CURRENT BYU 41

Based on the circuit topology, we have a battery $V$ connected to a series combination of two capacitors $C$ and $2C$. The switch is placed in parallel across the capacitor $C$.

Initial State (Switch Open):

The capacitors $C$ and $2C$ are in series across the battery $V$. The equivalent capacitance is:

$$ C_{eq} = \frac{C \cdot 2C}{C + 2C} = \frac{2}{3}C $$

The initial total charge supplied by the battery is:

$$ Q_i = C_{eq} V = \frac{2}{3}CV $$

The initial stored potential energy in the capacitors is:

$$ U_i = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} \left(\frac{2}{3}C\right) V^2 = \frac{1}{3}CV^2 $$

Final State (Switch Closed):

When the switch closes, it short-circuits the capacitor $C$. The potential difference across $C$ becomes zero, and its energy is dissipated. The circuit effectively becomes the battery $V$ connected directly to the capacitor $2C$.

The final charge on capacitor $2C$ is:

$$ Q_f = (2C)V = 2CV $$

The final stored potential energy is:

$$ U_f = \frac{1}{2} (2C) V^2 = CV^2 $$

Work Done and Heat Dissipated:

The battery has to supply additional charge to increase the charge on $2C$ from $Q_i$ to $Q_f$.

$$ \Delta Q = Q_f – Q_i = 2CV – \frac{2}{3}CV = \frac{4}{3}CV $$

The work done by the battery is:

$$ W_{\text{battery}} = \Delta Q \cdot V = \left(\frac{4}{3}CV\right)V = \frac{4}{3}CV^2 $$

Using the energy conservation principle ($W_{\text{battery}} + U_i = U_f + \text{Heat}$), the total heat dissipated is:

$$ H = W_{\text{battery}} – (U_f – U_i) $$ $$ H = \frac{4}{3}CV^2 – \left(CV^2 – \frac{1}{3}CV^2\right) $$ $$ H = \frac{4}{3}CV^2 – \frac{2}{3}CV^2 $$ $$ H = \frac{2}{3}CV^2 $$