Question

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required for the capacitor to lose the first one-ninth of its charge? b) How long is required for the capacitor to lose the first two-ninths of its charge?

Answer 1

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$.

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$.

Note that $\ln 9 = 2\,\ln 3$, and that$\ln 8 = 3\, \ln 2$.

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$, and the initial charge of the capacitor be $q_0$. Then at time $t$, the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$.

a)

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$.

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$:

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$.

The goal is to solve for $t$ in terms of $\tau$. Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$.

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$.

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$.

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$.

b)

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$.

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$:

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$.

The goal is to solve for $t$ in terms of $\tau$. Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$.

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$.

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$.

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$.