Neden kapasitörler seri olarak kapasitansı yitirir?

Şarj edilebilir pillerin aksine, kapasitörlerin seri olarak daha düşük kapasitansları vardır. Neden bu ve her kapağı ayrı olarak şarj edip seri halinde yerleştirdiğimde yine de daha düşük kapasitans olur mu?

Bunun cevabı kapasitansın ne olduğunu göz önüne almaktan gelir: kapasitörün üzerine bir voltaj (V) koyarsak depolayabileceğimiz dolum bombalarının (C) sayısıdır.

Etki 1: Kondansatörleri seri bağlarsak, kondansatörler arasında voltaj geliştirmeyi zorlaştırırız. Örneğin, seri olarak iki kapasitör 5V kaynağa bağlarsak, her kondansatör sadece yaklaşık 2,5V şarj edebilir. Tek başına bu etkiye göre, şarj (ve dolayısıyla kapasitans) aynı olmalıdır: seri olarak iki kapasitör bağlanır, her biri voltajın yarısına kadar şarj olur, ancak iki olduğundan kapasitemizin iki katına sahibiz: yani, kopma, doğru ? Yanlış!

Etki 2: İki kapasitörün yakın plakalarındaki yükler birbirini iptal eder. Sadece en dıştaki plakalar şarj taşır. Bu etki depolamayı yarıya indirir.

Aşağıdaki diyagramı göz önünde bulundurun. Sağdaki paralel dalda tek bir kapasitör yüklüdür. Şimdi seriye bir tane daha eklersek, soldaki dalı oluşturmak için hayal edin. Kapasitörler arasındaki bağlantı iletken olduğundan, iki plakayı aynı potansiyele getirerek -----, üst kapasitörün alt plakasındaki +++++yükler, alt kapasitörün üst plakasındaki yükleri yok edecektir .

Bu yüzden etkili bir şekilde sadece şarj deposu sağlayan iki plakamız var. Ancak, voltaj yarıya indirildi.

Bunu anlamanın bir başka yolu da, yüklenmekte olan iki plakanın birbirinden daha uzakta olmasıdır . Boş alanda, eğer plakaları birbirinden uzaklaştırırsak, alan kuvveti azaldığı için kapasite azalır. Seri kapasitörleri bağlayarak, plakları hemen hemen hareket ettiriyoruz. Elbette, kapasitörleri devre kartına daha yakın veya uzağa yerleştirebiliriz, ancak şimdi en üstteki plaka ve en alttaki plakanın arasına bir tane yerine iki boşluk var. Bu kapasitansı azaltır.

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.

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 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

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.

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.