Question

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

• A. $\frac{1}{ln2}\sec$
• B. $\frac{2}{ln2}\sec$
• C. $\frac{3}{ln2}\sec$
• D. $\frac{4}{ln2}\sec$

Answer 1

The correct option is B $\frac{2}{ln2}\sec$

$I=I_0(1-e^{-t/\tau })$
Where $t=\to$ time constant
$\therefore \frac{3}{4}{I}_0=I_0(I-e^{-t/\tau })$
$\Rightarrow \frac{3}{4}=I-e^{-t/\tau }\Rightarrow e^{-t/\tau }=\frac{1}{4}$
$\Rightarrow \frac{-t}{\tau }\ln e=\ln \frac{1}{4}$
$\frac{-4}{\tau }=-2\ln 2$
$\Rightarrow \tau =\frac{2}{\ln 2}$
$\therefore \tau =\frac{5\times 5890\times {10}^{10}}{0.45}=6.544\times {10}^{-4}$ cm