Solution
We analyze the circuit at steady state (long time after switches are closed). In steady state, the capacitors act as open circuits, so no current flows through the left or right branches ($I_{steady} = 0$). Consequently, there is no voltage drop across the resistor ($V_R = I \cdot R = 0$).
1. Potential Analysis:
Let the potential of the bottom rail be $0\,\text{V}$.
Since there is no current in the resistor, the potential of the top rail is determined solely by the middle battery:
$$ V_{top} = 550\,\text{V} $$
Now we calculate the potential difference across each capacitor.
Left Capacitor ($C_1$): The bottom plate is connected to the positive terminal of the 100V source (relative to the bottom rail). The potential at the point between $C_1$ and the 100V battery is $100\,\text{V}$. The top plate is at $550\,\text{V}$.
$$ \Delta V_1 = 550\,\text{V} – 100\,\text{V} = 450\,\text{V} $$
Right Capacitor ($C_2$): The bottom plate is connected to the positive terminal of the 200V source. The potential at the point between $C_2$ and the 200V battery is $200\,\text{V}$. The top plate is at $550\,\text{V}$.
$$ \Delta V_2 = 550\,\text{V} – 200\,\text{V} = 350\,\text{V} $$
2. Final Energy Stored ($U_f$):
$$ U_f = \frac{1}{2} C (\Delta V_1)^2 + \frac{1}{2} C (\Delta V_2)^2 $$
$$ U_f = \frac{1}{2} (10 \times 10^{-6}) (450^2 + 350^2) $$
$$ U_f = 5 \times 10^{-6} (202500 + 122500) = 5 \times 10^{-6} (325000) = 1.625\,\text{J} $$
3. Work Done by Batteries ($W_b$):
We analyze the charge flow.
Charge on $C_1$: $Q_1 = C \Delta V_1 = 10\mu\text{F} \times 450\text{V} = 4500\,\mu\text{C}$. (Top plate is positive).
Charge on $C_2$: $Q_2 = C \Delta V_2 = 10\mu\text{F} \times 350\text{V} = 3500\,\mu\text{C}$. (Top plate is positive).
Work by 550V source: The total positive charge on the top plates ($Q_1 + Q_2$) must have flowed from the 550V battery.
$$ W_{550} = (Q_1 + Q_2) \times 550 = (8000\,\mu\text{C}) \times 550\,\text{V} = 4.4\,\text{J} $$
Work by 100V source: The bottom plate of $C_1$ acquires a charge of $-Q_1$. This means $+Q_1$ flowed into the positive terminal of the 100V battery. The battery absorbs energy.
$$ W_{100} = -Q_1 \times 100 = -(4500\,\mu\text{C}) \times 100\,\text{V} = -0.45\,\text{J} $$
Work by 200V source: Similarly, $+Q_2$ flows into the positive terminal.
$$ W_{200} = -Q_2 \times 200 = -(3500\,\mu\text{C}) \times 200\,\text{V} = -0.70\,\text{J} $$
$$ W_{total} = 4.4 – 0.45 – 0.70 = 3.25\,\text{J} $$
4. Heat Dissipated ($H$):
Using the energy conservation principle (Work-Energy Theorem):
$$ W_{total} = \Delta U + H $$
$$ H = W_{total} – U_f $$
$$ H = 3.25\,\text{J} – 1.625\,\text{J} = 1.625\,\text{J} \approx 1.6\,\text{J} $$
Answer: 1.6 J