Capacitors in series and parallel

Why join capacitors?

A shop sells capacitors with fixed values, such as $10\ \mu\text{F}$ or $47\ \mu\text{F}$. Each one is marked with the greatest voltage it can take. What if you need a value that is not on the shelf? What if you need a safe voltage that is higher than any one capacitor allows?

You join capacitors. There are two simple ways, and every network is built from them.

Capacitors in parallel

In a parallel group, all the left plates are joined by one wire and all the right plates by another. The battery is joined across the two wires.

Two things follow at once.

• The voltage is the same. Every capacitor is joined across the same two wires. So each has the voltage $V$ of the battery.
• The charges add. Each capacitor holds $Q_1 = C_1V$, $Q_2 = C_2V$ and so on. The battery has supplied the total.

The group acts as one capacitor with $Q = CV$. So we read off

Why is it bigger? Joining capacitors side by side is like making one capacitor with a bigger plate area. A bigger area holds more charge for the same voltage.

A worked example. Capacitors of $4.0\ \mu\text{F}$, $6.0\ \mu\text{F}$ and $2.0\ \mu\text{F}$ are joined in parallel across a $12\ \text{V}$ battery.

1. The equivalent. $C = 4.0\ \mu\text{F} + 6.0\ \mu\text{F} + 2.0\ \mu\text{F} = 12\ \mu\text{F}$.
2. The charges. $Q_1 = 4.0\ \mu\text{F} \times 12\ \text{V} = 48\ \mu\text{C}$, $Q_2 = 72\ \mu\text{C}$ and $Q_3 = 24\ \mu\text{C}$.
3. Check. The total is $144\ \mu\text{C}$, and $12\ \mu\text{F} \times 12\ \text{V} = 144\ \mu\text{C}$.

Capacitors in series

In a series chain, the right plate of one capacitor is joined to the left plate of the next. There is only one path for the charge.

Look at the two plates that are joined together. They form one piece of metal that is not joined to anything else. Its total charge was zero at the start, and it stays zero. So if the left plate of the first capacitor holds $+Q$, its right plate holds $-Q$. That pulls $+Q$ onto the left plate of the next capacitor, and so on down the chain.

• The charge is the same. Every capacitor holds the same $Q$.
• The voltages add. The battery voltage is split between the capacitors.

The chain acts as one capacitor with $V = Q/C$. Divide both sides by $Q$:

Why is it smaller? Capacitors in a chain are like one capacitor with its plates pulled further apart. The area is the same, but the gap has grown.

For two capacitors only, this becomes $C = \dfrac{C_1C_2}{C_1 + C_2}$, the product over the sum. For $n$ identical capacitors, it is $C/n$. Never use the product over the sum for three or more.

Who takes the voltage? The charge is the same, so $V = Q/C$ is larger across the smaller capacitor. For two capacitors,

The numerator is the other capacitor. The smaller capacitor takes the larger share.

A worked example. A $6.0\ \mu\text{F}$ capacitor and a $3.0\ \mu\text{F}$ capacitor are joined in series across $18\ \text{V}$.

1. The equivalent. $C = \dfrac{6.0\ \mu\text{F} \times 3.0\ \mu\text{F}}{6.0\ \mu\text{F} + 3.0\ \mu\text{F}} = 2.0\ \mu\text{F}$.
2. The charge. $Q = CV = 2.0\ \mu\text{F} \times 18\ \text{V} = 36\ \mu\text{C}$, on each capacitor.
3. The voltages. $V_1 = \dfrac{36\ \mu\text{C}}{6.0\ \mu\text{F}} = 6.0\ \text{V}$ and $V_2 = \dfrac{36\ \mu\text{C}}{3.0\ \mu\text{F}} = 12\ \text{V}$.
4. Check. $6.0\ \text{V} + 12\ \text{V} = 18\ \text{V}$. The $3.0\ \mu\text{F}$ capacitor, the smaller one, has the larger voltage.

Start with the defaults. Which curve is higher, and by how much? Now slide the second capacitor to $20\ \mu\text{F}$. The series value never rises above one number. What is that number, and why?

A mixed network

Most networks are a mix. The method is always the same.

1. Find two or more capacitors that are plainly in series, or plainly in parallel.
2. Replace them by their equivalent. Redraw.
3. Repeat until one capacitor is left.
4. Then go back. Use the charge and the voltage at each step to find what happens in each capacitor.

A worked example. A $15\ \mu\text{F}$ capacitor is joined in series with a parallel pair, $4.0\ \mu\text{F}$ and $6.0\ \mu\text{F}$. The three are across $25\ \text{V}$.

