# Intuitively, why does putting capacitors in series decrease the equivalent capacitance?

Can someone please explain, intuitively (without any formula, I understand the formulas), why the equivalent capacitance of capacitors in series is less than the any individual capacitor's capacitance?

Let's take a simple case. Say we have 2 capacitors with Capacitance 2 (ignoring units), and we place them in series. A voltage $$V$$ develops in both, and a charge $$+Q$$ acculumates on one of their plates. Using the capacitance formulas, the equivalent capacitance is $$1/2$$ the original. Indeed we get $$Q/2V$$, where $$Q/V$$ is the original capacitance. But why? aren't we in total acculumating a charge of $$2Q$$ over a potential difference of $$2V$$? Why just $$1Q$$? (again, I'm speaking intuitively)

• Comments are not for extended discussion; this conversation has been moved to chat. – David Z Jun 22 '19 at 22:48

Can someone please explain, intuitively (without any formula, I understand the formulas), why the equivalent capacitance of capacitors in series is less than the any individual capacitor's capacitance?

I assume you know that the larger the capacitor plates are, the greater the capacitance, all other things being equal. Also I assume you know the greater the separation of the plates (the thicker the dielectric between the plates) the less the capacitance all other things being equal. Given these assumptions, consider the diagrams below.

The top diagram to the left shows two capacitors in parallel. It is equivalent to the diagram to the top right. If two or more capacitors are connected in parallel, the overall effect is that of a single (equivalent) capacitor having a total plate area equal to the sum of the plate areas of the individual capacitors. Thus for parallel capacitors the equivalent capacitance is the sum of the capacitances.

The bottom middle diagram shows two capacitors in series. It is equivalent to the diagram to the bottom right. If two or more capacitors are connected in series, the overall effect is that of a single (equivalent) capacitor having the sum total of the plate spacings of the individual capacitors. Thus for series capacitors the equivalent capacitor is less than the individual capacitors. If the capacitors are the same and equal $$C$$, the equivalent capacitance is $$C/2$$

For reference the diagram includes the relevant equations for capacitance based on the physical parameters ($$A$$, $$d$$, $$e$$) and electrical parameters ($$Q$$, $$V$$).

This is starting to make sense. But do you mind elaborating a teeny bit more on why the total charge for the series case is $$Q$$ not $$2Q$$?

The total charge on the equivalent series capacitance is $$Q/2$$ and not $$Q$$. There is less charge on the two capacitors in series across a voltage source than if one of the capacitors is connected to the same voltage source. This can be shown by either considering charge on each capacitor due to the voltage on each capacitor, or by considering the charge on the equivalent series capacitance.

The bottom left diagram shows one capacitor of capacitance $$C$$ connected to a voltage $$V$$. The charge on the capacitor is $$Q=CV$$ after it is fully charged as shown.

The bottom middle diagram shows two capacitors of the same capacitance $$C$$ in series across the same voltage source. The voltage across each is $$V/2$$. Since $$Q=CV$$ this means the charge on each will be $$Q=C\frac{V}{2}$$. However, as pointed out by @Kaz, the conductor and plates between the two capacitors don’t contribute to charge separation. To put it another way, the net charge on the plates and conductor between the capacitors is zero. This results in the charge on the equivalent capacitance equal to $$Q=C\frac{V}{2}$$ as shown on the bottom right diagram.

The same conclusion can be reached by considering that the equivalent capacitance of two equal capacitors in series is one half the capacitance of each, or $$C_{equiv}=\frac{C}{2}$$. Consequently the charge on the equivalent series capacitance is the same as the charge on each of the series capacitors, or $$\frac{C}{2}V$$ as shown on the bottom right diagram.

Hope this helps • Comments are not for extended discussion; this conversation has been moved to chat. – Chris Jun 23 '19 at 13:10

One way to look at it -- though perhaps more from an electronics than a physics perspectice -- is to not think of a capacitor as a thing that stores charge. Sine the entire component is electrically neutral when viewed from outside, the total amount of charge inside it is always the same; it just gets redistributed in ways that need not concern us at a higher level of abstraction.

