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During the process of charging of a capacitor, how long does it take for the current to decay to half its value ?

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$1.386 τ$

$0.693 τ$

$0.346 τ$

$0.231 τ$

answer is B.

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Detailed Solution

The current in a charging series RC circuit is given by
$I = I_{0} e^{\frac{1}{R C}} W h e n t h e c u r r e n t i s h a l v e d$

$\frac{I_{0}}{2} = I_{0} e^{\frac{1}{R C}} \Rightarrow \frac{1}{2} e^{\frac{1}{R C}}$

Taking the natural logarithmas on both sides

$\log_{e} 1 - \log_{e} 2 = \frac{t}{R C}$

On solving for t, we get

$t = R C \log_{e} 2 = τ \log_{e} 2 = 0.693 τ$

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