Capacitance determination from charging curve – Rankers Physics

Charging and Discharging of Capacitors: Practice Problem & Solution

A capacitor is connected to a \(12~\text{V}\) battery through a resistance of \(10~Omega\). It is found that the potential difference across the capacitor rises to \(4.0~\text{V}\) in \(1~\mu\text{s}\). Find the capacitance of the capacitor. (Take : \(ln \frac{3}{2} = 0.4\))
\(0.5~\mu\text{F}\)
\(0.25~\mu\text{F}\)
\(5.0~\mu\text{F}\)
\(1.0~\mu\text{F}\)

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Charging and Discharging of Capacitors. Understanding the underlying formula is key to arriving at the correct answer below:

Using \(V = V_0(1 - e^{-t/RC})\), we get \(4 = 12(1 - e^{-t/RC})⇒ e^{-t/RC} = 2/3\). Taking the natural logarithm, \(\frac{t}{RC} = \ln(1.5) = 0.4\), which yields \(C = \frac{10^{-6}}{10 \times 0.4} = 0.25~\mu\text{F}\).

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