Capacitors in Series and Parallel

Q: Is there a quick rule for two capacitors in series?

Yes: $C_{\text{ser}} = \frac{C_1 C_2}{C_1 + C_2}$, which is the product of the two values divided by their sum. For 10 $\mu$F and 30 $\mu$F, that is $300 / 40 = 7.5\ \mu\text{F}$. Use this shortcut only for exactly two capacitors. For three or more, use the reciprocal rule.

In one sentence: Capacitors connected side by side add their capacitances together, while capacitors connected in a chain give a smaller total that you find with a special rule.

$C_{\text{par}} = C_1 + C_2 + \cdots \qquad \frac{1}{C_{\text{ser}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$

Read it out loud: “C par, the total capacitance in parallel, equals C one plus C two, and so on. For a series chain, one divided by C ser equals one divided by C one, plus one divided by C two, and so on. The small letters par and ser stand for parallel and series. The three dots mean the pattern keeps going for as many capacitors as you have.”

Before you start

The ideas and signs this page uses, explained from zero. For powers of ten, Greek letters and other notation in general, see Math basics.

Capacitor and capacitance ( $C$ ): A capacitor is two metal plates with an insulator between them, a material that electric charge cannot flow through. Connect it to a voltage and one plate collects positive charge while the other collects an equal amount of negative charge. Its capacitance, written $C$ , says how much charge it stores for each volt across it: $C = Q / V$ . The unit is the farad, written F, which means one coulomb of charge per volt. A farad is a very large unit, so real capacitors are usually given in microfarads or smaller units.
Prefixes: micro, nano and pico ( $\mu$ F, nF, pF): A prefix is a short word placed in front of a unit to scale it. Micro, written $\mu$ (the Greek letter mu, said ‘mew’), means one millionth, $10^{-6}$ . So 10 microfarads, written $10\ \mu$ F, is $10 \times 10^{-6}$ F, which is 0.00001 F. Nano (n) means one billionth, $10^{-9}$ , and pico (p) means one trillionth, $10^{-12}$ . A 100 pF capacitor is therefore $100 \times 10^{-12}$ F, a very small capacitance.
Powers of ten, like $10^{-6}$ and $10^{3}$: Electronics uses numbers that are far too big or too small to write out easily, so they are written as a number times a power of ten. The small raised number, the exponent, says how many times to multiply by ten. So $10^{3}$ means $10 \times 10 \times 10$ , which is 1,000. A negative exponent means divide by ten instead: $10^{-6}$ is one divided by ten six times, which is 0.000001, a decimal point, five zeros and then a 1. When two powers of ten are multiplied, add their exponents; when one is divided by another, subtract them. The sign $\approx$ means approximately equal. It is used when a result has been rounded, so the number after it is close to the exact value but not exact.
Reciprocal, and flipping a fraction: The reciprocal of a number is one divided by that number. The reciprocal of 10 is $\frac{1}{10}$ , which is 0.1, and the reciprocal of 0.1 is 10. Taking a reciprocal flips a number upside down. The series rule adds the reciprocals of the capacitances, and then one more reciprocal turns that sum back into the total. Taking the reciprocal twice always gives back the number you started with.
Subscripts, like $C_1$ and $C_2$ , and the three dots $\cdots$: A small lowered label, a subscript, tells similar quantities apart. $C_1$ is read ‘C one’ and means the capacitance of the first capacitor, and $C_2$ is the second. The subscript is only a name, not a multiplication, so $C_2$ does not mean C times 2. The three dots, $\cdots$ , are read ‘and so on’ and mean the same pattern continues for as many capacitors as are in the group.
Parallel and series wiring: In a parallel connection, the capacitors sit side by side, with both of their ends joined to the same two wires, so every capacitor sees the same voltage. In a series connection, the capacitors are joined end to end in one chain, so the same charge ends up on each of them. The way the capacitors are wired changes the total, and the two cases use different rules.

What each symbol means

Symbol	Meaning
$C_{\text{par}}$	The total capacitance of the capacitors wired in parallel, in farads (F). It acts like one single capacitor that holds the same charge as the whole parallel group at the same voltage. Typical values range from picofarads in radio circuits to thousands of microfarads in power supplies.
$C_{\text{ser}}$	The total capacitance of the capacitors wired in series, in farads (F). It is always smaller than the smallest capacitor in the chain. It is found by adding reciprocals, so the answer must be flipped at the end to come out in farads.
$C_1$	The capacitance of the first capacitor in the group, in farads (F). In the worked example it is 10 microfarads, which is $10 \times 10^{-6}$ F. The same meaning applies to any other numbered capacitor in the group.
$C_2$	The capacitance of the second capacitor in the group, in farads (F). In the worked example it is 30 microfarads. Any further capacitors are labeled $C_3$ , $C_4$ and so on, in the same way.

