Capacitors in Series and Parallel

Spec mapping: OCR H556 Module 6.1 — Capacitors (combination of capacitors in parallel, $C_\text{total} = \sum C_i$ ; combination of capacitors in series, $1/C_\text{total} = \sum 1/C_i$ ; the contrast with resistor combination rules; the principle that series capacitors carry equal charge while parallel capacitors share the same potential difference). Refer to the official OCR H556 specification document for exact wording.

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 the single most common slip in the entire Module 6.1 topic.

This lesson continues Module 6.1, building on the definition $C = Q/V$ from lesson 1. The two combination rules unlock the analysis of every multi-capacitor circuit you will meet — from charge-redistribution problems, to camera-flash designs with multiple capacitors in series for voltage rating, to the timing networks that drive every microprocessor on every computer.

Capacitors in Parallel

Consider two capacitors $C_1$ and $C_2$ connected in parallel to a battery of e.m.f. $V$ :

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

$Q_1 = C_1 V \quad\text{and}\quad Q_2 = C_2 V.$

The battery delivers a total charge

$Q_\text{total} = Q_1 + Q_2 = (C_1 + C_2)V.$

Comparing with $Q_\text{total} = C_\text{total} V$ , the combined capacitance is

$\boxed{\,C_\text{parallel} = C_1 + C_2 + C_3 + \dots\,}$

Capacitors in parallel add. A clean physical picture: wiring two parallel-plate capacitors in parallel is equivalent to enlarging the plate area. The effective area is the sum $A_1 + A_2$ , and since $C = \varepsilon_0 A/d$ , capacitances add.

Capacitors in Series

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

This is more subtle. The two capacitors must carry the same charge $Q$ . Why? The wire and plates joining them at the "mid" node were uncharged before the battery was connected, and they remain electrically isolated from everything else. Any positive charge that appears on the left plate of $C_2$ must have come from the right plate of $C_1$ , so those two plates carry equal and opposite charges. By symmetry — and because charge is conserved on the isolated mid-node — the magnitude of charge on every plate in the chain must be the same.

The voltage across each capacitor is therefore

$V_1 = \frac{Q}{C_1} \quad\text{and}\quad V_2 = \frac{Q}{C_2}.$

Kirchhoff's voltage law (Module 4.1) gives

$V = V_1 + V_2 = Q\left(\frac{1}{C_1} + \frac{1}{C_2}\right).$

Comparing with $V = Q/C_\text{total}$ ,

$\boxed{\,\frac{1}{C_\text{series}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots\,}$

Capacitors in series: reciprocals add. Note also that the combined capacitance is smaller than the smallest individual capacitance — the opposite of resistors in series. The geometric picture: wiring two parallel-plate capacitors in series is equivalent to increasing the effective plate separation $d_1 + d_2$ while keeping the area $A$ the same, and $C = \varepsilon_0 A/d$ falls as $d$ rises.

Special case — two capacitors in series

For exactly two capacitors in series, the "product over sum" version is often more convenient:

$C_\text{series} = \frac{C_1 C_2}{C_1 + C_2}.$

Decision flow — series or parallel first?

Before plugging numbers into a multi-capacitor network, decide what is in series with what and what is in parallel with what. The decision flow is the same as for resistors but with reversed rules:

graph TD
    A[Identify nodes in the network] --> B{Branches share both<br/>endpoints?}
    B -->|Yes| C[Parallel: C_eq = ΣC<br/>same V on each]
    B -->|No| D{Single charge path<br/>through the chain?}
    D -->|Yes| E[Series: 1/C_eq = Σ1/C<br/>same Q on each]
    D -->|No| F[Reduce sub-networks first,<br/>then re-apply this flow]

The reduction is iterative: identify the innermost pair, replace it with its equivalent, and repeat until you have a single capacitor.

Summary table — resistors vs capacitors

Configuration	Resistors	Capacitors
Series	$R = R_1 + R_2$	$1/C = 1/C_1 + 1/C_2$
Parallel	$1/R = 1/R_1 + 1/R_2$	$C = C_1 + C_2$

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.

The deeper reason for the reversal is that resistors share current in series (the same $I$ flows through each) while capacitors share charge in series (the same $Q$ sits on each). Voltages then sum the same way in both cases. In parallel the situation is reversed: same $V$ across each capacitor or resistor, but currents/charges add. The asymmetry of the rules is a consequence of capacitance being defined by $C = Q/V$ rather than $R = V/I$ .

Worked Example 1 — Parallel Combination

Three capacitors of $2.0\ \mu$ F, $3.0\ \mu$ F and $5.0\ \mu$ F 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 $5.0\ \mu$ F capacitor.

