# Why do capacitors lose capacitance in series?

Unlike rechargeable batteries, capacitors have a lower capacitance in series. Why is this and if I charge each cap separately and then put them in series, will it still be a lower capacitance?

The answer to this comes from considering what is capacitance: it is the number of coulombs (C) of charge that we can store if we put a voltage (V) across the capacitor.

Effect 1: If we connect capacitors in series, we are making it harder to develop a voltage across the capacitors. For instance if we connect two capacitors in series to a 5V source, then each capacitor can only charge to about 2.5V. According to this effect alone, the charge (and thus capacitance) should be the same: we connect two capacitors in series, each one charges to just half the voltage, but we have twice the capacity since there are two: so break even, right? Wrong!

Effect 2: The charges on the near plates of the two capacitors cancel each other. Only the outer-most plates carry charge. This effect cuts the storage in half.

Consider the following diagram. In the parallel branch on the right, we have a single capacitor which is charged. Now imagine that if we add another one in series, to form the branch on the left. Since the connection between the capacitors is conductive, bringing the two plates to the same potential, the ----- charges on the bottom plate of the top capacitor will annihilate the +++++ charges on the top plate of the bottom capacitor.

So effectively we just have two plates providing the charge storage. Yet, the voltage has been cut in half.

Another way to understand this is that the two plates being charged are farther apart. In free space, if we move plates farther apart, the capacitance is reduced, because the field strength is reduced. By connecting capacitors in series, we are virtually moving plates apart. Of course we can place the capacitors closer or farther on the circuit board, but we have now have two gaps instead of one between the top-most plate and the bottom-most plate. This reduces capacitance.

• Instead of thinking of capacitors in terms of charged plates, I like to think of them as devices that build up voltage as charge is pushed through them. When two caps are in series, every coulomb of charge that goes through one goes through all, and the amount of voltage that builds up with each coulomb will be equal to the sum of the voltage that builds-up in the caps. Thus, the number of coulombs that can be pushed through for each additional volt will be reduced. – supercat May 7 '13 at 4:44
• @supercat Charges are not pushed through the capacitors. Electrons are either added or removed from the plates through an external circuit. Electrons collecting on the bottom of the top plate push away electrons on the bottom plate, and vice versa. With two capacitors in series, the total number of electrons in the middle stays constant. The electrons redistribute themselves according to the voltage applied across the elements. – Juan May 7 '13 at 19:47
• @Juan: I know that the electrons that enter one plate are not the same electrons that leave the other, but each electron which enters a plate will push an electron out the other, and every electron that leaves a plate will draw an electron into the other. If one views a capacitor as a black box, it will behave as though electrons move through it. Pushing 0.000001 coulombs into one leg of a 1uF cap while pulling 0.000001 coulombs out the other is many orders of magnitude easier than pushing electrons in without taking any out, or vice versa. – supercat May 7 '13 at 20:04
• Coulomb-pushing doesn't adequately explain it. A coulomb cannot be in two devices at once. So if we regard it like that, what happens in the end that we have half the overall capacitance, and that is further divided between the two devices that are at half voltage, so each one holds a quarter of the charge. – Kaz May 7 '13 at 21:55
• When you say the inner charges 'cancel each other out,' do you mean that the + & - charges spread themselves evenly out between the two inner plates? Why wouldn't they be separated and pulled apart to each outer plate? – T3db0t Feb 25 '18 at 16:17

The formula for capacitance is defined as:

$C = \epsilon_r \epsilon_0 \frac {A}{d}$

where

$C$ is the capacitance;
$A$ is the area of overlap of the two plates;
$\epsilon_r$ is the relative static permittivity (sometimes called the dielectric constant) of the material between the plates (for a vacuum, $\epsilon_r = 1$);
$\epsilon_0$ is the electric constant ($\epsilon_0 \approx 8.854 \times 10^{−12} \text{F m}^{–1}$); and
$d$ is the separation between the plates.

When you place multiple capacitors in series, you are effectively increasing its plate separation. As d goes up, C goes down.

This picture illustrates the equation, assuming $\epsilon$ and A remain constant throughout, and the distance of the plates in the series-connected capacitors just adds up:

You seem to be confusing capacitance and battery capacity. These concepts are somewhat related, so that is understandable.

Battery capacity is how much charge your battery can provide when fully charged until it discharges completely. When a battery is fully charged, its voltage will be high, and this value will remain somewhat stable until its charge is almost over:

If you place two identical batteries in series, the current will go through two batteries instead of one. That will be equivalent to a battery with double the voltage and same capacity as each one of the originals.

