Let C be the capacitance of a capacitor discharging though 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 of its initial value. Then, the ratio $t_{2} : t_{1}$ will be

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Detailed Solution

$q = q_{0} e^{- t / R C}$

and

$U = \frac{q^{2}}{2 C} = \frac{{q_{0}}^{2}}{2 C} e^{\frac{- 2 t}{C R}} = U_{0} e^{\frac{- 2 t}{C R}}$

$According to given problem$

$\frac{U_{0}}{2} = U_{0} e^{- 2 t_{1} / R C}$

$and \frac{q_{0}}{4} = q_{0} e^{- t_{2} / R C}$

Thus, $t_{1} = \frac{R C \ln 2}{2}$

and $t_{2} = R C \ln 4 = 2 R C \ln 2$

So, $\frac{t_{1}}{t_{2}} = \frac{R C \ln 2}{2} \times \frac{1}{2 R C \ln 2} = \frac{1}{4}$

$\frac{t_{2}}{t_{1}} = 4$

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