Question

A charged capacitor of capacitance C can be discharged through a resistor R. At time t after the discharge has started, the charge Q remaining on the capacitor can be given by the expression:
Q = Qoe^(-t/RC)
(i) Use this equation to obtain an expression for the half-life T1/2 of the discharge process.
(ii) Use this equation to define the time constant t of the discharge
(iii) Compare the values of T1/2 and t.

Answer 1

(a) At the half-life $Q=\dfrac{1}{2}Q_0, t=t_{1/2}$ and we can write:

$$0.5Q_0=Q_0e^{-\dfrac{t_{1/2}}{RC}},$$ $$0.5=e^{-\dfrac{t_{1/2}}{RC}},$$ $$ln(0.5)=ln(e^{-\dfrac{t_{1/2}}{RC}}),$$ $$ln(0.5)=-\dfrac{t_{1/2}}{RC},$$ $$t_{1/2}=-ln(0.5)RC=0.693\cdot RC.$$

(b) From this equation we can define the time constant:

$$\tau=RC.$$

(c)

$$\dfrac{t_{1/2}}{\tau}=\dfrac{0.693\cdot RC}{RC}=0.693.$$

Therefore, the half-life is 69.3% of the time constant.