Solution:
Energy is \(U \propto q^2 \propto e^{-2t/RC}\), so \(e^{-2t_1/RC} = 1/2 ⇒ t_1 = \frac{RC\ln 2}{2}\). Charge is \(q \propto e^{-t/RC}\), so \(e^{-t_2/RC} = 1/4 t_2 = 2RC\ln 2\). Thus, the ratio \(t_1/t_2 = 1/4\).
Let C be the capacitance of a capacitor discharging through a resistor R. Suppose \(t_1\) is the time taken for the energy stored in the capacitor to reduce to half its initial value and \(t_2\) is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio \(t_1/t_2\) will be :
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