1. Analysis of Potentials and Charges:
We first determine the initial potential at each node and the charge stored on each capacitor plate.
Bottom Wire: Ground potential = $0 \, \text{V}$.
Middle Node ($V_{\text{mid}}$): The $2.0 \, \mu\text{F}$ capacitor is connected between the middle node and ground. With a potential difference of $6.0 \, \text{V}$, the middle node is at $6.0 \, \text{V}$.
Top Node ($V_{\text{top}}$): The $1.0 \, \mu\text{F}$ capacitor is stacked on top. It has a potential difference of $4.0 \, \text{V}$. Thus, the top node is at $6.0 \, \text{V} + 4.0 \, \text{V} = $ $10.0 \, \text{V}$.
Initial Charges ($Q = CV$):
Capacitor $C_1$ ($1.0 \, \mu\text{F}$): $Q_1 = 1.0 \, \mu\text{F} \times 4.0 \, \text{V} = 4.0 \, \mu\text{C}$.
Top plate has $+4.0 \, \mu\text{C}$. Bottom plate has $-4.0 \, \mu\text{C}$.
Capacitor $C_2$ ($2.0 \, \mu\text{F}$): $Q_2 = 2.0 \, \mu\text{F} \times 6.0 \, \text{V} = 12.0 \, \mu\text{C}$.
Top plate has $+12.0 \, \mu\text{C}$. Bottom plate has $-12.0 \, \mu\text{C}$.
2. Calculation of Heat Dissipated (Method of Average Potential):
Since the potential across a discharging capacitor decreases linearly with the charge flown ($V \propto q$), we can use the average potential method to calculate the energy dissipated (Heat, $H$).
$$ H = \langle V \rangle \times q_{\text{flow}} $$
where $\langle V \rangle = \frac{V_{\text{initial}} + V_{\text{final}}}{2}$. Since the capacitors discharge completely, $V_{\text{final}} = 0$.
A. For the $10 \, \text{k}\Omega$ Resistor:
This resistor connects the Top Node to Ground.
Average Potential: $\langle V_{10k} \rangle = \frac{10 \, \text{V} + 0 \, \text{V}}{2} = 5 \, \text{V}$.
Charge Flow: The total charge stored on the Top Node is the charge on the top plate of $C_1$, which is $+4.0 \, \mu\text{C}$. All of this charge flows through the $10 \, \text{k}\Omega$ resistor to ground.
$$ H_{10k} = 5 \, \text{V} \times 4.0 \, \mu\text{C} = 20 \, \mu\text{J} $$
B. For the $3 \, \text{k}\Omega$ Resistor:
This resistor connects the Middle Node to Ground.
Average Potential: $\langle V_{3k} \rangle = \frac{6 \, \text{V} + 0 \, \text{V}}{2} = 3 \, \text{V}$.
Charge Flow: We must calculate the net charge residing on the Middle Node. The node connects the bottom plate of $C_1$ and the top plate of $C_2$.
$$ Q_{\text{net}} = Q_{\text{bottom}, C1} + Q_{\text{top}, C2} $$
$$ Q_{\text{net}} = -4.0 \, \mu\text{C} + 12.0 \, \mu\text{C} = +8.0 \, \mu\text{C} $$
This net charge of $+8.0 \, \mu\text{C}$ flows through the $3 \, \text{k}\Omega$ resistor to neutralize the system.
$$ H_{3k} = 3 \, \text{V} \times 8.0 \, \mu\text{C} = 24 \, \mu\text{J} $$
Final Answer:
Heat dissipated in the $10 \, \text{k}\Omega$ resistor: 20 μJ
Heat dissipated in the $3 \, \text{k}\Omega$ resistor: 24 μJ
