Question

An inductor of inductance $5.0\text{H}$ having a negligible resistance is connected in series with a $100\Omega$ resistor and a battery of emf $2\text{V}$. Find the potential difference across the resistor $20\text{ms}$ after the circuit is switched on.

$[$Use $(1-e^{-0.4})=0.33]$

• A. $0.66\text{V}$
• B. $0.33\text{V}$
• C. $0.11\text{V}$
• D. $0.06\text{V}$

Answer 1

The correct option is A $0.66\text{V}$

Given:
$L=5\text{H};R=100\Omega E=2\text{V};t=20\text{ms}$


At any instant time $t$ current through the $L-R$ circuit is,

$i=i_0(1-e^{-t/\tau })$

Here, $i_0=\frac{E}{R}=\frac{2}{100}=0.02\text{A}$ and

The time constant is,

$\tau =\frac{L}{R}=\frac{5}{100}=0.05\text{s}$

Putting these values in $\left(1\right)$ we get,

$i=0.02(1-e^{\frac{-20\times {10}^{-3}}{0.05}})$

$i=0.02(1-e^{-0.4})=0.02\times 0.330$

$\therefore i=0.0066\text{A}$

$\therefore$ Potential difference across the resistor at $t=20\text{ms}$ is,

$V=iR=0.0066\times 100=0.66\text{V}$

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