Capacitor combinations — series and parallel with charge and voltage distribution

Question

Three capacitors of $2 \mu F$ , $3 \mu F$ , and $6 \mu F$ are connected first in series, then in parallel, to a 12 V battery. Find the equivalent capacitance, total charge, and voltage across each capacitor in both cases.

(CBSE 12 + JEE Main + NEET)

Solution — Step by Step

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

Key: Charge is same on all capacitors. Voltage divides.

\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = 1

$C_{eq} = \mathbf{1 \mu F}$

Total charge: $Q = C_{eq} \times V = 1 \times 12 = 12 \mu C$

Each capacitor has the same charge ( $12 \mu C$ ), and voltages are: $V_1 = 12/2 = 6$ V, $V_2 = 12/3 = 4$ V, $V_3 = 12/6 = 2$ V. Check: $6 + 4 + 2 = 12$ V.

C_{eq} = C_1 + C_2 + C_3

Key: Voltage is same across all capacitors. Charge divides.

$C_{eq} = 2 + 3 + 6 = \mathbf{11 \mu F}$

Total charge: $Q = 11 \times 12 = 132 \mu C$

Each has 12 V across it, and charges are: $Q_1 = 2 \times 12 = 24 \mu C$ , $Q_2 = 3 \times 12 = 36 \mu C$ , $Q_3 = 6 \times 12 = 72 \mu C$ . Check: $24 + 36 + 72 = 132 \mu C$ .

Property	Series	Parallel
Same quantity	Charge ( $Q$ )	Voltage ( $V$ )
Divides	Voltage	Charge
$C_{eq}$	Smaller than smallest	Larger than largest
Formula pattern	Like resistors in parallel	Like resistors in series

The last row is crucial: capacitor series formula looks like resistor parallel formula, and vice versa. This is because $C = Q/V$ while $R = V/I$ — they are “reciprocal” quantities.

flowchart TD
    A["Capacitor Problem"] --> B{"Series or Parallel?"}
    B -- Series --> C["Same Charge Q on all"]
    C --> D["Find C_eq using 1/C formula"]
    D --> E["Q = C_eq × V_total"]
    E --> F["V across each = Q/C_i"]
    B -- Parallel --> G["Same Voltage V across all"]
    G --> H["C_eq = C1 + C2 + C3"]
    H --> I["Q_total = C_eq × V"]
    I --> J["Q on each = C_i × V"]
    A --> K{"Mixed series-parallel?"}
    K --> L["Simplify innermost group first"]
    L --> B

Why This Works

In series, capacitors share the same charge because they are connected end-to-end — the charge leaving one plate has nowhere to go except onto the next capacitor’s plate. The voltage splits because the total voltage is the sum of individual drops (Kirchhoff’s voltage law).

In parallel, capacitors share the same voltage because both ends are connected to the same two nodes. Each capacitor independently stores charge based on its own capacitance, so charges add up.

Alternative Method

For two capacitors in series, use the shortcut: $C_{eq} = \frac{C_1 \times C_2}{C_1 + C_2}$ (product over sum — same as resistors in parallel). For JEE, if all $n$ capacitors are identical with capacitance $C$ : series gives $C/n$ , parallel gives $nC$ .

Common Mistake

Students swap the series and parallel formulas for capacitors because they memorised the resistor formulas first. Remember: capacitors are the OPPOSITE of resistors. Capacitors in series use the reciprocal formula (like resistors in parallel), and capacitors in parallel simply add (like resistors in series). A mnemonic: “Capacitors are Contrary.”

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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