Physics | VITEEE 2016 | Discussion Page

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

A) $\frac{1}{In 2}sec$

B) $\frac{2}{In 2}sec$

C) $\frac{3}{In 2}sec$

D) $\frac{4}{In 2}sec$

Option B

$I= I_{0}(1-e^{-t / \tau}$)

Where $\tau$ → time constant

$\therefore$ $\frac{3}{4}I_{0}=I_{0}(1-e^{-t/\tau})$

$\Rightarrow\frac{3}{4}=1-e^{-t/\tau}\Rightarrow e^{-t/\tau}=\frac{1}{4}$

$\Rightarrow\frac{t}{\tau}In e=In \frac{1}{4}\Rightarrow\frac{-4}{\tau}=-2 In 2$

$\Rightarrow $ $\tau= \frac{2}{In 2}$

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