Question

The current in an $L-R$ circuit builds up to $\left(3 / 4^{\text {th }}\right)$ of its steady state value in $4$ seconds. The time constant of this circuit is

• A. $\frac{1}{\ln 2} \sec$
• B. $\frac{2}{\ln 2} \sec$
• C. $\frac{3}{\ln 2} \sec$
• D. $\frac{4}{\ln 2} \sec$

Answer 1

The Correct Option is B

$I = I _{0}\left(1- e ^{-t / \tau}\right)$
Where $t=\rightarrow$ time constant
$\therefore \frac{3}{4} I _{0}= I _{0}\left( I - e ^{- t \tau}\right) $
$\Rightarrow \frac{3}{4}= I - e ^{- t \tau}$
$ \Rightarrow e ^{- t \tau}=\frac{1}{4}$
$\Rightarrow \frac{- t }{\tau} \ln e =\ln \frac{1}{4}$
$\frac{-4}{\tau}=-2 \ln 2 $
$\Rightarrow \tau=\frac{2}{\ln 2} $
$\therefore \tau=\frac{5 \times 5890 \times 10^{10}}{0.45}$
$=6.544 \times 10^{-4} \,cm$