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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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$\frac{1}{4}$
$\frac{1}{4}$
$\mathrm{U}=\frac{1}{2} \frac{\mathrm{q}^2}{\mathrm{C}}=\frac{1}{2 \mathrm{C}}\left(\mathrm{q}_0 \mathrm{e}^{-\mathrm{t} / \mathrm{T}}\right)^2=\frac{\mathrm{q}_0^2}{2 \mathrm{C}} \mathrm{e}^{-2 \mathrm{t} / \mathrm{T}} \quad($ where $\tau=\mathrm{CR})$
$\mathrm{U}=\mathrm{U}_{\mathrm{i}} \mathrm{e}^{-2 \mathrm{t} / \mathrm{\tau}}$
$\frac{1}{2} \mathrm{U}_{\mathrm{i}}=\mathrm{U}_{\mathrm{i}} \mathrm{e}^{-2 \mathrm{t}_1 / \mathrm{\tau}}$
$\frac{1}{2}=\mathrm{e}^{-2 \mathrm{t}_1 / \tau} \Rightarrow \mathrm{t}_1=\frac{\mathrm{T}}{2} \ln 2$
Now $\quad \mathrm{q}=\mathrm{q}_0 \mathrm{e}^{-\mathrm{t} / \mathrm{T}}$
$\frac{1}{4} \mathrm{q}_0=\mathrm{q}_0 \mathrm{e}^{-\mathrm{t} / 2 \mathrm{~T}}$
$\mathrm{t}_2=\operatorname{T} \ln 4=2 \operatorname{T} \ln 2$
$\therefore \quad \frac{\mathrm{t}_1}{\mathrm{t}_2}=\frac{1}{4}$
$\mathrm{U}=\mathrm{U}_{\mathrm{i}} \mathrm{e}^{-2 \mathrm{t} / \mathrm{\tau}}$
$\frac{1}{2} \mathrm{U}_{\mathrm{i}}=\mathrm{U}_{\mathrm{i}} \mathrm{e}^{-2 \mathrm{t}_1 / \mathrm{\tau}}$
$\frac{1}{2}=\mathrm{e}^{-2 \mathrm{t}_1 / \tau} \Rightarrow \mathrm{t}_1=\frac{\mathrm{T}}{2} \ln 2$
Now $\quad \mathrm{q}=\mathrm{q}_0 \mathrm{e}^{-\mathrm{t} / \mathrm{T}}$
$\frac{1}{4} \mathrm{q}_0=\mathrm{q}_0 \mathrm{e}^{-\mathrm{t} / 2 \mathrm{~T}}$
$\mathrm{t}_2=\operatorname{T} \ln 4=2 \operatorname{T} \ln 2$
$\therefore \quad \frac{\mathrm{t}_1}{\mathrm{t}_2}=\frac{1}{4}$
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