In this view, a capacitor is a dependent voltage source, where the voltage at any point in time is proportional to the net amount of charge that has passed through the capacitor in its lifetime.

The capacitance measures how much charge we need to push through the capacitor to change its voltage by a given amount.

If we have two capacitors in series, any charge we push through the entire complex will pass through both capacitors at once, but the voltage we measure across it will be the sum of the individual capacitor voltages. So it takes less charge to create any desired change in total voltage -- that is, the capacitance is less.

Here's an explanation in terms of how charges are pushed around in a circuit.

In the case of one capacitor, the equation $$Q = CV$$ means that when a voltage $$V$$ is applied to a capacitor system with capacitance $$C$$, the amount of charge moved from one place to another is $$Q$$. See the below diagram. When a voltage $$V_1$$ is applied to a capacitor $$C_1$$, a charge of $$Q$$ is removed from one plate and deposited on the opposite plate.

Now, let's consider what happens with two capacitors in series: The source puts out a voltage $$V_1 = V$$ onto capacitors $$C_1$$ and $$C_2$$. Our goal is to determine the total amount of charge moved once the current drops to zero after connecting the voltage source.

The top plate of capacitor $$C_1$$ is at voltage $$+V$$ and the bottom plate of capacitor $$C_2$$ is at voltage $$0$$ (the absolute values don't matter, just their differences). In between, the bottom plate of capacitor $$C_1$$ must be at the same voltage $$V'$$ as the top plate of capacitor $$C_2$$ since they are connected by a wire. What is this voltage? It can't be $$V$$, since then there would be no voltage difference across $$C_1$$ and it wouldn't gain any charge--like it wasn't there. Similarly, it can't be zero since then $$C_2$$ would have no voltage across it. In the case that $$C_1 = C_2 = C,$$ we can conclude that $$V' = V/2$$ since identical components in series should have equal voltage drops. Thinking another way, the two capacitors will have the same charge after the circuit is closed since any charge moving off of $$C_1$$ must end up on $$C_2$$ and vice versa. Since they are identical capacitors with equal charges, they must have the same voltage drop across them.

So, how much charge is on each capacitor, we use the first equation to find that $$Q' = CV/2$$ since there is now half the voltage across the capacitor. The two capacitors are identical, so the second capacitor gets the same charge $$Q' = CV/2.$$

Finally, how much charge was moved? Charge can only move across wires, so if the top plate of the top capacitor gained a charge of $$Q/2$$, that charge must have come from the bottom plate of the bottom capacitor. Even though both capacitors gained a charge of $$Q/2$$ (where $$Q$$ is the charge moved in the first circuit), the charges moved from one capacitor to the other. So, the total charge moved was Q/2.

From all this, we find that the capacitance of two identical capacitors in series is half of a single capacitor because half the amount of charge is moved with the same voltage.

### It's simple (but not obvious):

1. Assume parallel-plate: $$C = \epsilon A/d$$

2. Assume infinitely thin plates

3. Insert an infinitely thin plate in the middle

4. Is C affected? No, because the plate is disconnected, and permittivity ($$\epsilon$$) is unaffected.

5. But: you now have 2 capacitors of distance $$d/2$$, each therefore with capacitance $$2C$$.

QED.

Aren't we in total acculumating a charge of $$2Q$$ over a potential difference of $$2V$$? Why just $$1Q$$?

So this is the crux of the issue here. The relevant charge here is $$Q$$, not $$2Q$$. So why is that?

It's because in a series circuit, all of the components see all of the charge. (Equivalently, all of the components see all of the current.) If I push 1 coulomb of charge through the entire circuit, that means I'm pushing 1 coulomb of charge through each capacitor.

It's a lot like how, if I have a pair of 2 doors in series, and I send 100 people through the doors, then each door has 100 people walking through it, not 50.

Please take a look at an analogy in a field of water pressure. Let introduce a device built with a piston in a tube connected with a spring. When pressure difference increases, the piston moves and pulls the spring until forces caused by pressure and spring would balance. If we consider pressure difference as an analogy of a voltage, then a volume of water that flows through the pipe would be the electric charge and a ratio of both of them will be analogy of the capacitance.

