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A capacitor is charged and then made to discharge through a resistance. The time constant is $ \tau$. In what time will the potential difference across the capacitor decrease by 10%

$\begin {array} {1 1} (A)\;\tau \: ln (0.1) & \quad (B)\;\tau \: ln (0.9) \\ (C)\;\tau \: ln \bigg(\large\frac{10}{9}\bigg) & \quad (D)\;\tau \: ln \bigg(\large\frac{11}{10}\bigg) \end {array}$

 

1 Answer

$\begin{align*}Q = Q_o e^{-\frac{t}{\tau}} \end{align*}$
$\begin{align*}V = V_o e^{-\frac{t}{\tau}} \end{align*}$
Given $V = 0.9 V_o$
$ \therefore \begin{align*}V = \frac{9}{10} V_o \end{align*}$
$\begin{align*}\frac{9}{10} V_o = V_o e^{-\frac{T}{\tau}} \end{align*}$
$\begin{align*}\implies ln \begin{pmatrix} \frac{9}{10} \end{pmatrix} = -\frac{t}{\tau} \end{align*}$
$\begin{align*} ln \begin{pmatrix} \frac{9}{10} \end{pmatrix}& = -\frac{t}{\tau} \\ ln \begin{pmatrix} \frac{10}{9} \end{pmatrix}&= \frac{t}{\tau} \end{align*}$
$\therefore t = \tau \;ln\; \begin{pmatrix}\frac{10}{9} \end{pmatrix}$
answered Feb 10, 2014 by thanvigandhi_1
edited Nov 28, 2017 by priyanka.c
 

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