Capacitance, however, is not a measure of maximum charge: it measures the charge/voltage ratio in a component. A 2F capacitor will show 1V across it's terminals when charged with 2C. This makes capacity and capacitance uncomparable, since you can always (assuming a undestructible capacitor) put more charge in a capacitor by increasing it's voltage. The maximum charge you can actually get from a capacitor is C*V, where V is the maximum voltage at which you can charge the capacitor.

So when capacitors are building up charge, their voltage is constantly increasing, while in batteries it remains relatively stable. In a system of two identical capacitors in series, then, current will make both capacitors build up voltage. The result is a greater total voltage and, by definition (C = Q/V), a smaller capacitance for the system. However, that does not affect the total charge that can go through the system, as this smaller capacitance can be charged to a higher voltage, since each capacitor only "takes" half the voltage.

• +1 "it measures the charge/voltage ratio in a component." By that definition, two batteries in series would also have only half the capacitance of one. Actually, I'd rather say capacitance measures the derivative of the charge w.r.t. voltage, which means that an idealised battery has always infinite capacitance – which doesn't change if you put two in series (or parallel, for that matter). — Capaciᴛʏ, on the other hand, is merely total charge. That stays the same for serial, doubles for parallel batteries as well as capacitors. – leftaroundabout May 7 '13 at 14:21

From a different perspective than any of the other answers (at the time of my writing this), consider the problem in the phasor domain. Recall first, the fundamental time domain relationship:

$i_C = C \dfrac{dv_C}{dt}$

This defines the ideal capacitor circuit element.

Now, recall that a time derivative becomes multiplication by the complex frequency in the phasor domain, thus:

$\vec I_C = j \omega C \ \vec V_C$

Series connected components have identical currents so, for two series connected capacitors:

$\vec V_{C_{eq}} = \vec V_{C_1} + \vec V_{C_2} = \vec I \dfrac{1}{j \omega C_1} + \vec I \dfrac{1}{j \omega C_2} = \dfrac{\vec I}{j \omega} (\dfrac{1}{C_1} + \dfrac{1}{C_2}) = \vec I \dfrac{1}{j \omega C_{eq}}$

Where

$C_{eq} = (C_1 || C_2)$

So, for series capacitors, capacitance "combines" like the resistance of parallel resistors, i.e., the equivalent capacitance of two series capacitors is less than the smallest individual capacitance.

I think you almost answered your own question. Imagine two parallel plate capacitors each carrying charge Q and charged to a voltage V. Now, when you connect them in series, the voltage across the combination is 2V but the total charge is Q (the charges on the sides connected together cancel out). Since capacitance is the ratio of Q and V, it is halved.

• If the charge on one side of each plate was neutralized, then I would have thought that the voltage across each plate would be halved, since half the charge is gone and V ∝ q. I might attempt an answer in the same vein as yours. – Elliot Aug 26 '15 at 19:51

If you attach two capacitors in series, with the bottom plate of the second attached to ground: $$C_1 (V_1 -V_2) = Q_1\\ C_2 (V_2) = Q_2$$

If you solve these equations, you get: $$V_1 = \frac{Q_1}{C_1} + \frac{Q_2}{C_2}$$ The net charge where the capacitors connect (bottom plate, top plate) is : $$-Q_1 + Q_2 = 0\\ Q_1 = Q_2$$

The equivalent capacitance is then: $$C_{eq} = \frac{1}{\frac{1}{C_1}+\frac{1}{C_2}}$$ and so it looks like a capacitor $$C_{eq}V_1 = Q_1$$

If you charge both capacitors before connecting them: $$Q_1 \neq Q_2$$ and you can find the voltage across each of them using the first 2 equations.

If you assume that: $$Q_1-Q_2 = Q_0$$ where $$Q_0$$ is the excess charge when putting the charged capacitors in series, then the equation is: $$V_1 = \frac{Q_1}{C_{eq}} - \frac{Q_0}{C_2}$$ so that it now looks like a capacitor with a fixed charge. It will still kind of look like a capacitor, but the voltage will be offset.

Skyler,

I'd love to hear someone else chime in on this. I don't have a good explanation, but I believe efox29's explanation is inadequate (if not wholely incorrect). If it was true, then 'd' would be a hard-known constant that could be computed and used for capacitors of equal size in series. It doesn't matter how far apart you put the capacitors; what matters is the topology of the circuit (the mere fact that they are in series). This holds true, of course, assuming the inductance and capacity of wire connecting them and environment factors are all neglible. The formula for series capacitance is the reciprocal sum of the reciprocal values of the capacitors. Such as this:

Known values C1, C2, and C3 Series total capacitance = C 1/C = 1/C1 + 1/C2 + 1/C3

efox29's explanation is probably what some folks teach in school, but I think it fails to properly explain the mechanics of what's actually happening.

