Question

In a decaying $L-R$ circuit, the time after which energy stored in the inductor reduces to one-fourth of its initial value is :

• A. $\left(\ln 2\right)\frac{L}{R}$
• B. $0.5\frac{L}{r}$
• C. $\sqrt{2}\frac{L}{R}$
• D. $\left(\frac{\sqrt{2}}{\sqrt{2}-1}\right)\frac{L}{R}$

Answer 1

Answer: (A)

$\left(\ln 2\right)\frac{L}{R}$

$\frac{1}{2}L{i}^2=\frac{1}{4}\left[\frac{1}{2}L{i}_0^2\right]$
So, $i=\frac{i_0}{2}$, half value
$\therefore$ $t=t_{1/2}=\left(\ln 2\right){\tau }_L=\left(\ln 2\right)\left(\frac{L}{R}\right)$