Capacitor problems — charging, discharging, energy stored, combinations

Question

Three capacitors of 2 µF, 3 µF, and 6 µF are connected in parallel across a 12 V battery. Find: (a) the equivalent capacitance, (b) total charge stored, (c) total energy stored, and (d) charge on each capacitor.

(CBSE Class 12 / JEE Main pattern)

Solution — Step by Step

flowchart TD
    A["Capacitor Combination"] --> B{"How connected?"}
    B -->|"Same voltage across all\n(parallel)"| C["C_eq = C₁ + C₂ + C₃\nVoltage same, charge splits"]
    B -->|"Same charge on all\n(series)"| D["1/C_eq = 1/C₁ + 1/C₂ + 1/C₃\nCharge same, voltage splits"]
    C --> E["Q_total = C_eq × V"]
    D --> F["Q = C_eq × V_total"]

In parallel, all capacitors have the same voltage (12 V each). The charges are different.

$C_{\text{eq}} = C_1 + C_2 + C_3 = 2 + 3 + 6 = \mathbf{11 \text{ }\mu\text{F}}$

$Q_{\text{total}} = C_{\text{eq}} \times V = 11 \times 10^{-6} \times 12 = \mathbf{132 \text{ }\mu\text{C}}$

$U = \frac{1}{2}C_{\text{eq}}V^2 = \frac{1}{2} \times 11 \times 10^{-6} \times 144 = \mathbf{7.92 \times 10^{-4} \text{ J} = 792 \text{ }\mu\text{J}}$

In parallel, $V$ is the same for each:

$Q_1 = C_1 V = 2 \times 12 = 24$ µC

$Q_2 = C_2 V = 3 \times 12 = 36$ µC

$Q_3 = C_3 V = 6 \times 12 = 72$ µC

Check: $24 + 36 + 72 = 132$ µC = $Q_{\text{total}}$ . Correct.

Why This Works

In a parallel combination, all capacitors are directly connected to the battery terminals, so each one “sees” the full battery voltage. The charges are different because each capacitor has a different capacitance — more capacitance means more charge stored at the same voltage.

The energy formula $U = \frac{1}{2}CV^2$ comes from integrating the work done to charge the capacitor: as charge builds up, the voltage increases, so the work per unit charge increases. The integral gives the $\frac{1}{2}$ factor.

Three equivalent energy formulas: $U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{QV}{2}$ . Use whichever form has the known quantities.

Alternative Method — Series Combination

For series: $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

All capacitors carry the same charge $Q$ , but the voltage divides: $V_i = Q/C_i$ . The smallest capacitor gets the largest voltage share — this is the one most likely to break down, a point that appears in JEE problems about dielectric breakdown.

For JEE Main, combination problems often involve mixed series-parallel networks. Strategy: start from the innermost pair, reduce step by step, and keep track of which capacitors share voltage (parallel) vs which share charge (series). Redrawing the circuit at each step helps enormously.

Common Mistake

The formulas for series and parallel are opposite to resistors. Capacitors in parallel add directly ( $C_{\text{eq}} = C_1 + C_2$ ), while resistors in parallel use the reciprocal formula. Students who have just studied resistors often apply the resistor rules to capacitors and get inverted answers. A quick sanity check: parallel should always give a larger equivalent capacitance than any individual one, and series should give a smaller one.

Question

Solution — Step by Step

Why This Works

Alternative Method — Series Combination

Common Mistake

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