Question

An inductor $L=20\text{mH}$, a resistor $R=100\Omega$ and a battery $E=10\text{V}$ are connected in series. After a long time the battery is short-circuited. Find the current in the circuit, $1\text{ms}$ after short-circuiting.
$[$Use $e^{-5}=0.00673]$

• A. $6.7\times {10}^{-4}\text{A}$
• B. $2.5\times {10}^{-4}\text{A}$
• C. $0.5\times {10}^{-4}\text{A}$
• D. $2\times {10}^{-3}\text{A}$

Answer 1

The correct option is A $6.7\times {10}^{-4}\text{A}$

Given:
$L=20\text{mH};R=100\Omega ;E=10\text{V};t=1\text{ms}$

The steady state current in the circuit is,

$i_0=\frac{E}{R}=\frac{10}{100}=0.1\text{A}$

The time constant of the circuit is,

$\tau =\frac{L}{R}=\frac{20\times {10}^{-3}}{100}$

$\tau =2\times {10}^{-4}\text{s}$

Using, the relation for current decay we get,

$i=i_0{e}^{-t/\tau }$

$i=0.1e^{-({10}^{-3}/2\times {10}^{-4})}$

$\therefore i=0.1\times e^{-5}=6.7\times {10}^{-4}\text{A}$

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

Why this question ? This question gives you a basic understanding of decay of current in $L-R$ circuit.