1. The pair. $4.0\ \mu\text{F} + 6.0\ \mu\text{F} = 10\ \mu\text{F}$.
2. The series. $C = \dfrac{15\ \mu\text{F} \times 10\ \mu\text{F}}{15\ \mu\text{F} + 10\ \mu\text{F}} = 6.0\ \mu\text{F}$.
3. The charge. $Q = 6.0\ \mu\text{F} \times 25\ \text{V} = 150\ \mu\text{C}$. This is the charge on the $15\ \mu\text{F}$ capacitor, and the total charge on the pair.
4. The voltages. The $15\ \mu\text{F}$ capacitor has $\dfrac{150\ \mu\text{C}}{15\ \mu\text{F}} = 10\ \text{V}$. The pair has $\dfrac{150\ \mu\text{C}}{10\ \mu\text{F}} = 15\ \text{V}$.
5. Inside the pair. Both have $15\ \text{V}$. So $Q = 4.0\ \mu\text{F} \times 15\ \text{V} = 60\ \mu\text{C}$ and $6.0\ \mu\text{F} \times 15\ \text{V} = 90\ \mu\text{C}$. Together that is $150\ \mu\text{C}$.

The voltage rating of a stack

A capacitor marked "$100\ \text{V}$" has a thin insulating layer. Above $100\ \text{V}$ the layer breaks down and the capacitor is spoiled. When we put capacitors in series, the voltage is shared. But it is shared unequally, so one capacitor may reach its limit first.

A worked example. A $2.0\ \mu\text{F}$ capacitor and an $8.0\ \mu\text{F}$ capacitor are each rated $100\ \text{V}$. They are joined in series. What is the greatest voltage across the pair?

1. The charge limits. The $2.0\ \mu\text{F}$ capacitor reaches $100\ \text{V}$ at $Q = 2.0\ \mu\text{F} \times 100\ \text{V} = 200\ \mu\text{C}$. The $8.0\ \mu\text{F}$ capacitor would reach it only at $800\ \mu\text{C}$.
2. The smaller limit wins. The same charge is on both, so the pair can hold $200\ \mu\text{C}$ at most.
3. The voltages. The $2.0\ \mu\text{F}$ capacitor has $100\ \text{V}$. The $8.0\ \mu\text{F}$ capacitor has $\dfrac{200\ \mu\text{C}}{8.0\ \mu\text{F}} = 25\ \text{V}$.
4. The pair. $100\ \text{V} + 25\ \text{V} = 125\ \text{V}$. That is less than $200\ \text{V}$, which equal capacitors would give.

The lesson: in series, the small capacitor sets the limit. Equal capacitors share the voltage equally and give the best use of the ratings.

Practise with new numbers

Try it in Python

The program joins three capacitors in parallel and then in series, and shows how the voltage is shared.

Common mistakes

• Mixing up the two rules. Parallel capacitors share the voltage. Series capacitors share the charge.
• Adding reciprocals for a parallel group. A parallel group simply adds the capacitances.
• Using the product over the sum for three capacitors. It works for two only. For more, add the reciprocals.
• Forgetting the last step of a series sum. Adding $\dfrac{1}{C_1} + \dfrac{1}{C_2}$ gives $\dfrac{1}{C}$, not $C$. Turn it over.
• Giving the larger voltage to the larger capacitor. In series it is the smaller capacitor that has the larger voltage.

Remember these

• Parallel: same $V$, $Q = Q_1 + Q_2 + \dots$, and $C = C_1 + C_2 + \dots$
• Series: same $Q$, $V = V_1 + V_2 + \dots$, and $\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dots$
• Two in series: $C = \dfrac{C_1C_2}{C_1 + C_2}$, and $V_1 = V\,\dfrac{C_2}{C_1 + C_2}$.
• Identical capacitors: $n$ in series give $C/n$, and $n$ in parallel give $nC$.
• A rating: in series, the capacitor that reaches its limit first sets the limit for the whole chain.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Capacitors of $2.0\ \mu\text{F}$, $3.0\ \mu\text{F}$ and $5.0\ \mu\text{F}$ are joined in parallel. What is the equivalent capacitance?

2. Four capacitors of $8.0\ \mu\text{F}$ each are joined in series. What is the equivalent capacitance?

3. Why is the equivalent capacitance of a series chain smaller than the smallest capacitor in it?

4. A $9.0\ \mu\text{F}$ capacitor and an $18\ \mu\text{F}$ capacitor are joined in series across $27\ \text{V}$. Which one has the larger voltage, and what is it?

5. A student says: "These two capacitors are in parallel, so each holds the same charge." Spot the mistake.