Capacitors in Series and Parallel

Like resistors, capacitors are often combined in networks. To analyse a capacitor circuit you need to know how the overall capacitance of a combination depends on the individual capacitances. Perhaps surprisingly, the rules are exactly the opposite of those for resistors: capacitors in parallel simply add, while in series the reciprocals add. Getting the correct rule in the correct place is a standard exam check.

This lesson continues OCR H556 Module 6.1, building on the definition C = Q/V from Lesson 1.

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Capacitors in Parallel

Consider two capacitors C₁ and C₂ connected in parallel to a battery of e.m.f. V:

graph LR
    B((+V)) --- A
    A --> C1[C₁]
    A --> C2[C₂]
    C1 --> G
    C2 --> G
    G((0 V))

Each capacitor has the same potential difference V across it, because both are wired between the same two nodes. The charges on the two capacitors are

Q₁ = C₁V    Q₂ = C₂V

The battery delivers a total charge

Q_total = Q₁ + Q₂ = (C₁ + C₂)V

Comparing with Q_total = C_total V, the combined capacitance is

C_parallel = C₁ + C₂ + C₃ + …

Capacitors in parallel add. This is easy to remember if you think of a parallel combination as two capacitors physically side by side: the effective plate area is the sum, and capacitance is proportional to area, so capacitances add.

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Capacitors in Series

Now connect the same two capacitors in series across the battery:

graph LR
    B((+V)) --- C1[C₁] --- M((mid)) --- C2[C₂] --- G((0 V))

This is more subtle. The two capacitors must carry the same charge Q. Why? The wire and plate between them (the "mid" node) was uncharged before the battery was connected. Whatever positive charge appears on the left plate of C₂ must have come from the right plate of C₁, so those two plates carry equal and opposite charges. By symmetry, and because the mid node is insulated, the charge on every plate in the chain has the same magnitude.

The voltage across each capacitor is therefore

V₁ = Q / C₁    V₂ = Q / C₂

Kirchhoff's voltage law gives

V = V₁ + V₂ = Q(1/C₁ + 1/C₂)

Comparing with V = Q / C_total,

1 / C_series = 1/C₁ + 1/C₂ + 1/C₃ + …

Capacitors in series: reciprocals add. Note also that the combined capacitance is smaller than the smallest individual capacitance — the opposite of resistors in series. This is because effectively you are increasing the separation between plates while the area stays the same.

Special Case — Two Capacitors in Series

For exactly two capacitors in series it is often convenient to use the "product over sum" version:

C_series = (C₁ × C₂) / (C₁ + C₂)

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Summary Table — Resistors vs Capacitors

Configuration	Resistors	Capacitors
Series	R = R₁ + R₂	1/C = 1/C₁ + 1/C₂
Parallel	1/R = 1/R₁ + 1/R₂	C = C₁ + C₂

Exam Tip: Swap "series" and "parallel" in your head when you go from resistors to capacitors. OCR's markers know this is the number-one slip and will not award the wrong rule applied to the right numbers.

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Worked Example — Parallel Combination

Three capacitors of 2.0 μF, 3.0 μF and 5.0 μF are connected in parallel across a 12 V battery. Calculate (a) the combined capacitance, (b) the total charge drawn from the battery, and (c) the charge on the 5.0 μF capacitor.

(a) Combined capacitance

C_total = 2.0 + 3.0 + 5.0 = 10.0 μF

(b) Total charge

Q_total = C_total × V
       = 10.0 × 10⁻⁶ × 12
       = 1.20 × 10⁻⁴ C
       = 120 μC