Question

In a discharging RC circuit, at what time the electrical potential energy will become the half of its initial value? [in terms of time constant of RC circuit,]

Answer 1

Correct Answer: 0.34

Explanation:
Let $q_0$ be the initial charge on the capacitor.
Thus, the initial electrical potential energy of the capacitor is $U_0=\frac{q_0^2}{2C}$
Let in time $t$, the electrical potential energy will become half of its initial value.
The electrical potential energy of the capacitor at time $t$ is
$U=\frac{q^2}{2C}$
where $q$ is the charge on the capacitor at time $t$.
In a discharging RC circuit, charge $q$ at any time $t$ is given as
$q=q_0{e}^{-t/\tau }$
where $\tau$ is the time constant of the $RC$ circuit.
$\Rightarrow U=\frac{1}{2C}{q}_0^2{e}^{-2t/\tau }$
Now, according to the given condition
$U=\frac{1}{2}{U}_0$or, $\frac{1}{2C}{q}_0^2{e}^{-2t\tau }=\frac{1}{2}\frac{q_0^2}{2C}$or, $\frac{1}{2}=e^{-2t\tau }$
Taking natural logarithms of both sides, we get$$
ln⁡(12)=−2tτor, $-0.693=-\frac{2t}{T}$or, $t=0.34\tau$
This is the time in which the electrical potential energy becomes half of its initial value.
Hence, the correct answer is $0.34\tau$.