Es ist die gesamte Kapazität von allen in Reihe (seriell) geschalteten \(n\) Kondensatoren. In einer Reihenschaltung von Kondensatoren addieren sich die Kapazitäten der Kondensatoren reziprok.

Die Kapazität ist eine charakteristische Größe des Kondensators und sagt aus, wie viele Ladungen auf den Kondensator gebracht werden müssen, um den Kondensator auf die Spannung \( 1 \, \text{V} \) aufzuladen.

Capacitances of individual series-connected capacitors.

Example: Two capacitors are connected in series. They have the capacitances \(C_1 = 100 \, \mu\mathrm{F}\) and \(C_2 = 300 \, \mu\mathrm{F}\). The total capacitance is thus:
\begin{align}
\frac{1}{C} &= \frac{1}{C_1} + \frac{1}{C_2} \\\\
&= \frac{1}{100 \, \mu\mathrm{F}} + \frac{1}{300 \, \mu\mathrm{F}} \\\\
&= \frac{1}{75 \, \mu\mathrm{F}}
\end{align}
Consequently, the total capacitance is the reciprocal of the resulting value: \( 75 \, \mu\mathrm{F} \).

Here we want to derive the total capacitance \(C\) of a circuit in which two capacitors are connected in series or parallel. One capacitor has the capacitance \(C_1\) and the other capacitor has the capacitance \(C_2\). The total capacitance is also called equivalent capacitance because we can replace individual capacitances \(C_1\) and \(C_2\) with \(C\) without changing the functionality of the circuit.

Total Capacitance of a Series Connection of Capacitors

Let us consider a circuit with applied voltage. For this we take two capacitors with the capacitances \(C_1\) and \(C_2\). We apply an voltage \( U \) to these two capacitors. In this way we have constructed a series of capacitors.

Charge amount on the capacitor plates:
Applied voltage causes a electric current \( \class{red}{I} \) to flow. Since the series circuit has no nodes at which the current could split, the same current \( \class{red}{I} \) flows through the two capacitors. The current is defined as charge \(\class{red}{Q}\) per time \(t\):
1
$$\class{red}{I} ~=~ \frac{\class{red}{Q}}{t}$$

At time \(t\), in both capacitors the amount of charge \(\class{red}{Q} ~=~ \class{red}{I} \, t \) is stored. Since the current through both capacitors is the same, the amount of charge \( \class{red}{Q} \) at time \(t\) is the same on both capacitors.

Voltage on the capacitors:
The applied voltage \( U \) is the total voltage which drops across both capacitors. It is the sum of the voltage \(U_1\) applied between the electrodes of the first capacitor and the voltage \(U_2\) applied between the electrodes of the second capacitor:
2
$$U ~=~ U_1 ~+~ U_2$$

We bring the capacitance into play by using the relationship between the charge and the voltage (\(\class{red}{Q} = C\, U\)). For the first and second capacitor we have the following equations:
3
$$\begin{align}\class{red}{Q} &~=~ C_1 \, U_1 \\\\
\class{red}{Q} &~=~ C_2 \, U_2\end{align}$$

The total capacitance \(C\) of the series circuit is related to the total voltage \( U \) in the same way as the individual capacitances in 3:
4
$$\class{red}{Q} ~=~ C \, U$$

These equations state that the charge \( \class{red}{Q} \) on the capacitor plates is proportional to the corresponding voltage between the capacitor plates, where the constant of proportionality is the capacitance. Rearrange both equations 3 and4 for the voltage:
5
$$\begin{align}U_1 &~=~ \frac{\class{red}{Q}}{C_1} \\\\
U_2 &~=~ \frac{\class{red}{Q}}{C_2} \\\\
U &~=~ \frac{\class{red}{Q}}{C}\end{align}$$

Now you can replace the voltages in 3 and 4 with those in 5:
6
$$\begin{align}U &~=~ U_1 ~+~ U_2 \\\\
\frac{\class{red}{Q}}{C} &~=~ \frac{\class{red}{Q}}{C_1} ~+~ \frac{\class{red}{Q}}{C_2}\end{align}$$

Divide both sides by the charge \(\class{red}{Q}\) to eliminate it:
7
$$\frac{1}{C} ~=~ \frac{1}{C_1} ~+~ \frac{1}{C_2}$$

This equation can be rearranged for \(C\):
8
$$C ~=~ \frac{C_1\, C_2}{C_1 ~+~ C_2}$$

If you have more than two capacitors, you can repeat all the steps above to derive the following formula:
9
$$\frac{1}{C} ~=~ \frac{1}{C_1} ~+~ \frac{1}{C_2} ~+~ ... ~+~ \frac{1}{C_n}$$

Here \(C_n\) denotes the capacitance of the \(n\)th capacitor.

Total Capacitance of a Parallel Connection of Capacitors

Let's consider a slightly different circuit. Again we take two capacitors with the capacitances \(C_1\) and \(C_2\). We apply an voltage \( U \) to these two capacitors, but this time as shown in the following Illustration. In this way, we have constructed a parallel circuit of capacitors.

