Question

An e.m.f. of $15\text{V}$ is applied in a circuit containing $5\text{H}$ inductance and $10\Omega$ resistance. The ratio of the currents in the circuit at time $t=1\text{s}$ and $t=\infty$ is

• A. $1-e^{-1}$
• B. $\frac{e^2-1}{e^2}$
• C. $\frac{e^{\left(\frac{1}{2}\right)}-1}{e^{\left(\frac{1}{2}\right)}}$
• D. $\frac{1}{e}$

Answer 1

The correct option is B $\frac{e^2-1}{e^2}$

For a d.c. series $LR$ circuit, rising current at any instant is,

$I=I_0(1-e^{(-\frac{t}{\tau })})[\because \tau =\frac{L}{R}]$

$I=I_0(1-e^{(-\frac{Rt}{L})})$

When $t=1\text{s}$ current in the circuit is,

$I_1=I_0[1-e^{-\frac{R}{L}}].......\left(1\right)$

When $t=\infty$ current in the circuit is,

$I_{\infty }=I_0[1-e^{-\frac{R\infty }{L}}]$

$I_{\infty }=I_0.......\left(2\right)[\because e^{-\infty }=0]$

Now, from $\left(1\right)$ and $\left(2\right)$ we get,

$\frac{I_1}{I_{\infty }}=\frac{I_0(1-e^{-\frac{R}{L}})}{I_0}$

$\frac{I_1}{I_{\infty }}=1-e^{-\frac{10}{5}}=1-e^{-2}$

$\therefore \frac{I_1}{I_{\infty }}=1-\frac{1}{e^2}=\frac{e^2-1}{e^2}$

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