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 : \(\ln \left(\frac{10}{9}\right)=0.105\)
- A \(9.52\)
- B \(0.95\)
- C \(0.105\)
- D \(1.85\)
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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