Current through capacitor:
In a parallel circuit, the total current \( \class{red}{I} \) splits at the nodes to the capacitors. Now we cannot assume that the current through both capacitors is the same. Therefore, we denote the current through the first capacitor as \( \class{red}{I_1} \) and through the second capacitor as \( \class{red}{I_2} \). The total current must of course be the sum of the two individual currents because of charge conservation:
10
$$\class{red}{I} ~=~ \class{red}{I_1} ~+~ \class{red}{I_2}$$

Charge on the capacitor plates:
We can use the definition of current (as charge per time) in 2 to get an equation for the total charge \(\class{red}{Q}\):
11
$$\frac{\class{red}{Q}}{t} ~=~ \frac{\class{red}{Q_1}}{t} ~+~ \frac{\class{red}{Q_2}}{t}$$

Divide both sides by the time \(t\):
12
$$\class{red}{Q} ~=~ \class{red}{Q_1} ~+~ \class{red}{Q_2}$$

In contrast to a series circuit, the charge on the capacitors is different in a parallel circuit.

Now use the relationship \(\class{red}{Q} = C\, U\) between the charge and the voltage to bring the capacitance into play:
13
$$\begin{align}\class{red}{Q_1} &~=~ C_1 \, U \\\\
\class{red}{Q_2} &~=~ C_2 \, U \\\\
\class{red}{Q} &~=~ C \, U\end{align}

Replace the charges in 12 with those from 13:
14
$$C \, U ~=~ C_1 \, U ~+~ C_2 \, U$$

Divide both sides by the voltage \(U\) to eliminate it:
15
$$C ~=~ C_1 ~+~ C_2$$

If you have more than two capacitors, you can follow the same steps above to derive the following formula:
16
$$C ~=~ C_1 ~+~ C_2 ~+~ ... ~+~ C_n$$

Here \(C_n\) denotes the capacitance of the \(n\)th capacitor.

Exercise #1: Capacitor with and without dielectric

Consider a plate capacitor with the plate surface \(A\) and distance \(d\) between the electrodes. Half of the plate capacitor contains a dielectric with the relative permittivity \(\varepsilon_1\) and the other half contains a dielectric with the relative permittivity \(\varepsilon_2\).

What is the capacitance of the capacitor with the two dielectrics?

By what factor does the voltage change with dielectrics compared to the voltage without dielectrics?

By what factor does the electrical energy change with dielectrics compared to the energy without dielectrics?

Solution to the exercise #1.1

Since one half of the plate capacitor is filled with a dielectric and the other half with another dielectric, the problem can be considered as a parallel connection of two capacitors, each having a plate area \(A/2\) (because the dielectric fills only half of the capacitor).

In a parallel circuit of capacitors, the total capacitance is the sum of individual capacitances:
1
$$ C ~=~ C_1 + C_2 $$
Here \(C_1\) is the capacitance of one capacitor and \(C_2\) is the capacitance of the other capacitor, which still have to be determined. The capacitance of the first plate capacitor with relative permittivity \(\varepsilon_1\) and plate area \(\frac{A}{2}\) is:
2
$$ C_1 ~=~ \varepsilon_0 \, \varepsilon_1 \, \frac{A}{2d} $$

Similarly, the capacitance of the other capacitor with relative permittivity is \(\varepsilon_2\):
3
$$ C_2 ~=~ \varepsilon_0 \, \varepsilon_2 \, \frac{A}{2d} $$

The total capacitance is therefore the sum of Eq. 2 and Eq. 3:
4
$$ C ~=~ \varepsilon_0 \, \frac{A}{2d} \, ( \varepsilon_1 + \varepsilon_2 ) $$

Solution to the exercise #1.2

The voltage and charge are proportional:
5
$$ Q ~=~ C \, U $$

Assuming that the charge \(Q\) is kept constant, the voltage \( U_{\text v} \) on the capacitor with plate area \(A\) and with vacuum in between is given by:
6
\begin{align}
U_{\text v} &~=~ \frac{Q}{C} \\\\
&~=~ \frac{d \, Q}{\varepsilon_0 \, A}
\end{align}

We only used the formula for the capacitance of the plate capacitor with the relative permittivity = 1 and the area \(A\).

And the voltage \( U_{\text d} \) with the two dielectrics results by substituting the total capacitance 4 into Eq. 5:
7
\begin{align}
U_{\text d} &~=~ \frac{Q}{C} \\\\
&~=~ \frac{Q \, d}{\varepsilon_0 \, A} \, \frac{2}{\varepsilon_1 ~+~ \varepsilon_2}
\end{align}

The comparison of q>6 and 7 shows that the voltage at the plate capacitor with the two dielectrics differs by the following factor:
8
$$ \frac{ U_{\text v} }{ U_{\text d} } ~=~ \frac{2}{\varepsilon_1 ~+~ \varepsilon_2} $$

Solution to the exercise #1.3

The electrical energy \(W\) stored in the plate capacitor is given by:
9
$$ W ~=~ \frac{1}{2} \, C \, U^2 $$

Inserting the capacitance of the plate capacitor with vacuum between the plates (like in exercise b) gives the following energy:
10
$$ W_{\text v} ~=~ \frac{1}{2} \, \varepsilon_0 \, \frac{A}{d} \, U^2 $$

Substitute the capacitance 4 for a capacitor with the two dielectrics into equation 9:
11
$$ W_{\text d} ~=~ \frac{1}{2} \, \varepsilon_0 \, \frac{A}{d} \, \frac{\varepsilon_1 ~+~ \varepsilon_2}{2} \, U^2 $$

The comparison of 10 and 11 shows that the electric energy at the plate capacitor with the two dielectrics differs by the following factor:
12
$$ \frac{ W_{\text v} }{ W_{\text d} } ~=~ \frac{\varepsilon_1 ~+~ \varepsilon_2}{2} $$