Question

An $L-R$ circuit has $L=1\text{H}$ and $R=2\Omega$. It is connected across an emf of $2\text{V}$ at $t=0$. Find the rate of change of current in the circuit at $t=100\text{ms}.$ $(e^{1/5}=1.22)$

• A. $1.63\text{A/s}$
• B. $3.26\text{A/s}$
• C. $4.81\text{A/s}$
• D. $6\text{A/s}$

Answer 1

The correct option is A $1.63\text{A/s}$

Given:
$L=1\text{H};R=2\Omega ;E=2\text{V}$

The current in an $L-R$ circuit is given by

$i=\frac{E}{R}(1-e^{\frac{-t}{\tau }})$

$\therefore$ Rate of change of current is,

$\frac{di}{dt}=\frac{E}{R}\times \frac{1}{\tau }\left(e^{\frac{-t}{\tau }}\right)$

$\Rightarrow \frac{di}{dt}=\frac{E}{R\tau }{e}^{\frac{-t}{\tau }}$

Here, $\tau =\frac{L}{R}=\frac{1}{2}\text{s}=0.5\text{s}$

$\therefore \text{At}t=100\text{ms}=0.1\text{s}$ ,

$\frac{di}{dt}=\frac{2}{2\times 0.5}\times e^{(-\frac{0.1}{0.5})}=2e^{-\frac{1}{5}}=\frac{2}{e^{\frac{1}{5}}}$
$\therefore \frac{di}{dt}=1.63\text{A/s}$

Hence, $\left(\text{A}\right)$ is the correct answer.