A capacitor is connected to a 20 V battery through a resistance

Question:

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 : $\ln \left(\frac{10}{9}\right)=0.105$

  1. $9.52$

  2. $0.95$

  3. $0.105$

  4. $1.85$

Correct Option: , 2

Solution:

$\mathrm{V}=\mathrm{V}_{0}\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{RC}}\right)$

$2=20\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{RC}}\right)$

$\frac{1}{10}=1-\mathrm{e}^{-\mathrm{t} / \mathrm{RC}}$

$\mathrm{e}^{-\mathrm{t} / \mathrm{RC}}=\frac{9}{10}$

$\mathrm{e}^{\mathrm{t} / \mathrm{RC}}=\frac{10}{9}$

$\frac{\mathrm{t}}{\mathrm{RC}}=\ln \left(\frac{10}{9}\right) \Rightarrow \mathrm{C}=\frac{\mathrm{t}}{\mathrm{R} \ln \left(\frac{10}{9}\right)}$

$\mathrm{C}=\frac{10^{-6}}{10 \times .105}=.95 \mu \mathrm{F}$

Option (2)

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