Example 10.4: The RC circuit

Question: A capacitor of capacitance $C= 15\,\mu{\rm F}$ is charged up to a voltage of $V = 800\,{\rm V}$, and then discharged by connecting a resistor of resistance $R=8\,{\rm M}\Omega$ across its terminals. How long does it take for the charge on the positive plate of the capacitor to be reduced to $10\%$ of its original value?
 
Answer: Suppose that the resistor is first connected across the capacitor at $t=0$. The charge $q$ on the positive plate of the capacitor is given by

$$q(t) = Q\,{\rm e}^{-t/R\,C},$$

which can be rearranged to give

$$\frac{Q}{q} = {\rm e}^{\,t/R\,C}.$$

Taking the natural logarithm of both sides, we obtain

$$\ln\left(\frac{Q}{q}\right) = \frac{t}{R\,C}.$$

Hence,

$$t = \tau\,\ln\left(\frac{Q}{q}\right),$$

where

$$\tau = R\,C = (8)\,(15) = 120\,\,{\rm s}$$

is the $RC$ time. Since $q/Q = 0.1$, in this case, it follows that

$$t = (120) \,(\ln 10)= 276.3\,\,{\rm s}.$$