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If I connect a circuit with charged capacitor that has charge $q$ to an uncharged capacitor and resistor, how would I set up a differential equation that can be used to find the time constant? I know that at any time $q_1 + q_2 = q$ where $q_1$ is the charge on the initially charged capacitor and $q_2$ is the charge on the initially uncharged capacitor. I thought that $(\frac{q_1}{C_1})-(\frac{q-q_1}{C_2})-R(-\frac{dq_1}{dt})=0$ might work, but I am not sure that this correctly describes the circuit. Are the signs on the second capacitor and resistor supposed to match? Is $\frac{dq_1}{dt}$ representative of the circuit current? Once I have that differential equation made, how can I obtain the time constant by comparing to the basic RC-circuit $\frac{dq}{dt}=\frac{CV-q}{RC}$?

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  • $\begingroup$ Can we assume the resistor is in series with the uncharged capacitor and that there is a switch between the charged capacitor and the series combination and that you want to find the time constant upon closing the switch? $\endgroup$
    – Bob D
    Nov 28, 2020 at 21:15
  • $\begingroup$ Yes, those are both true $\endgroup$ Nov 28, 2020 at 21:17

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Your diff. eq. looks perfectly fine, based on your sign conventions for $q_1$ and $q_2$. Note that $\frac{\mathrm{d}q_1}{\mathrm{d}t}=-\frac{\mathrm{d}q_2}{\mathrm{d}t}$ under this convention. Both of these are representative of the circuit current, just with different signs, since everything is connected in series.

Let's define $C\equiv\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}}$ as the equivalent capacitance of the two capacitors; and $V\equiv\frac{q}{C_2}$ as a sort of voltage constant. Then, the diff. eq. takes the form:

$$\frac{\mathrm{d}q_1}{\mathrm{d}t}=\frac{CV-q_1}{RC}$$

So, as usual, we can find the time constant:

$$\tau=RC=\frac{R}{\frac{1}{C_1}+\frac{1}{C_2}}$$

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  • $\begingroup$ Thank you. Would the time constant of both capacitors be the same? $\endgroup$ Nov 28, 2020 at 22:13
  • $\begingroup$ @physicsaficionado Well, as you yourself observed, $q_1+q_2=q$. This implies that the charges on both capacitors change with the same magnitude, and that they'll have the same "time constant". $\endgroup$
    – DanDan面
    Nov 28, 2020 at 22:50

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