The formula piece by piece

$C_1$: The starting point is one capacitor, with its capacitance $C_1$ in farads. Capacitance says how much charge the capacitor holds for each volt across it, so a bigger number means a bigger store of charge. Every capacitor in the group has its own value of $C$ .
$C_1 + C_2$: Adding two capacitances gives the parallel total. This is the rule for capacitors placed side by side on the same voltage, where each one adds its own charge store, so the group holds the sum of their charges.
$\frac{1}{C_1}$: The reciprocal of one capacitance, which is one divided by it. A small capacitance has a large reciprocal, and a large capacitance has a small reciprocal. Series wiring uses reciprocals because the same charge goes onto each part, and the voltages add up.
$\frac{1}{C_1} + \frac{1}{C_2}$: Adding the reciprocals of the series capacitances gives one divided by the total. One more reciprocal then turns that sum into the total capacitance itself, which is the rule for capacitors in a single chain.
$C_{\text{par}} = C_1 + C_2 + \cdots \qquad \frac{1}{C_{\text{ser}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$: The two rules in one sentence: the total capacitance of capacitors in parallel equals their capacitances added together, and one divided by the total capacitance of capacitors in series equals the sum of the reciprocal of each capacitance.
$C_{\text{ser}} = \frac{C_1 C_2}{C_1 + C_2}$: A standard shortcut for exactly two capacitors in series. Multiply the two capacitances, then divide by their sum. It gives the same answer as the reciprocal rule, so it is a useful check when you have only two parts.

What the formula is doing, and why

You have two or more capacitors and want one number that acts like the whole group. That number is the total capacitance, written $C_{\text{par}}$ for parallel wiring and $C_{\text{ser}}$ for series wiring. The inputs are the separate capacitances $C_1$ , $C_2$ and so on, all in farads. The output is the total, also in farads. The two rules differ because the wiring changes what is shared across each capacitor: in parallel it is the voltage, and in series it is the charge.

Why add in parallel? In a parallel group, every capacitor is connected across the same voltage $V$ . Each one stores its own charge, $Q_i = C_i V$ , because charge equals capacitance times voltage. The total charge is the sum of those charges: $C_1 V + C_2 V = (C_1 + C_2) V$ . Dividing the total charge by the shared voltage gives the total capacitance, $C_{\text{par}} = C_1 + C_2$ . Every added capacitor adds room for more charge, so the total is always bigger than the largest single capacitor.

Why reciprocals in series? In a series chain, charge flows onto one plate and an equal charge is pushed off the other plate of the same capacitor. The two plates joined at each connection have no other route for charge, so they must carry equal and opposite charges. The same charge $Q$ therefore ends up on every capacitor in the chain. The supply voltage is shared among them, $V = V_1 + V_2 + \cdots$ , and each capacitor has $V_i = Q / C_i$ . Putting these together gives $V = Q\left(\frac{1}{C_1} + \frac{1}{C_2} + \cdots\right)$ . Since the total satisfies $V = Q / C_{\text{ser}}$ , the reciprocal rule follows. Each capacitor takes its own share of the voltage, so the chain needs more total voltage for the same charge, which makes its total smaller than its smallest part.

Opposite to resistors. The rules are the reverse of the rules for resistors. Resistances add in series, and reciprocals are used in parallel. This is easy to mix up, so check the wiring first: parallel capacitors add directly, and series capacitors use reciprocals.

When the rules apply. The rules describe ideal capacitors wired in simple parallel or series groups. Real circuits often mix both kinds of connection. Solve them one small group at a time, starting with the innermost group. Replace that group with its single total, then work outward until one capacitor is left.

An everyday picture

Think of a capacitor as a tank that holds charge for each volt of push. Tanks placed side by side and filled to the same level hold their amounts added together, which is the parallel case. Tanks joined in a chain, where the same water flows through each one, share the push between them. The chain needs a bigger total push for the same water, so it holds less per volt than its smallest tank, which is the series case.

Worked example: 10 microfarad and 30 microfarad capacitors, in parallel and in series

Imagine a hobbyist’s parts drawer with two capacitors, one with $C_1 = 10\ \mu\text{F}$ and one with $C_2 = 30\ \mu\text{F}$ . These are made-up values with easy arithmetic. First wire them side by side, in parallel, and find the total. Then wire them end to end, in series, and find that total. Every answer is in microfarads, the same unit as the values.

