac{L}{R}=\frac{1}{100}\text{s}$ $U=2.8\text{mJ}$