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 (\frac {10}{9})=0.105 $]
(A) 1.85
(B) 0.95
(C) 0.105
(D) 9.52
Solution
In RC charging circuit, $V=E (1-e^{-\frac {t}{\tau}}) $
$\Rightarrow 2=20 (1-e^{-\frac {1}{\tau})}$ where $\tau $ is in $\mu s $
$\Rightarrow 0.9=e^{-\frac {1}{\tau}} $
$\Rightarrow ln (\frac {10}{9})=\frac {1}{\tau} =0.105=\frac {1}{RC}$ where C is in $\mu F $
$\Rightarrow 10C=\frac {1}{0.105}$
$\Rightarrow C=\frac {1}{1.05} \mu F=0.95 \mu F$
Answer: (B)