Question

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

• A. $\frac{\sqrt{e}}{\sqrt{e}-1}$
• B. $\frac{e^2}{e^2-1}$
• C. $1-\frac{1}{e}$
• D. $\frac{1}{e}$

Answer 1

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

Current in the growing circuit is given by

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

When, $t=\infty ,I=I_0$

$I_{t=\infty }=\frac{E}{R}=\frac{15}{10}=1.5\text{A}$

At $t=1\text{sec}$

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

$=1.5\times [1-e^{(-\frac{10}{5})}]$

$\therefore I_{t=1}=1.5(1-\frac{1}{e^2})=1.5\left(\frac{e^2-1}{e^2}\right)$

Now, $\frac{I_{t=\infty }}{I_{t=1}}=\frac{e^2}{e^2-1}$

Hence, $\left(\text{B}\right)$ is the correct option.