If we connect both such devices in paralel, then the amounts of water it can 'accumulate' under certain pressure will sum up. So the capacitance will be obviously the sum.

If we connect them in series, the amount of water that manage to flow through them must be the same in all devices in series. The pressure will spread equally between two springs which will cause pulling force to be lower and Δx to be lower. Eventually water volume per pressure will be lower.

Turning back to the field of electricity, it is worth note that electric charge in a link between capacitors will never change because we consider it as isolated. Instead, an electric potential will balance somewhere between potentials of voltage source.

Think of two simple capacitors each consisting of two plates each with area A separated by distance D. If they are connected in series the bottom plate of the upper capacitor is shorted to the top plate of the lower capacitor creating the equivalent of a single plate that is not really connected to anything else so it can be neglected. This results in the equivalent of a single capacitor with area A and plate separation 2D.

Since capacitance is inversely proportional to the plate separation the capacitance will be one half that of a single capacitor. This can be generalized to more than two capacitors. (note- no formulas used)

• What about 3 caps, seems that would only effect the two outer? – marshal craft Jun 22 '19 at 4:18
• Well, you'd presumably end up with two irrelevant plates, and a gap of 3D. It's a good analogy. – Sod Almighty Jun 22 '19 at 17:24
• +1 was going to post this answer if nobody else had. In parallel, the area adds. In series, the distances add, like one cap with the plates farther apart. – Peter Cordes Jun 22 '19 at 21:14

why the equivalent capacitance of capacitors in series is less than the any individual capacitor's capacitance?

To help build your intuition, place in series with another capacitor an ideal parallel plate capacitor (vacuum dielectric for simplicity), with plate spacing $$d$$. Now consider the limit as the distance $$d$$ goes to zero (assume zero initial charge).

Intuitively, the capacitance of the ideal parallel plate capacitor increases without bound and what's left in the limit is, effectively, an ideal short circuit when $$d = 0$$ (the two plates touch).

That is, the series combination of two capacitors has become a capacitor in series with an ideal short circuit. Clearly, the combined capacitance is just the capacitance of the remaining capacitor.

Now, hold the spacing $$d$$ constant and let the plate area $$A$$ go to zero instead. Intuitively, the capacitance of the ideal plate capacitor goes to zero and what's left in the limit is, effectively, an ideal open circuit.

That is, the series combination of two capacitors has become a capacitor in series with an ideal open circuit. Clearly, the combined capacitance is zero.

It follows that the total capacitance for two series capacitors with finite capacitance is less than the smaller of the two capacitances

Capacitance is charge per voltage. Two equal-valued capacitors in series containing the same charge will have the same charge available at the two outer capacitor plates as a single capacitor does, but the voltage will be double. So the capacitance is half.

• @EricDuminil, KVL. – The Photon Jun 21 '19 at 20:36

Can someone please explain, intuitively ...

First of all:

The definition $$C = \frac{Q}{U}$$ is a random definition made in the 18th or 19th century.

Back then, they could have defined the word $$\text {"capacitance"} = C^* = \frac{U}{Q}$$ instead. (Indeed this value is named "elastance" today. Thanks Alfred Centauri for your comment.)

If they had done this, the "capacitance" (which would be $$\frac {U}{Q}$$) of capacitors in series would increase and not decrease!

For this reason I doubt that it is possible to explain the phenomenon "intuitively" without at least referring to the formula $$C=\frac {Q}{U}$$.

I'd also like to give a generic answer that also applies to "real" capacitors not having plates...

... why the equivalent capacitance of capacitors in series is less than the any individual capacitor's capacitance?

First you should remember what you are talking about if you talk about the "capacitance of a series connection":

You are talking about the voltage measured over both ends of the series connection and the charge that was flowing into one end of the series connection.

You are not talking about the voltages measured inside the series connection and/or charges somewhere inside the series connection.

If some electrons are flowing into one end of a capacitor or one end of the series connection, the same amount of electrons will flow out of the capacitor or series connection at the other end. This amount of electrons is the "charge of the capacitor" $$Q$$.

In the series connection the electrons that flow out of the first capacitor will flow into the second capacitor. This means that if some charge $$Q$$ flows into one end of the series connection, all capacitors will be charged with a charge of $$Q$$.

