Why do capacitors lose capacitance in series?

Question

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?

Answer 1

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 arefarther 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.

Answer 2

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:

Answer 3

你似乎令人困惑的电容和电池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.

Answer 4

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}\$

Thisdefinesthe 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.

Answer 5

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.

Answer 6

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.

Answer 7

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

Etc. for additional capacitors.

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.

Answer 8

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.

Answer 9

"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"

Coming to your question about capacitors in series

Now the opposite happens in case of two capacitors connected in series. The second plate of the first capacitor will lose electrons but the electrons will get accumulated on the top plate of the second capacitor. These electrons will not go anywhere. So the first plates electrons which is connected to the battery will have enough charges to reach the batteries potential level as they are still evenly spread enough to develop a potential. Hence that plate will require less charges to built to the batteries potential than it would require if it was not in series with the second one. The electrons that get accumulated on the top plate of the second capacitors in series has an electric field which effects the amount of charges that get deposited on the first plate.

The result is less charges and hence not the complete use of the capacitors space. Thus we can say that capacitance has decreased. Basically capacitance is the same but the charges required to reach the batteries potential are less, which is as good as saying less capacitance.

Coming to why voltage will reduce is because the opposite plate has charges as well and hence the difference in the opposite plate and the plate connected to battery is less. Less difference means less potential.