Write down the two values and the unit. The first capacitor is $C_1 = 10\ \mu\text{F}$ and the second is $C_2 = 30\ \mu\text{F}$ . Both are already in microfarads, so we can do the whole calculation in microfarads and skip any conversion. Keeping one unit from start to finish stops mistakes that come from mixing scales.
Parallel: add the two capacitances. Side by side, the two capacitors sit on the same voltage, so their charges add. The total capacitance is the sum: $C_{\text{par}} = C_1 + C_2 = 10\ \mu\text{F} + 30\ \mu\text{F} = 40\ \mu\text{F}$ . The total is bigger than either part, which makes sense because the group can hold more charge than either capacitor alone.
Series: take the reciprocal of each value. Series wiring uses one divided by each capacitance. The reciprocal of 10 is one tenth, and the reciprocal of 30 is one thirtieth: $\frac{1}{C_1} = \frac{1}{10}$ and $\frac{1}{C_2} = \frac{1}{30}$ , both in units of one over microfarads. Each reciprocal is a small fraction, and the next step adds them together.
Add the reciprocals. A common denominator lets us add fractions. One tenth is the same as three thirtieths, so $\frac{1}{10} + \frac{1}{30} = \frac{3}{30} + \frac{1}{30} = \frac{4}{30} \approx 0.1333$ . This sum is one divided by the total, not the total itself, so one more step is needed to get the capacitance back.
Flip the answer back to get the total. The total is the reciprocal of 0.1333, so $C_{\text{ser}} = \frac{1}{0.1333} \approx \frac{1}{\frac{4}{30}} = \frac{30}{4} = 7.5\ \mu\text{F}$ . Dividing by a fraction is the same as multiplying by its flip, so one divided by four thirtieths is thirty divided by four, which is 7.5.
Convert to farads to see the powers of ten. Micro means $10^{-6}$ , so $7.5\ \mu\text{F} = 7.5 \times 10^{-6}$ F. Converting the values first gives the same answer. Each value becomes $10 \times 10^{-6}$ F or $30 \times 10^{-6}$ F, and their reciprocals add to about 133,333 per farad. Flipping that number gives $7.5 \times 10^{-6}$ F again, so the powers of ten move with the numbers and the capacitance stays the same.
Check the answer against the smallest part. A series total must be smaller than the smallest capacitor in the chain. Here 7.5 µF is smaller than the 10 µF capacitor, so the result passes this test. If the answer had come out as 40 µF, that is the parallel total, and it would be a clear sign that the wrong rule was used.
Use the shortcut for two capacitors. When only two capacitors are in series, the reciprocal rule simplifies to $C_{\text{ser}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{10 \times 30}{10 + 30} = \frac{300}{40} = 7.5\ \mu\text{F}$ . Multiplying the two values and dividing by their sum gives the same answer, which is a quick check when you have only two parts.

Answer: Wired in parallel, the two capacitors act like one 40 µF capacitor. Wired in series, they act like one 7.5 µF capacitor, which is smaller than either part. The parallel total is bigger than either capacitor, and the series total is smaller than either capacitor.

The numbers in this example are recomputed by script every time the site is built.

Common mistakes

Using the series rule for parallel capacitors, or the parallel rule for series ones. The rules are the reverse of the resistor rules, so it is easy to swap them. Parallel capacitors add directly, and series capacitors use reciprocals.
Forgetting to flip the answer back. The number 0.1333 is one divided by the total, not the total itself. The total is $1 / 0.1333 = 7.5\ \mu\text{F}$ .
Expecting a series total to be bigger than its parts. A series total is always smaller than the smallest capacitor in the chain, so a series answer that is bigger than its smallest part is a sign of a mistake.
Mixing units partway through. Work in one unit, such as microfarads, for the whole calculation, or convert every value to farads before you start.

Where you will meet it

Power supplies, where several filter capacitors are wired in parallel so that their capacitances add up to a large value that smooths the output.
High-voltage banks, where capacitors are wired in series so the group can handle a higher voltage than any one part. Balancing resistors are often added so the voltage is shared evenly.
Filter and timing circuits, where the right total capacitance is built from standard part values.
Decoupling on circuit boards, where capacitors of several sizes wired in parallel each help with a different range of frequencies.

Questions people ask

Why do parallel capacitors add, when parallel resistors do not?

In parallel, every capacitor sits on the same voltage, and each one holds its own charge. The total charge is the sum of the charges, and capacitance is charge divided by voltage, so capacitances add. Resistors work the other way around: in parallel the currents add, and the rule uses reciprocals of the resistances.

Why is a series total always smaller than the smallest capacitor?

Each capacitor in the chain takes its own part of the total voltage, and the charge is the same on every one. So the chain needs a bigger total voltage for that charge. Capacitance is charge divided by voltage, so a bigger voltage for the same charge means a smaller capacitance than any single part.

Is there a quick rule for two capacitors in series?

Yes: $C_{\text{ser}} = \frac{C_1 C_2}{C_1 + C_2}$ , which is the product of the two values divided by their sum. For 10 $\mu$ F and 30 $\mu$ F, that is $300 / 40 = 7.5\ \mu\text{F}$ . Use this shortcut only for exactly two capacitors. For three or more, use the reciprocal rule.

Capacitance: Capacitance tells you how much electric charge a capacitor stores for each volt across it.
Resistors in Series and Parallel: Resistors in a line add up, and resistors side by side always give a total smaller than the smallest one.
Parallel-Plate Capacitor: The parallel-plate formula says a capacitor holds more charge per volt when its plates are bigger, closer together, or separated by a material with a higher permittivity.