The current in an LR circuit builds up to $\frac{3^{\text{th }}}{4}$of its steady state value in 4 s. The time constant of this circuit is

We know that $\mathrm{i}={\mathrm{i}}_{\mathrm{o}}\left[1-{\mathrm{e}}^{\frac{-\mathrm{Rt}}{\mathrm{L}}}\right]\text{ or }\frac{3}{4}{\mathrm{i}}_{\mathrm{o}}={\mathrm{i}}_{\mathrm{o}}\left[1-{\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }}\right]$ 

$\text{ (where }\left.\mathrm{\tau }=\frac{\mathrm{L}}{\mathrm{R}}=\text{ time constant }\right)$

$\frac{3}{4}=1-{\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }}\text{ or }{\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }}=1-\frac{3}{4}=\frac{1}{4}$

${\mathrm{e}}^{\mathrm{t}/\mathrm{\tau }}=4\text{ or }\frac{\mathrm{t}}{\mathrm{\tau }}=\ln 4$

$\Rightarrow \mathrm{\tau }=\frac{\mathrm{t}}{\ln 4}=\frac{4}{2\ln 2}\Rightarrow \mathrm{\tau }=\frac{2}{\ln 2} \mathrm{s}$