Question

An $L-R$ circuit is connected to a battery at time $t=0$. The energy stored in the inductor reaches half its maximum value at time

• A. $\frac{R}{L}\log \left[\frac{\sqrt{2}}{\sqrt{2}-1}\right]$
• B. $\frac{L}{R}\log \left[\frac{\sqrt{2}-1}{\sqrt{2}}\right]$
• C. $\frac{L}{R}\log \left[\frac{\sqrt{2}}{\sqrt{2}-1}\right]$
• D. $\frac{R}{L}\log \left[\frac{\sqrt{2}-1}{\sqrt{2}}\right]$

Answer 1

The correct option is C $\frac{L}{R}\log \left[\frac{\sqrt{2}}{\sqrt{2}-1}\right]$

$\frac{1}{2}L{i}^2=\frac{1}{2}\left[\frac{1}{2}L{i}_0^2\right]$

$\therefore i=\frac{i_0}{\sqrt{2}}=i_0(1-e^{-t/{\tau }_L})$

$e^{-t/{\tau }_L}=1=1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}$

$\therefore \frac{t}{{\tau }_L}=\ln \left(\frac{\sqrt{2}}{\sqrt{2}-1}\right)$

or $t={\tau }_L\ln \left(\frac{\sqrt{2}}{\sqrt{2}-1}\right)=\frac{L}{R}\ln \left(\frac{\sqrt{2}}{\sqrt{2}-1}\right)$