Question:
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 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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Use \(V = V_0\left(1 - e^{-t/RC}\right)\) and solve for \(RC\) using the given value of \(\ln(10/9)\).
Updated On: Nov 4, 2026
- 9.52
- 0.95
- 0.105
- 1.85
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The Correct Option is B
Solution and Explanation
While charging, \(V = V_0\left(1 - e^{-t/RC}\right)\).
\(2 = 20\left(1 - e^{-t/RC}\right) \Rightarrow e^{-t/RC} = 0.9 \Rightarrow e^{t/RC} = \frac{10}{9}\).
\(\frac{t}{RC} = \ln\frac{10}{9} = 0.105\).
\(RC = \frac{t}{0.105} = \frac{1\,\mu\text{s}}{0.105} = 9.52\,\mu\)s.
\(C = \frac{9.52\times10^{-6}}{10}\) F \(= 0.952\times 10^{-6}\) F \(\approx 0.95\,\mu\)F.
Answer: (2) 0.95
\(2 = 20\left(1 - e^{-t/RC}\right) \Rightarrow e^{-t/RC} = 0.9 \Rightarrow e^{t/RC} = \frac{10}{9}\).
\(\frac{t}{RC} = \ln\frac{10}{9} = 0.105\).
\(RC = \frac{t}{0.105} = \frac{1\,\mu\text{s}}{0.105} = 9.52\,\mu\)s.
\(C = \frac{9.52\times10^{-6}}{10}\) F \(= 0.952\times 10^{-6}\) F \(\approx 0.95\,\mu\)F.
Answer: (2) 0.95
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