Capacitors: Diagram-Based Questions (4)

Question

In a circuit, three capacitors of $2 \mu\text{F}$, $3 \mu\text{F}$, and $6 \mu\text{F}$ are connected: the $2 \mu\text{F}$ and $3 \mu\text{F}$ are in parallel, and that combination is in series with the $6 \mu\text{F}$, all connected across a $30 \text{ V}$ battery. Find the equivalent capacitance, the charge on each capacitor, and the voltage across each.

Solution — Step by Step

Why This Works

Capacitor combination rules are the opposite of resistor rules:

• Series capacitors: $1/C_{\text{eq}} = \sum 1/C_i$ (like resistors in parallel)
• Parallel capacitors: $C_{\text{eq}} = \sum C_i$ (like resistors in series)

The reason: in series, the same charge sits on each capacitor (charge can’t disappear at the junctions), so voltages add. In parallel, the same voltage appears across each, so charges add.

Alternative Method

Use energy: total energy stored is $\tfrac{1}{2}C_{\text{eq}}V^2 = \tfrac{1}{2} \times (30/11) \times 900 \approx 1227 \mu\text{J}$. Then verify by summing individual energies: $\tfrac{1}{2} \times 2 \times 16.4^2 + \tfrac{1}{2} \times 3 \times 16.4^2 + \tfrac{1}{2} \times 6 \times 13.6^2$. Useful as a sanity check on long capacitor problems.

Common Mistake

Final answer: $C_{\text{eq}} \approx 2.73 \mu\text{F}$, $Q_6 \approx 81.8 \mu\text{C}$, $V_6 \approx 13.6 \text{ V}$, $V_p \approx 16.4 \text{ V}$, $Q_2 \approx 32.7 \mu\text{C}$, $Q_3 \approx 49.1 \mu\text{C}$.