For a discharging capacitor in a series RC circuit, its potential energy is one- half of its initial value at time t. This time t in terms of time constant $\tau$ is
Answer & explanation
Correct answer: option 4
For Potential energy to become half the charge in capacitor becomes $\frac{1}{\sqrt 2}$ times.
$ q = q_0 e^{-\frac{t}{RC}} = \frac{q_0}{\sqrt 2}$
$ \frac{1}{\sqrt 2} = e^{-\frac{t}{RC}}$
$ \frac{t}{RC} = \frac{1}{2}ln2$
$ t = \frac{\tau}{2} ln2$