Charge sharing and discharge Joule heating between two capacitors

Explanation

State 1: When the switch is connected to terminal $\text{a}$

When the switch $\text{S}$ is connected to terminal $\text{a}$ , a series charging circuit is formed connecting the power supply, resistor $\text{R}_1$ , capacitor $\text{C}_1$ , and ground. At this time, capacitor $\text{C}_2$ is disconnected from the switch, and its charge remains $0$ .

After a sufficiently long time has elapsed, the charging of capacitor $\text{C}_1$ is complete, and no current flows. At this time, the voltage across the plates of capacitor $\text{C}_1$ becomes equal to the electromotive force $V$ of the power supply, so the stored charge $Q_0$ and electrostatic energy $U_1$ are as follows:

$Q_0 = C_1 V$ $U_1 = \frac{1}{2} C_1 V^2$

State 2: When the switch is switched to terminal $\text{b}$

When switch $\text{S}$ is switched to terminal $\text{b}$ , the path including the power supply and resistor $\text{R}_1$ becomes completely isolated and no longer participates in the operation of the circuit. The newly formed circuit is a closed circuit (discharge and charge sharing circuit) with a capacitor $\text{C}_1$ with charge $Q_0$ , a resistor $\text{R}_2$ , and a capacitor $\text{C}_2$ with no charge connected in series.

After a sufficiently long time, the current settles and the potential difference between the two capacitors becomes equal, reaching an equilibrium state. If the common potential difference at this time is $V_f$ , the total amount of charge in the isolated plate system is conserved (law of conservation of charge), and the following relationship holds: $Q_0 = C_1 V = (C_1 + C_2) V_f$

From this, the common potential difference $V_f$ after the change can be found as follows: $V_f = \frac{C_1}{C_1 + C_2} V$

Change in Electrostatic Energy and Joule Heating

In equilibrium, the total electrostatic energy stored in the two capacitors $U_2$ is given by: $U_2 = \frac{1}{2} C_1 V_f^2 + \frac{1}{2} C_2 V_f^2 = \frac{1}{2} (C_1 + C_2) V_f^2$

Substituting the previously calculated $V_f$ into this equation and simplifying, we get:

$U_2 = \frac{1}{2} (C_1 + C_2) \left( \frac{C_1}{C_1 + C_2} V \right)^2 = \frac{1}{2} \frac{C_1^2}{C_1 + C_2} V^2$

In this switching process, the only resistance in the circuit is resistor $\text{R}_2$ . Therefore, according to the law of conservation of energy (all the decreased electrostatic energy is converted into Joule heat), the total amount of Joule heat $Q_2$ generated by resistor $\text{R}_2$ can be found as the difference between the initial electrostatic energy $U_1$ and the final electrostatic energy $U_2$ .

$Q_2 = U_1 - U_2 = \frac{1}{2} C_1 V^2 - \frac{1}{2} \frac{C_1^2}{C_1 + C_2} V^2$ $Q_2 = \frac{1}{2} C_1 V^2 \left( 1 - \frac{C_1}{C_1 + C_2} \right) = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} V^2$

Numerical Substitution

Substitute the given numerical constraints into the equation and calculate. The resistance values $R_1, R_2$ do not depend on the energy change after sufficient time has elapsed, so they are not directly used in the calculation of the total amount of Joule heat.

$Q_2 = \frac{1}{2} \times \frac{3.00 \times 2.00}{3.00 + 2.00} \times 50.0^2$ $Q_2 = \frac{1}{2} \times \frac{6.00}{5.00} \times 2500$ $Q_2 = \frac{1}{2} \times 1.20 \times 2500 = 0.600 \times 2500 = 1500 \text{ J}$

The natural number we are looking for is $1500$ .

Answer: 1500

Charge sharing and discharge Joule heating between two capacitors