(a) Combined capacitance. In parallel, capacitances add:

$C_\text{total} = 2.0 + 3.0 + 5.0 = 10.0\ \mu\text{F}.$

(b) Total charge.

\begin{aligned} Q_\text{total} &= C_\text{total}\, V \\ &= 10.0 \times 10^{-6} \times 12 \\ &= 1.20 \times 10^{-4}\ \text{C} \\ &= 120\ \mu\text{C}. \end{aligned}

(c) Charge on the $5.0\ \mu$ F capacitor. In parallel each capacitor has the full $12$ V across it:

$Q_5 = C_5\, V = 5.0 \times 10^{-6} \times 12 = 60\ \mu\text{C}.$

Check: $2\ \mu\text{F} \times 12\ \text{V} = 24\ \mu$ C, $3\ \mu\text{F} \times 12\ \text{V} = 36\ \mu$ C, $5\ \mu\text{F} \times 12\ \text{V} = 60\ \mu$ C. Sum $= 120\ \mu$ C. The "share of $Q$ " goes in the same ratio as the capacitances — the bigger capacitor takes the bigger share of $Q$ .

Worked Example 2 — Series Combination

A $2.0\ \mu$ F capacitor and a $6.0\ \mu$ F capacitor are connected in series across a $9.0$ V battery. Calculate (a) the combined capacitance, (b) the charge on each capacitor, and (c) the voltage across each.

(a) Combined capacitance. Reciprocals add:

$\frac{1}{C} = \frac{1}{2.0} + \frac{1}{6.0} = \frac{3}{6} + \frac{1}{6} = \frac{4}{6}\ \mu\text{F}^{-1},$

$C = \frac{6}{4} = 1.5\ \mu\text{F}.$

Note: $C$ is smaller than either individual capacitor, as expected in series. The product-over-sum shortcut gives the same answer:

$C = \frac{C_1 C_2}{C_1 + C_2} = \frac{2.0 \times 6.0}{2.0 + 6.0} = \frac{12}{8} = 1.5\ \mu\text{F}.\ \checkmark$

(b) Charge on each. In series, every capacitor has the same charge, equal to $Q = C_\text{total}\, V$ :

$Q = (1.5 \times 10^{-6})(9.0) = 1.35 \times 10^{-5}\ \text{C} = 13.5\ \mu\text{C}.$

Both capacitors therefore hold $13.5\ \mu$ C.

(c) Voltage across each.

\begin{aligned} V_1\ (2\ \mu\text{F}) &= \frac{Q}{C_1} = \frac{13.5\ \mu\text{C}}{2.0\ \mu\text{F}} = 6.75\ \text{V}, \\ V_2\ (6\ \mu\text{F}) &= \frac{Q}{C_2} = \frac{13.5\ \mu\text{C}}{6.0\ \mu\text{F}} = 2.25\ \text{V}. \end{aligned}

Check: $6.75 + 2.25 = 9.00$ V. $\checkmark$

Notice how the smaller capacitor develops the larger voltage — the opposite of resistors, where larger $R$ means larger $V$ . This is sometimes called the "voltage-stress" rule: the smaller capacitor in a series chain takes the biggest fraction of the applied voltage, and is therefore most at risk of dielectric breakdown.

Worked Example 3 — Mixed Network

A $4\ \mu$ F capacitor is in series with a parallel combination of two $6\ \mu$ F capacitors. The network is connected to a $10$ V supply. Find (a) the equivalent capacitance, (b) the voltage across the $4\ \mu$ F capacitor, and (c) the charge on each of the $6\ \mu$ F capacitors.

graph LR
    B((+10 V)) --- C4[4 μF] --- M((mid))
    M --> C6a[6 μF]
    M --> C6b[6 μF]
    C6a --> G((0 V))
    C6b --> G

Step 1: combine the parallel pair.

$C_\text{parallel} = 6 + 6 = 12\ \mu\text{F}.$

Step 2: combine with the $4\ \mu$ F in series.

$\frac{1}{C} = \frac{1}{4} + \frac{1}{12} = \frac{3}{12} + \frac{1}{12} = \frac{4}{12},$

$C = 3\ \mu\text{F}.$

Step 3: find the charge on the $4\ \mu$ F capacitor. The combination is in series with the $4\ \mu$ F, so it carries the same charge $Q$ as the $4\ \mu$ F capacitor on its own:

$Q = CV = (3 \times 10^{-6})(10) = 30\ \mu\text{C}.$

Step 4: voltage across the $4\ \mu$ F capacitor.

$V_4 = \frac{Q}{C_4} = \frac{30\ \mu\text{C}}{4\ \mu\text{F}} = 7.5\ \text{V}.$

Step 5: voltage across the parallel pair, and charge on each $6\ \mu$ F capacitor. The remaining $2.5$ V drops across the parallel pair (consistent with $12\ \mu\text{F} \times 2.5\ \text{V} = 30\ \mu$ C, the same charge as enters from the $4\ \mu$ F branch). Each $6\ \mu$ F capacitor has $2.5$ V across it, so

$Q_6 = (6 \times 10^{-6})(2.5) = 15\ \mu\text{C} \text{ per capacitor}.$

Total charge into the parallel pair $= 2 \times 15 = 30\ \mu$ C — consistent with what came out of the $4\ \mu$ F. $\checkmark$

Capacitors in Series and Parallel

Capacitors in Series and Parallel

Capacitors in Parallel

Capacitors in Series

Special case — two capacitors in series

Decision flow — series or parallel first?

Summary table — resistors vs capacitors

Worked Example 1 — Parallel Combination

Worked Example 2 — Series Combination

Worked Example 3 — Mixed Network

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