JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2021
- Hard
- MCQ
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 : \(\ln \left(\frac{10}{9}\right)=0.105\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(0.95\)
Step-by-step Solution
Detailed explanation
\({V}={V}_{0}\left(1-{e}^{-t / {RC}}\right)\)
\(2=20\left(1-{e}^{-t / {RC}}\right)\)
\(\frac{1}{10}=1-{e}^{-t / {RC}}\)
\({e}^{-t / R C}=\frac{9}{10}\)
\({e}^{{t} / {RC}}=\frac{10}{9}\)
\(\frac{{t}}{{RC}}=\ln \left(\frac{10}{9}\right) \Rightarrow {C}=\frac{{t}}{{R} \ell {n}\left(\frac{10}{9}\right)}\)
\({C}=\frac{10^{-6}}{10 \times .105}=.95\,\mu {F}\)
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