Two or more capacitors are often joined together. FIGURE 26.22 illustrates two basic combinations: parallel capacitors and series capacitors. Notice that a capacitor is represented in circuit diagrams by two parallel lines.

FIGURE 26.22 Parallel and series capacitors.

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As we’ll show, parallel or series capacitors (or, as is sometimes said, capacitors “in parallel” or “in series”) can be represented by a single equivalent capacitance. We’ll demonstrate this first with the two parallel capacitors $C_1$ and $C_2$ of FIGURE 26.23a. Because the two top electrodes are connected by a conducting wire, they form a single conductor in electrostatic equilibrium. Thus the two top electrodes are at the same potential. Similarly, the two connected bottom electrodes are at the same potential. Consequently, two (or more) capacitors in parallel each have the same potential difference $\text{Δ}V{}_{\text{C}}$ between the two electrodes.

FIGURE 26.23 Replacing two parallel capacitors with an equivalent capacitor.

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The charges on the two capacitors are $Q_1=C_1\text{Δ}{V}_{\text{C}}$ and $Q_2=C_2\text{Δ}{V}_{\text{C}}$. Altogether, the battery’s charge escalator moved total charge $Q=Q_1+Q_2$ from the negative electrodes to the positive electrodes. Suppose, as in FIGURE 26.23b, we replaced the two capacitors with a single capacitor having charge $Q=Q_1+Q_2$ and potential difference $\text{Δ}{V}_{\text{C}}$. This capacitor is equivalent to the original two in the sense that the battery can’t tell the difference. In either case, the battery has to establish the same potential difference and move the same amount of charge.

By definition, the capacitance of this equivalent capacitor is

This analysis hinges on the fact that parallel capacitors each have the same potential difference $\text{Δ}{V}_{\text{C}}$. We could easily extend this analysis to more than two capacitors. If capacitors $C_1\text{,}{C}_2\text{,}{C}_3\text{, }\text{…}$ are in parallel, their equivalent capacitance is

Neither the battery nor any other part of a circuit can tell if the parallel capacitors are replaced by a single capacitor having capacitance $C_{\text{eq}}$.

Now consider the two series capacitors in FIGURE 26.24a. The center section, consisting of the bottom plate of $C_1$, the top plate of $C_2$, and the connecting wire, is electrically isolated. The battery cannot remove charge from or add charge to this section. If it starts out with no net charge, it must end up with no net charge. As a consequence, the two capacitors in series have equal charges $\pm Q$. The battery transfers Q from the bottom of $C_2$ to the top of $C_1$. This transfer polarizes the center section, as shown, but it still has $Q_{\text{net}}=0$.

FIGURE 26.24 Replacing two series capacitors with an equivalent capacitor.

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The potential differences across the two capacitors are $\text{Δ}{V}_1=Q\text{/}{C}_1$ and $\text{Δ}{V}_2=Q\text{/}{C}_2$. The total potential difference across both capacitors is $\text{Δ}{V}_{\text{C}}=\text{Δ}{V}_1+\text{Δ}{V}_2$. Suppose, as in FIGURE 26.24b, we replaced the two capacitors with a single capacitor having charge Q and potential difference $\text{Δ}V{}_{\text{C}}=\text{Δ}{V}_1+\text{Δ}{V}_2$. This capacitor is equivalent to the original two because the battery has to establish the same potential difference and move the same amount of charge in either case.

By definition, the capacitance of this equivalent capacitor is $C_{\text{eq}}=Q\text{/}\text{Δ}{V}_{\text{C}}$. The inverse of the equivalent capacitance is thus

This analysis hinges on the fact that series capacitors each have the same charge Q. We could easily extend this analysis to more than two capacitors. If capacitors $C_1\text{,}{C}_2\text{,}{C}_3\text{, }\text{…}$ are in series, their equivalent capacitance is

Let’s summarize the key facts before looking at a numerical example:

• Parallel capacitors all have the same potential difference $\text{Δ}{V}_{\text{C}}$. Series capacitors all have the same amount of charge $\pm Q$.

• The equivalent capacitance of a parallel combination of capacitors is larger than any single capacitor in the group. The equivalent capacitance of a series combination of capacitors is smaller than any single capacitor in the group.

EXAMPLE 26.7 A capacitor circuit

Find the charge on and the potential difference across each of the three capacitors in FIGURE 26.25.

FIGURE 26.25 A capacitor circuit.

FIGURE 26.25 Full Alternative Text

MODEL Assume the battery is ideal, with $\text{Δ}V{}_{\text{bat}}=\epsilon =12\text{V}$. Use the results for parallel and series capacitors.

SOLVE The three capacitors are neither in parallel nor in series, but we can break them down into smaller groups that are. A useful method of circuit analysis is first to combine elements until reaching a single equivalent element, then to reverse the process and calculate values for each element. FIGURE 26.26 shows the analysis of this circuit. Notice that we redraw the circuit after every step. The equivalent capacitance of the $3\mu \text{F}$ and $6\mu \text{F}$ capacitors in series is found from

FIGURE 26.26 Analyzing the capacitor circuit.

FIGURE 26.26 Full Alternative Text

$$C_{\text{eq}}=({\displaystyle \frac{1}{3\mu \text{F}}}+{\displaystyle \frac{1}{6\mu \text{F}}}{)}^{-1}={({\displaystyle \frac{2}{6}}+{\displaystyle \frac{1}{6}})}^{-1}\mu \text{F}=2\mu \text{F}$$

Once we get to the single equivalent capacitance, we find that $\text{Δ}V{}_{\text{C}}=\text{Δ}{V}_{\text{bat}}=12\text{V}$ and $Q=C\text{Δ}{V}_{\text{C}}=24\mu \text{C}$. Now we can reverse direction. Capacitors in series all have the same charge, so the charge on $C_1$ and on $C_{2+3}$ is $\pm 24\mu \text{C}$. This is enough to determine that $\text{Δ}V{}_1=8\text{V}$ and $\text{Δ}V{}_{2+3}=4\text{V}$. Capacitors in parallel all have the same potential difference, so $\text{Δ}V{}_2=\text{Δ}{V}_3=4\text{V}$. This is enough to find that $Q_2=20\mu \text{C}$ and $Q_3=4\mu \text{C}$. The charge on and the potential difference across each of the three capacitors are shown in the final step of Figure 26.26.

REVIEW Notice that we had two important checks of internal consistency. $\text{Δ}{V}_1+\text{Δ}{V}_{2+3}=8\text{V}+4\text{V}$ add up to the 12 V we had found for the $2\mu \text{F}$ equivalent capacitor. Then $Q_2+Q_3=20\mu \text{C}+4\mu C$ add up to the $24\mu \text{C}$ we had found for the $6\mu \text{F}$ equivalent capacitor. We’ll do much more circuit analysis of this type in Chapter 28, but it’s worth noting now that circuit analysis becomes nearly foolproof if you make use of these checks of internal consistency.