A capacitor is connected to a $20 \mathrm{~V}$ battery through a resistance of $10 \Omega$. It is found that the potential difference across the capacitor rises to $2 \mathrm{~V}$ in $1 \mu \mathrm{s}$. The capacitance of the capacitor is___________ $\mu \mathrm{F} .$
Given : $\ln \left(\frac{10}{9}\right)=0.105$
Correct Option: , 2
$\mathrm{V}=\mathrm{V}_{0}\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{RC}}\right)$
$2=20\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{RC}}\right)$
$\frac{1}{10}=1-\mathrm{e}^{-\mathrm{t} / \mathrm{RC}}$
$\mathrm{e}^{-\mathrm{t} / \mathrm{RC}}=\frac{9}{10}$
$\mathrm{e}^{\mathrm{t} / \mathrm{RC}}=\frac{10}{9}$
$\frac{\mathrm{t}}{\mathrm{RC}}=\ln \left(\frac{10}{9}\right) \Rightarrow \mathrm{C}=\frac{\mathrm{t}}{\mathrm{R} \ln \left(\frac{10}{9}\right)}$
$\mathrm{C}=\frac{10^{-6}}{10 \times .105}=.95 \mu \mathrm{F}$
Option (2)
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