Logic and Approach

The problem asks for energy stored and energy dissipated in the first 100 seconds. Given the time constant is of the order of milliseconds, 100s represents the steady state. The charging process occurs in two logical stages:

1. Stage 1 (Rapid Initial Charging): Immediately after the switch is closed, current flows to charge the capacitor plates. During this brief transient, the resistance $R$ (100k$\Omega$) is very high compared to the uncharged capacitors (which act as short circuits), so we can approximate the initial charge distribution as if $R$ were an open circuit.
2. Stage 2 (Redistribution): Once charges build up, a potential difference develops between the midpoints of the two branches. A small current flows through $R$ until these potentials equalize. In the final steady state, the resistor acts as a short circuit (equipotential connection) between the branches.

The heat dissipated in the resistance corresponds to the energy lost during this redistribution phase (Stage 2).

Stage 1: Initial Charging (R is Open)

With R open, the circuit consists of two independent parallel branches connected to $V_0 = 200V$.

Top Branch ($C_1, C_3$ in series):

$$ C_{top} = \frac{3 \times 6}{3 + 6} = 2 \mu F $$ $$ Q_{top} = C_{top} V_0 = 2 \times 200 = 400 \mu C $$

Bottom Branch ($C_2, C_4$ in series):

$$ C_{bot} = \frac{6 \times 3}{6 + 3} = 2 \mu F $$ $$ Q_{bot} = C_{bot} V_0 = 2 \times 200 = 400 \mu C $$

Totals for Stage 1:

$$ Q_{initial} = 400 + 400 = 800 \mu C $$ $$ U_{initial} = \frac{1}{2} C_{top} V_0^2 + \frac{1}{2} C_{bot} V_0^2 = \frac{1}{2}(4 \mu F)(200)^2 = 0.08 \, J $$

Stage 2: Final Steady State (R is Short/Equipotential)

In steady state, the potential at the midpoint of the top branch equals the potential at the midpoint of the bottom branch. We can treat the connection points as being shorted together. This effectively places $C_1 || C_2$ and $C_3 || C_4$.

Equivalent Capacitance:

$$ C_{left} = C_1 + C_2 = 3 + 6 = 9 \mu F $$ $$ C_{right} = C_3 + C_4 = 6 + 3 = 9 \mu F $$

Now $C_{left}$ and $C_{right}$ are in series:

$$ C_{final} = \frac{9 \times 9}{9 + 9} = 4.5 \mu F $$

Totals for Stage 2 (Final):

$$ Q_{final} = C_{final} V_0 = 4.5 \times 200 = 900 \mu C $$ $$ U_{final} = \frac{1}{2} C_{final} V_0^2 = \frac{1}{2} (4.5 \times 10^{-6}) (40000) = 0.09 \, J $$

(b) Total Energy Dissipated in Resistance

The energy dissipated in the resistor is the heat generated during the redistribution of charge from Stage 1 to Stage 2. We use the Work-Energy Theorem for the transition.

Additional Work done by Battery:

$$ \Delta Q = Q_{final} – Q_{initial} = 900 \mu C – 800 \mu C = 100 \mu C $$ $$ W_{batt} = \Delta Q \cdot V_0 = (100 \times 10^{-6}) (200) = 0.02 \, J $$

Energy Balance Equation:

$$ W_{batt} + U_{initial} = U_{final} + H_{resistor} $$ $$ 0.02 + 0.08 = 0.09 + H_{resistor} $$ $$ 0.10 = 0.09 + H_{resistor} $$ $$ H_{resistor} = 0.01 \, J $$