Several capacitors can be replaced by an equivalent capacitance when their external terminals have the same charge-voltage relationship as one effective capacitor.

The combination rule depends on how the capacitors are connected.

Capacitors in parallel

Capacitors are in parallel when both terminals of every capacitor connect to the same two nodes. They therefore share the same potential difference $V$.

For capacitor $i$,

$$Q_i=C_iV.$$

The source supplies total charge

$$Q_{\rm total}=Q_1+Q_2+\cdots.$$

Thus

$$Q_{\rm total}=(C_1+C_2+\cdots)V.$$

Defining

$$Q_{\rm total}=C_{\rm eq}V,$$

we obtain

$$\boxed{C_{\rm eq}=C_1+C_2+\cdots}.$$

Parallel connection therefore increases capacitance.

Capacitors in series

Consider capacitors connected end-to-end, with the internal junctions initially neutral and isolated from any external source of charge.

When the series chain is charged, the same charge magnitude $Q$ appears on each capacitor. This follows from charge conservation at each isolated internal node: charge cannot accumulate there except as equal and opposite charges on the two connected plates.

The total potential difference is the sum of the individual voltage drops:

$$V=V_1+V_2+\cdots.$$

Since

$$V_i=\frac{Q}{C_i},$$

$$V=Q\left(\frac1{C_1}+\frac1{C_2}+\cdots\right).$$

Using

$$V=\frac{Q}{C_{\rm eq}},$$

we obtain

$$\boxed{\frac1{C_{\rm eq}} =\frac1{C_1}+\frac1{C_2}+\cdots}.$$

For two capacitors,

$$\boxed{C_{\rm eq}=\frac{C_1C_2}{C_1+C_2}}.$$

A series combination has equivalent capacitance smaller than any individual capacitor in the chain.

Worked example

Let

$$C_1=6.0,\mu\mathrm F,\qquad C_2=3.0,\mu\mathrm F.$$

In parallel,

$$C_{\rm eq}=6.0+3.0=\boxed{9.0,\mu\mathrm F}.$$

In series,

$$C_{\rm eq} =\frac{(6.0)(3.0)}{6.0+3.0},\mu\mathrm F =\boxed{2.0,\mu\mathrm F}.$$

Suppose the series combination is connected across $12,\mathrm V$. The common charge magnitude is

$$Q=C_{\rm eq}V =(2.0,\mu\mathrm F)(12,\mathrm V) =24,\mu\mathrm C.$$

The individual voltage magnitudes are

$$V_1=\frac{Q}{C_1}=4.0,\mathrm V,$$

$$V_2=\frac{Q}{C_2}=8.0,\mathrm V,$$

and indeed

$$V_1+V_2=12,\mathrm V.$$

The smaller capacitance receives the larger voltage in this series pair because both capacitors carry the same charge magnitude.

Mixed networks

Networks containing recognizable series and parallel groups can be simplified step by step. At each step, use the actual topology rather than the visual appearance of the drawing:

• parallel elements share both terminal nodes;
• series capacitors share an internal node that has no other branch attached and, under the usual ideal assumptions, carry the same charge magnitude.

These rules are consequences of the capacitor relation $Q=CV$, charge conservation, and potential differences—not independent empirical formulas.