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(A) The equation for the discharging of a capacitor is given by $V = V_{0} e^{-t/	au}$,where $	au = RC$ is the time constant.At $t = 	au$,the poten

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$100\; sec$ and $150\; sec$

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(A) The equation for the discharging of a capacitor is given by $V = V_{0} e^{-t/	au}$,where $	au = RC$ is the time constant.At $t = 	au$,the potential difference becomes $V = V_{0} / e \approx 0.37 V_{0}$.From the graph,the initial potential difference $V_{0} = 25\; V$.Therefore,at $t = 	au$,$V = 0.37 	imes 25\; V = 9.25\; V$.Looking at the graph,at $t = 100\; sec$,the potential is slightly above $10\; V$,and at $t = 150\; sec$,the potential is slightly below $10\; V$ (specifically around $8\; V$ to $9\; V$).Since $9.25\; V$ corresponds to the time constant $	au$,and this value lies between $100\; sec$ and $150\; sec$ on the time axis,the correct range is $100\; sec$ and $150\; sec$.

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

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