Question

An inductor, a resistor and a battery are connected in series as shown in the figure. Find the time elapsed before the current reaches $99\%$ of the maximum value.

• A. $3\ln \left(10\right)\text{ms}$
• B. $2.5\ln \left(10\right)\text{ms}$
• C. $\ln \left(10\right)\text{ms}$
• D. $0.4\ln \left(10\right)\text{ms}$

Answer 1

The correct option is D $0.4\ln \left(10\right)\text{ms}$

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

The time constant of the given circuit is,

$\tau =\frac{L}{R}=\frac{20\times {10}^{-3}}{100}=0.2\text{ms}$

The maximum current in the circuit is,

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

Using the relation for current growth,

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

$\frac{99}{100}{i}_0=i_0(1-e^{-t/\tau })(i=99\%of{i}_0)$

$e^{-t/\tau }=\frac{1}{100}\Rightarrow e^{t/\tau }=100$

Taking log on both sides, we get,

$t=\tau \times \ln \left(100\right)=0.2\times 2\ln \left(10\right)$

$\therefore t=0.4\ln \left(10\right)\text{ms}$

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

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