A capacitor is discharging through a resistor $$R$$. Consider in time $$t_1$$, the energy stored in the capacitor reduces to half of its initial value and in time $$t_2$$, the charge stored reduces to one eighth of its initial value. The ratio $$\frac{t_1}{t_2}$$ will be

A capacitor discharges through resistance $$R$$. In time $$t_1$$, energy reduces to half. In time $$t_2$$, charge reduces to one-eighth. Find $$t_1/t_2$$.

$$q = q_0 e^{-t/RC}$$

Energy $$U = \frac{q^2}{2C} = \frac{q_0^2}{2C}e^{-2t/RC}$$

So energy decays as $$U = U_0 e^{-2t/RC}$$.

Energy reduces to half: $$\frac{1}{2} = e^{-2t_1/RC}$$

$$\ln 2 = \frac{2t_1}{RC}$$

$$t_1 = \frac{RC \ln 2}{2}$$

Charge reduces to $$\frac{1}{8}$$: $$\frac{1}{8} = e^{-t_2/RC}$$

$$\ln 8 = \frac{t_2}{RC}$$

$$t_2 = RC \ln 8 = 3RC \ln 2$$

$$\frac{t_1}{t_2} = \frac{RC\ln 2/2}{3RC\ln 2} = \frac{1}{6}$$

Hence, the correct answer is Option D: $$\dfrac{1}{6}$$.