# Time constant of a $RC$ circuit is $2/ln(2) s$. Capacitor is discharged at time $t = 0$. The ratio of charge on the capacitor at time $t = 2s$ and $t = 6s$ is

$\begin {array} {1 1} (A)\;3 : 1 & \quad (B)\;8:1 \\ (C)\;4:1 & \quad (D)\;2:1 \end {array}$

## 1 Answer

Time constant $\tau = \frac{2}{In(2)}$
$I = I_o e^{-t/\tau}$
\begin{align*}\frac{I_1}{I_2} = \frac{e^{^{-2/(2/In2)}}}{e^{^{-6/(2/In2) }}} \end{align*}
\begin{align*}= \frac{e^{3In2}}{e^{In2}} = \frac{2^3}{2} = \frac{4}{1} \end{align*}
Ans : (c) 4 : 1
answered Feb 13, 2014
edited Feb 28

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