Q.
The figure shows an experimental plot discharging of a capacitor in an $RC$ circuit. The time constant $\tau$ of this circuit lies between
A) $150\sec{and\ 200\ sec}$
B) $0\ and\ 50\ sec$
C) $50\sec{and\ 100\ sec}$
D) $100\sec{and\ 150\ sec}$
Explanation
(d) For discharging of an $RC$ circuit,
$V = V_{0}e^{- t/\tau}$
So, when $V = \frac{V_{0}}{2}$
$\frac{V_{0}}{2} = V_{0}e^{- t/\tau}$
$\mathcal{l}n\frac{1}{2} = - \frac{t}{\tau} \Rightarrow \tau = \frac{1}{\mathcal{l}n2}$
From graph when $V = \frac{V_{0}}{2},t = 100s$
$\therefore\tau = \frac{100}{\mathcal{l}n2} = 144.3\ \sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.