# How do you set up Kirchhoff's loop rule for a circuit with a charged capacitor, uncharged capacitor, and resistor?

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 $$\frac{dq}{dt}=\frac{CV-q}{RC}$$?

• 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? Commented Nov 28, 2020 at 21:15
• Yes, those are both true Commented Nov 28, 2020 at 21:17

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:
$$\frac{\mathrm{d}q_1}{\mathrm{d}t}=\frac{CV-q_1}{RC}$$
$$\tau=RC=\frac{R}{\frac{1}{C_1}+\frac{1}{C_2}}$$
• @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". Commented Nov 28, 2020 at 22:50