Q62 / 90 · JEE Main 2012 (Offline) · Physics

The figure shows an experimental plot for discharging of a capacitor in an $R-C$ circuit. The time constant $\tau$ of this circuit lies between: Figure

  • A
    $150 \mathrm{~sec}$ and $200 \mathrm{~sec}$
  • B
    $0$ and $50 \mathrm{~sec}$
  • C
    $50 \mathrm{~sec}$ and $100 \mathrm{~sec}$
  • Correct answer
    D
    $100 \mathrm{~sec}$ and $150 \mathrm{~sec}$

Solution

For discharging of an RC circuit, $V=V_0 e^{-t / \tau}$ So, when $V=\frac{V_0}{2}$ $\frac{\mathrm{V}_0}{2}=\mathrm{V}_0 \mathrm{e}^{-\mathrm{t} / \tau}$ $\ln \frac{1}{2}=-\frac{\mathrm{t}}{\tau} \Rightarrow \tau=\frac{\mathrm{t}}{\ln 2}$ From graph when $\mathrm{V}=\frac{\mathrm{V}_0}{2}, \mathrm{t}=100 \mathrm{~s} \quad \therefore \tau=\frac{100}{\ln 2}=144.3 \mathrm{~sec}$

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

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