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(A) The potential difference across a discharging capacitor is given by $V(t) = V_0 e^{-t/	au}$,where $	au$ is the time constant.At $t = 	au$,the p

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$100 \ s$ and $150 \ s$

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(A) The potential difference across a discharging capacitor is given by $V(t) = V_0 e^{-t/	au}$,where $	au$ is the time constant.At $t = 	au$,the potential difference is $V(	au) = V_0 e^{-1} \approx 0.37 V_0$.From the graph,the initial potential difference $V_0 = 25 \ V$.Therefore,at $t = 	au$,$V(	au) = 0.37 	imes 25 \ V = 9.25 \ V$.Looking at the graph,at $t = 100 \ s$,$V \approx 13 \ V$ and at $t = 150 \ s$,$V \approx 9 \ V$.Since $9.25 \ V$ lies between $13 \ V$ and $9 \ V$,the time constant $	au$ must lie between $100 \ s$ and $150 \ s$.

The figure shows an experimental graph of the discharging of a capacitor in an $RC$ circuit. The time constant of this circuit lies between which of the following time intervals?

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