Question

A solenoid of inductance $50\text{mH}$ and resistance $10\Omega$ is connected to a battery of $6\text{V}.$ The time elapsed before the current acquires half of its steady state value from zero is-
$(\ln 2=0.693)$

• A. $10\text{ms}$
• B. $3.5\text{ms}$
• C. $0.25\text{ms}$
• D. $7\text{ms}$

Answer 1

The correct option is B $3.5\text{ms}$

Given, $L=50\text{mH};R=10\Omega ;E=6\text{V}$

Time constant of the circuit is,

$\tau =\frac{L}{R}=\frac{50\times {10}^{-3}}{10}=5\text{ms}$

Using, the relation for current growth we get,

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

For, $i=\frac{i_0}{2}$

$\frac{i_0}{2}=i_0(1-e^{-t/\tau })$

$\Rightarrow e^{-t/\tau }=\frac{1}{2}$

$\therefore e^{t/\tau }=2$

Taking log on both sides we get,

$\frac{t}{\tau }=\ln 2$

$t=\tau \ln 2=5\times {10}^{-3}\times 0.693$

$\therefore t\approx 3.5\text{ms}$

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

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