As far as charging them first and putting in series, just do an experiment yourself. You'll retain and understand the information 4x better if you just test it. To get an idea of their capacity, charge them up and discharge them into another capacitor of known value and measure the voltage of the newly charged capacitor. You can compare that voltage to the measurements from different configurations to find out how things are actually behaving. Then, you'll understand what math formulas work and why.

• I don't know what 'standard' values for Er, A and d are, but lets just use the following. Er=2.6, E0=8.85e-12, A=1 and d = 1. If we use these values, C = 2.30e-11 Farads. If you use the series capacitance equation for two 2.30e-11 Farad capacitors, you get 1.15e-11 Farads (half of the capacitance as expended). All is good. If you use the equation in what I presented, and change d=2, you also get 1.15e-11 Farads. Ie. running caps in series, is the same as increasing their plate separation. – efox29 May 7 '13 at 5:42
• I agree with @efox29 - his explanation is perfectly sound – Andy aka May 7 '13 at 7:55
• Show how efox's explanation holds for two different capacitors – Scott Seidman Jun 2 '13 at 14:57
• @ScottSeidman, first observe that an equivalent capacitors can be made with uniform area (say 1 square metre) and dieletric (say a vacuum), by varying the plate separation. Perform these substitutions, and then sum the plate separations for the equivalent single capacitor. – sh1 Jun 26 '13 at 10:14

I think a lot of the explanations here are almost too detailed, in an ELI5 style:

The charge stored when capacitors are in series doesn't actually change, if you take two capacitors charged in parallel and connect them in series they don't suddenly hold less charge, they'll output the same current as before but at twice the voltage.

The "Capacitance" of the new capacitor created by the series connection is lower due to the equation for capacitance involving more than just the charge.

• Charge is $Q$ and the unit is the coulombs (C) Capacitance is $C$ (not F) and the unit is the farad (F). – Transistor Aug 17 '18 at 17:52
• I believe Kaz and efox do a decent job. Your answer is not informative, the punctuation is terrible, and you mix up variables (Q, C) with units (C, F). Reconsider answering an old question with many (and much better) existing answers. – calcium3000 Aug 17 '18 at 18:14
• I appreciate your correction on the units however I feel the overlapping use of C is confusing for those arriving here just looking for a simple answer so I have edited my reply to remove the units. They do a decent job for those that want to understand the equations, for those who don't fully understand what Capacitance represents or like myself use the units and names fairly interchangeably I feel a simple explanation added value, I'm not sure what your problem is with my punctuation, a couple of missing full stops? – Triff Aug 18 '18 at 18:16
• If anything Yuriy's answer is probably what mine should have been but I didn't see it till now as it's lost between the other posts, – Triff Aug 18 '18 at 18:23

"In working with electricity, it is sometimes convenient to try to place as much charge within a body as possible, with as little effort as possible. Suppose you have a metal plate, insulated in such a way that any electric charge added to it would remain. If you touch the plate with a negatively-charged rod, electrons will how into the metal plate and give it a negative charge.

You can continue this process as long as you can maintain a potential difference between rod and plate--that is, as long as you can keep the rod, by protracted rubbing, more negatively charged than the plate. Eventually, however, you will increase the negative charge of the plate to such a level that no amount of rubbing will make the rod more negatively charged than that. The potential difference between rod and plate will then be zero, and a charge will no longer spontaneously move.

Suppose, however, you next bring a second metal plate, one that is positively charged, down over the first and parallel to it, but not touching. The electrons in the first plate are pulled toward the positively-charged second plate and crowd into the surface facing the positive plate. (The electrons crowding into that surface are now closer together than they, were before, when they had been spread out evenly. They are "condensed" so to speak, and so this device of two flat plates held parallel and a short distance apart, may be called a condenser.)

With the electrons in the negative plate crowding into the surface facing the positive plate, the opposite surface has fewer electrons and a lower potential. There is once again a potential difference between the negatively charged rod and that surface of the first plate, which is away from the second plate. Electrons can once more pass from the rod into the plate, and the total charge on the plate can be built up considerably higher than would have been possible in the absence of the second plate."

The above explanation is from "Understanding Physics, Volume Two - Issac Asimov"