Because we defined the charge that was flowing into one end of the series connection as "charge of the series connection", the "charge of the series connection" is only $$Q$$, not $$N\times Q$$ if there are $$N$$ capacitors in series that all have a charge of $$Q$$.

On the other hand the voltage $$U$$ describes the energy that is needed to transport an electron from one point in a circuit to another one. To transport some electron from one end of the series connection to the other end, we need the energy to transport the electron from one end of the first capacitor to the other end of the first capacitor. And we need energy to transport it from one end of the second capacitor to the other end of it. This means that the voltage over the series connection is the sum of the voltages of the capacitors.

Now we are at the point where we need the formula $$C=\frac{Q}{U}$$ because things would be completely different if the "capacitance" was defined as $$C^{*}=\frac{U}{Q}$$:

Because the voltages sum up but the charge of the series connection is equal to the charge of each single capacitor, the capacitance of the series is:

$$\displaystyle{C = \frac{Q}{\sum U_\text{capacitor}}}$$

This means that the numerator of the fraction $$\frac{Q}{U}$$ is the same for the single capacitor and the series connection, but the denominator is larger in the series connection.

• Interesting point. If we measured caps like we measure resistors and inductors, where a higher value meant higher reactance, then they would add linearly in series like resistance and (ideal) inductance. The capacitive reactance at a given AC frequency does add in series. – Peter Cordes Jun 22 '19 at 21:23
• Elastance – Alfred Centauri Jun 23 '19 at 11:39

When you say "intuitively", the answer depends on what you already consider intuitive. To you, are the analogous rules for resistors intuitive? (Resistors in series have a total resistance equal the to sum of the two individual resistances). Do you find the relationship between capacitance and impedance to be intuitive? ( Xc = 1 / (2 * pi * f * c) ).

If so, then the explanation is simple:

Capacitors behave exactly like resistors. The impedance of two capacitors in series is equal to the sum of the individual impedances of the two capacitors. Since the impedance is proportional to the inverse of the capacitance, the larger impedance of the series circuit means a smaller capacitance.

The voltage change across a capacitor, or any region of space, is the negative path integral of field dot path. So if two capacitors are placed in series, you have twice the distance for the path integral, compared to one capacitor by itself. Now, since the distance is doubled, to get a final voltage equal to the battery that charges it, the electric field only needs to be 1/2 the electric field of the single capacitor. Since the electric field is proportional to the charge density, you only need 1/2 the charge density that would be on the single capacitor, so C = (Qsingle/2)/V.

In the Electronics StackExchange, I gave this answer six years ago which has more detailed discussion and images.

Capacitors store energy by charge separation.

If we put two identical capacitors in series, effectively we have only two plates that store the separated charges, not four:

   C1    C2
---| |---| |---
^
- these two plates and conductor don't contribute
to charge separation.
`

Furthermore, the outer plates are twice as far apart as in a single capacitor. The increased separation cuts the capacitance in half.

The larger the gap, the smaller the capacitance. Putting multiple capacitors in series puts multiple gaps in series, thus making the gaps larger.

Another interpretation is that it it a voltage divider, and thus the charge induced is only corresponding to a fraction of the voltage.

If you have a capacitor and you put a charge on one of the plates, on the other plate an opposite charge gathers by induction; in order to mantain that configuration, you have to do a certain effort (i.e. apply a certain potential). The capacity is defined as the charge you can keep on the plates using a "budget" of $$1$$ Volt.

However, if you have a series of capacitors, when you charge the first plate all the others charge up with the same or opposite charge -by induction- in a sort of chain reaction: you can imagine that the effort (that is the potential) to keep all that charge in place is magnified. Then the charge you can afford to keep on a series of capacitors with your fixed budget of $$1$$ Volt is smaller; and so is its capacity.

Because it takes the "charge density" available from the "potential difference" and must spread across more surfaces or or capacitors. Key point is there isn't continuity. So you have a limited supply of charge density for each cap, where with just one you more. You can see its kind of a compound situation and so may be difficult to describe

• Maybe I should say charge carriers, that mediate the field. – marshal craft Jun 22 '19 at 4:07