Solution:
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 \(C = \frac{10^{-6}}{10 \times 0.4} = 0.25~\mu\text{F}\).
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 : \(ln \frac{3}{2} = 0.4\))
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