Let C be the capacitance of a capacitor discharging through a… | JEE Main 2010 (AIEEE)

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Capacitors in DC Circuits & RC

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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Answer: (c)

Discharging: $q = q_0e^{-t/\tau}$ and $U = U_0e^{-2t/\tau}$ , with $\tau = RC$ .
Energy halves: $e^{-2t_1/\tau} = \frac{1}{2} \Rightarrow t_1 = \frac{\tau\ln 2}{2}$
Charge to one fourth: $e^{-t_2/\tau} = \frac{1}{4} \Rightarrow t_2 = 2\tau\ln 2$
$\frac{t_1}{t_2} = \frac{1/2}{2} = \frac{1}{4}$
So the answer is (c) $\frac{1}{4}$ .

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