JEE Main · 2012 · 29 Apr 2012 · Offline · Current Electricity

In a discharging $RC$ circuit the capacitor voltage falls as $V=V_0{e}^{-t/\tau }$, so the time constant $\tau$ is the time taken for the voltage to drop to $1/e$, about $37\%$, of its starting value. A convenient landmark is the half-value point: putting $V=V_0/2$ gives $\frac{1}{2}=e^{-t/\tau }$, so $\tau =\frac{t}{\ln 2}$. Reading the experimental graph, the voltage reaches half its initial value at about $t=100$ s, which gives $\tau =\frac{100}{\ln 2}\approx 144$ s. This value lies between $100$ and $150$ seconds.