Question

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

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

Answer 1

The correct option is B

$\frac{L}{R}\ln \left[\frac{\sqrt{n}}{\sqrt{n}-1}\right]$

Step 1. Given data :

Battery = $V$

Time =$t$

Inductance =$L$

Current =$I$

Time constant = $\tau$

Energy = $U$

Step 2. Find the time :

Energy stored $\left(U\right)$ in the induction

$U=\frac{1}{2}L{I}^2$

($L$ is inductance ,$I$is current flow through the circuit )

$$\begin{gathered} \frac{1}{2}L{I}^2=\frac{1}{n}\left(\frac{1}{2}L{I_{\circ }}^2\right) \\ I=\frac{I_{\circ }}{\sqrt{n}} \end{gathered}$$

Value of current for $L-R$circuit, where $\tau$ is time constant,

$I=I_{\circ }\left(1-e^{\frac{-t}{\tau }}\right)$

$$\begin{gathered} \frac{I_{\circ }}{\sqrt{n}}=I_{\circ }\left(1-e^{\frac{-t}{\tau }}\right) \\ {e}^{\frac{-t}{\tau }}=1-\frac{1}{\sqrt{n}}=\frac{\sqrt{n}-1}{\sqrt{n}} \end{gathered}$$

Taking log on both side,

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

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

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

Hence, option (B) is correct.