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  4. 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 t2, the charge stored reduces to one eighth of its initial value. The ratio t1 / t2 will be :

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Q. 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 $t_{1} / t_{2}$ will be :

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Solution:

In $t _{1}$ time energy becomes half so charge will become $\frac{1}{\sqrt{2}}$ time
$q = Q _{0} e ^{-\frac{ t _{1}}{ RC }}=\frac{ Q _{0}}{\sqrt{2}}$
and $q = Q _{0} e ^{-\frac{ t _{1}}{ RC }}=\frac{ Q _{0}}{8}=\left(\frac{ Q _{0}}{\sqrt{2}}\right)^{6}$
$t _{2}=6 t _{1}$
$\frac{ t _{1}}{ t _{2}}=\frac{1}{6}$