A capacitor is connected to a $20\, {V}$ battery through a resistance of $10\, \Omega .$ It is found that the potential difference across the capacitor rises to $2\, {V}$ in $1\, \mu {s}$. The capacitance of the capacitor is $....\,\mu {F}$ Given : $\ln \left(\frac{10}{9}\right)=0.105$
JEE MAIN 2021, Diffcult
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${V}={V}_{0}\left(1-{e}^{-t / {RC}}\right)$
$2=20\left(1-{e}^{-t / {RC}}\right)$
$\frac{1}{10}=1-{e}^{-t / {RC}}$
${e}^{-t / R C}=\frac{9}{10}$
${e}^{{t} / {RC}}=\frac{10}{9}$
$\frac{{t}}{{RC}}=\ln \left(\frac{10}{9}\right) \Rightarrow {C}=\frac{{t}}{{R} \ell {n}\left(\frac{10}{9}\right)}$
${C}=\frac{10^{-6}}{10 \times .105}=.95\,\mu {F}$
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$K(x) = K_0 + \lambda x$ ( $\lambda =$ constant)
The capacitance $C,$ of the capacitor, would be related to its vacuum capacitance $C_0$ for the relation