A capacitor with initial charge q_ois discharged through a resistor. What multiple of the time constant $\mathbf{\zeta }$ gives the time the capacitor takes to lose (a) the first one-third of its charge and (b) two-thirds of its charge?

A capacitor with initial charge q_o is discharged through a resistor.

Using the charging and discharging concept of the capacitor, we can get the required charge relation using the time constant of the RC circuit. Thus, the condition will give the multiplication value of the time constant.

Formula:

The charge equation of an RC circuit, $\mathrm{q}={\mathrm{q}}_{\mathrm{o}}{\mathrm{e}}^{-\mathrm{t}/\mathrm{RC}}$ (1)

For the given condition of the charge stored in the capacitor $\mathrm{q}=2{\mathrm{q}}_{\mathrm{o}}/3$, the multiple of the time constant $\mathrm{\zeta }$that the capacitor takes to lose the first one-third of its charge can be given using equation (1) as follows:

$\begin{array}{rcl}\mathrm{t}& =& \mathrm{\zeta ln}\left(\frac{{\mathrm{q}}_{\mathrm{o}}}{\mathrm{q}}\right)\\ & =& \mathrm{\zeta ln}\left(\frac{{\mathrm{q}}_{\mathrm{o}}}{2{\mathrm{q}}_{\mathrm{o}}/3}\right)\\ & =& \mathrm{\zeta ln}\left(\frac{3}{2}\right)\\ & =& 0.41\mathrm{\zeta }\end{array}$

Hence, the multiple values of the time constant are 0.41.

For the given condition of the charge stored in the capacitor $\mathrm{q}={\mathrm{q}}_{\mathrm{o}}/3$, the multiple of the time constant $\mathrm{\zeta }$that the capacitor takes to lose the first two-thirds of its charge can be given using equation (1) as follows:

$\begin{array}{rcl}\mathrm{t}& =& \mathrm{\zeta ln}\left(\frac{{\mathrm{q}}_{\mathrm{o}}}{\mathrm{q}}\right)\\ & =& \mathrm{\zeta ln}\left(\frac{{\mathrm{q}}_{\mathrm{o}}}{{\mathrm{q}}_{\mathrm{o}}/3}\right)\\ & =& \mathrm{\zeta ln}\left(3\right)\\ & =& 1.1\mathrm{\zeta }\end{array}$

Hence, the multiple values of the time constant are 1.1.