Question

A coil of inductance $8.4\text{mH}$ and resistance $6\Omega$ is connected to a $12\text{V}$ battery. The current in the coil is $1.0\text{A}$ at approximately is
[Use $\ln 2=0.693$]

• A. $10\text{s}$
• B. $10\text{ms}$
• C. $1\text{s}$
• D. $1\text{ms}$
  

Answer 1

The correct option is D $1\text{ms}$


Given, $L=8.4\text{mH};R=6\Omega V=12\text{V};I=1\text{A}$

The current growth in $RL-$ circuit is given by,

$I=I_0(1-e^{-\frac{t}{\tau }})$

Where, $I_0=\frac{V}{R}\text{and}\tau =\frac{L}{R}$

$\therefore I=\frac{V}{R}(1-e^{-\frac{Rt}{L}})$

$1=\frac{12}{6}(1-e^{-\frac{6t}{(8.4\times {10}^{-3})}})$

$\ln 2=\frac{6t}{8.4\times {10}^{-3}}$

$\Rightarrow t=0.97\times {10}^{-3}\text{s}=0.97\text{ms}$

$\Rightarrow t=1\text{ms}$

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