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  4. The current in a LR circuit builds up to 3 / 4 text th of its steady state value in 4 s. The time constant of this circuit is :-

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Q. The current in a LR circuit builds up to $3 / 4^{\text {th }}$ of its steady state value in $4\, s$. The time constant of this circuit is :-

Solution:

Current in $\frac{L}{ R }$ circuit
$I = I _{0}\left(1- e ^{-\frac{ t }{\tau}}\right)=\frac{3}{4} I _{0}$
$\Rightarrow e ^{-\frac{ t }{\tau}}=\frac{1}{4}$
$\Rightarrow e \frac{ t }{\tau}=4$
$\frac{ t }{\tau}=\ln 4=2 \ln 2, \tau=\frac{ t }{2 \ln 2}$
$t =4\, \sec , \tau=\frac{2}{\ln 2} \sec$