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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Aicharya · Accepted Answer

We know that in the discharging circuit the charge is given by,

$q = q_{0} e^{\frac{- t}{R C}}$ $\Rightarrow \frac{t}{R C} = \ln (\frac{q_{0}}{q})$

In time $t_{1}$ the energy stored reduces to half. Hence, $\frac{{q_{1}}^{2}}{2 C} = \frac{1}{2} \times \frac{{q_{0}}^{2}}{2 C} \Rightarrow \frac{q_{0}}{q_{1}} = \sqrt{2}$ . Therefore,

$\frac{t_{1}}{R C} = \ln (\sqrt{2}) = \frac{1}{2} \ln 2$ .

In time $t_{2}$ the charge stored reduces to ${\frac{1}{8}}^{th}$ of initial value. Hence, $\frac{q_{2}}{q_{0}} = \frac{1}{8}$ . Therefore,

$\frac{t_{2}}{R C} = \ln (8) = 3 \ln 2$ .

Hence,

$\frac{t_{1}}{t_{2}} = \frac{\frac{1}{2} \ln 2}{3 \ln 2} = \frac{1}{6}$

A capacitor is discharging through a resistor R . Consider in time t 1 , the energy stored in the capacitor…

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