Question

A coil of resistance $40\Omega$ is connected across a $4.0\text{V}$ battery. If $0.10\text{s}$ after the battery is connected, the current in the coil is $63\text{mA}$. Find the inductance of the coil.

[Use $\ln \left(2.70\right)=1$]

• A. $1\text{H}$
• B. $2\text{H}$
• C. $3\text{H}$
• D. $4\text{H}$

Answer 1

The correct option is D $4\text{H}$

Given:
$R=40\Omega ;E=4\text{V};i=63\text{mA};t=0.10\text{s}$

At any instant current through the $L-R$ circuits is,

$i=i_0(1-e^{\frac{-t}{\tau }})......\left(1\right)$

Here, $i_0=\frac{E}{R}=\frac{4}{40}=0.1\text{A}$ and

The time constant is, $\tau =\frac{L}{R}=\frac{L}{40}\text{s}$

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

$63\times {10}^{-3}=0.1[1-e^{\frac{-0.1}{\left(\frac{L}{40}\right)}}]$

$63\times {10}^{-2}=(1-e^{\left(-4/L\right)})$

$e^{\left(-4/L\right)}=0.37$

$\Rightarrow e^{\left(4/L\right)}=\frac{1}{0.37}=2.70$

Taking log on both sides,

$\frac{4}{L}=\ln \left(2.70\right)=1$

$\therefore L=4\text{H}$

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