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

Explanation:
Let q0 be the initial charge on the capacitor.
Thus, the initial electrical potential energy of the capacitor is U0=q022C
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=q22C
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=q0e−t/τ
where τ is the time constant of the RC circuit.
⇒U=12Cq02e−2t/τ
Now, according to the given condition
U=12U0or, 12Cq02e−2tτ=12q022Cor, 12=e−2tτ
Taking natural logarithms of both sides, we get
ln⁡(12)=−2tτor, −0.693=−2tTor, t=0.34τ
This is the time in which the electrical potential energy becomes half of its initial value.
Hence, the correct answer is 0.34τ.