How does a capacitor discharge through another capacitor and resistor?

So if a charged capacitor (say unit PD and charge), discharges across a resistor (say unit resistance) in series with an identical (uncharged) capacitor, how would I deduce the resultant pd? Because any charge moved from one capacitor to another would also lose some energy when moved across the capacitor, changing the pd.

• Are you writing “pd” as shorthand for “potential difference”? That’s not really a standard abbreviation.
– rob
Commented Jun 11, 2022 at 14:01

The resistance determines how fast the charge will transfer from one capacitor to the the othr. The final state is determined by the equality of the potential diference across the two capacitors. The final voltages and charges are not dependenton the value of the resistance. No charge is lost in the resistor. Same final charges will be found at equilibrium. And the sum of the final charges will be equal to the sum of the initial charges. These two conditions (equal voltage and conservation of charge) will provide the two equations for calculating the equilibrium values for voltage and charge.

What is lost in the process is electrostatic potential energy. And what is the most counter-intuitive fact is that the fraction of energy lost is also independent of the value of the resistor. There are threads here discussing this aspect. As an example, for two capacitors with equal capacitance, if one of them is initialy uncharged, the energy lost in the process is one half.

The resistor determines just the time constant of the charge transfer. Both the voltage and the charge on the capacitors are exponential functions of time containing the exponential factor $$e^{-t\tau}$$ where $$\tau = RC$$ is the time constant of the process. As you can see, larger the resistance, larger the time constant.
As an example, for two identical capacitors (capacitance C) with one of them (number 1) initialy charged at $$V_0$$ and the other (number 2) discharged, the potential difference as a function of time is $$V_1=\frac{V_0}{2}(1+e^{-t\tau})$$. The charge will be $$Q_1= \frac{V_0}{2}C(1+e^{-t\tau})$$ None of the final values (for $$t-> \infty$$ ) depend on the value of the resiance.