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

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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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