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JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance

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

  1. A \(9.52\)
  2. B \(0.95\)
  3. C \(0.105\)
  4. D \(1.85\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(0.95\)

Step-by-step Solution

Detailed explanation

\({V}={V}_{0}\left(1-{e}^{-t / {RC}}\right)\) \(2=20\left(1-{e}^{-t / {RC}}\right)\) \(\frac{1}{10}=1-{e}^{-t / {RC}}\) \({e}^{-t / R C}=\frac{9}{10}\) \({e}^{{t} / {RC}}=\frac{10}{9}\)…
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