A capacitor of capacitance \(C\) is connected to a \(6.0 \mathrm{~V}\) battery through a resistance of \(5…

A capacitor of capacitance \(C\) is connected to a \(6.0 \mathrm{~V}\) battery through a resistance of \(5 \Omega\). The potential difference between the plates rises from zero to \(3.0 \mathrm{~V}\) in \(6.9 \mu \mathrm{s}\). The value of \(C\) is
  1. \(1 \mu \mathrm{F}\)
  2. \(2 \mu \mathrm{F}\)
  3. \(3 \mu \mathrm{F}\)
  4. \(4 \mu \mathrm{F}\)

Solution

During charging, the charge at time \(t\) on the capacitor is given by
\(Q=Q_{0}\left(1-e^{-t / R C}\right)\)
Therefore, the potential difference at time \(t\) between the capacitor plates is
\(\begin{array}{l}
V=\frac{Q}{C}=\frac{Q_{0}}{C}\left(1-e^{-t / R C}\right) \\
\Rightarrow V=V_{0}\left(1-e^{-t / R C}\right)
\end{array}\)
where \(V_{0}=\frac{Q_{0}}{C}\) is the final potential difference \(=6.0 \mathrm{~V}\), the voltage of the battery. Given \(V=3.0 \mathrm{~V}=\frac{V_{0}}{2}\). Hence
\(\frac{V_{0}}{2}=V_{0}\left(1-e^{-t / R C}\right)\)
\(\Rightarrow \quad 1-e^{-t / R C}=\frac{1}{2}\)
\(\Rightarrow \quad e^{-t / R C}=\frac{1}{2}\)
\(\Rightarrow \quad \frac{t}{R C}=\ln (2)\)
\(\Rightarrow \quad C=\frac{t}{R \ln (2)}=\frac{6.9 \mu \mathrm{s}}{5 \times 0.69}=2 \mu \mathrm{F}\) *

Asked in: JEE Mains - Capacitance - Test 3

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