Q62 / 90 · JEE Main 2010 · Physics

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

  • A
    1
  • B
    $\frac{1}{2}$
  • Correct answer
    C
    $\frac{1}{4}$
  • D
    2

Solution

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

Hence, the correct answer is option: 'C'.

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