Question

A simple LR circuit is connected to a battery at time $t=0$. The energy stored in the inductor reaches half its maximum value at time

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

Answer 1

Answer: (C)

$\frac{L}{R}\ln \left[\frac{\sqrt{2}}{\sqrt{2}-1}\right]$

RL circuit analysis :-

Kirchhoff's Volatage Law;

$⟹V-iR-L\frac{di}{dt}=0$;

$⟹V-iR=L\frac{di}{dt}$

$⟹\frac{1}{L}dt=\frac{1}{V-iR}di$

$⟹{\int }_0^t\frac{1}{L}dt={\int }_0^i\frac{1}{V-iR}di$

$⟹\frac{-t}{L}R=\ln \left(\frac{V-iR}{V}\right)$

$⟹e^{\frac{-Rt}{L}}=\frac{V-iR}{V}$

$⟹iR=V(1-e^{\frac{-Rt}{L}})$;

$⟹i=\frac{V}{R}(1-e^{\frac{-Rt}{L}})$;

When $t\to \infty ,i\to \frac{V}{R}$

Energy stored in an inductor $=L{i}^2/2$

Energy stored reaches half ita maximum value when $i$ becomes $⟹\frac{V}{\sqrt{2}R}$.

thus,

$⟹\frac{V}{R\sqrt{2}}=\frac{V}{R}(1-e^{\frac{-Rt}{L}})$.

$⟹1-e^{\frac{-Rt}{L}}=\frac{1}{\sqrt{2}}$

$⟹1-\frac{1}{\sqrt{2}}=e^{\frac{-Rt}{L}}$

$⟹\ln (1-\frac{1}{\sqrt{2}})=\frac{-Rt}{L}$

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