Question

An $L-R$ circuit having $L=1\text{H}$ , $R=5\Omega$ and $E=2\text{V}$ is switched on at $t=0.$ Find the charge that passes through the battery in one time constant $\left(\tau \right)$.

• A. $0.872\text{C}$
• B. $1.215\text{C}$
• C. $0.0294\text{C}$
• D. $0.005\text{C}$

Answer 1

The correct option is C $0.0294\text{C}$

Given:
$L=1\text{H};R=5\Omega ;E=2\text{V}$

In the $L-R$ circuit, current at any time $t$ is,

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

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

$\because \tau =\frac{L}{R}=\frac{1}{5}\text{s}$

$i=\frac{2}{5}[1-e^{\frac{-t}{\left(1/5\right)}}]$

$i=\frac{2}{5}[1-e^{-5t}]$

The charge passing through the battery during period $t$ to $t+dt$ is,

$dq=idt......\left(1\right)$

Thus, total charge passed during $t=0$ to $t=\tau =\frac{1}{5}\text{s}$ is,

Integrating eq.$\left(1\right)$, with proper limits, we get,

${\int }_0^{q}dq={\int }_0^{\tau }\frac{2}{5}(1-e^{-5t})dt$

$\Rightarrow q=\frac{2}{5}{[t-\frac{e^{-5t}}{-5}]}_0^{\tau }=\frac{2}{5}{[t-\frac{e^{-5t}}{-5}]}_0^{\frac{1}{5}}$

$q=\frac{2}{5}(\frac{1}{5}+\frac{1}{5e}-\frac{1}{5})$

$q=\frac{2}{5}\times \frac{1}{5e}$

$q=\frac{2}{25e}=0.0294\text{C}$

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