Question

A series L-R circuit is connected to a battery of emf $V$. If the circuit is switched on at $t=0$, then the time at which the energy stored in the inductor reaches $\left(\frac{1}{n}\right)$ times of its maximum value is

• A. $\frac{L}{R}\ln \left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$
• B. $\frac{L}{R}\ln \left(\frac{\sqrt{n}+1}{\sqrt{n}-1}\right)$
• C. $\frac{L}{R}\ln \left(\frac{\sqrt{n}}{\sqrt{n}+1}\right)$
• D. $\frac{L}{R}\ln \left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)$

Answer 1

The correct option is A $\frac{L}{R}\ln \left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$

Potential energy stored in the inductor is,
$U=\frac{1}{2}L{I}^2\Rightarrow U\propto I^2$

During growth of current,
$I=I_{max}(1-e^{-\frac{RT}{L}})$

When, $U=\frac{U_{max}}{n},I=\frac{I_{max}}{\sqrt{n}}$

$\therefore \frac{I_{max}}{\sqrt{n}}=I_{max}(1-e^{-\frac{RT}{L}})$

$\Rightarrow e^{-\frac{RT}{L}}=1-\frac{1}{\sqrt{n}}=\frac{\sqrt{n}-1}{\sqrt{n}}$

$\Rightarrow -\frac{Rt}{L}=\ln \left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)$

$\Rightarrow t=\frac{L}{R}\ln \left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$

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