Derivation: Capacitance - Series and Parallel Connection of Capacitors

Table of contents

Total capacitance of a series connection of capacitors Here we derive a formula for the total capacitance (equivalent capacitance) when capacitors are connected in series.
Total capacitance of a parallel connection of capacitors Hier leiten wir eine Formel für die Gesamtkapazität (Ersatzkapazität) her, wenn Kondensatoren parallel geschaltet sind.

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

Series connection of two capacitors.

Let us consider a circuit with an alternating voltage. For this we take two capacitors with the capacitances $C_1$ and $C_2$. We apply an AC voltage $ U(t) $ to these two capacitors, as shown in Illustration 1. In this way we have constructed a series of capacitors.

Charge amount on the capacitor plates:
Applied AC voltage causes an alternating electric current $ \class{red}{I(t)} $, which charges and discharges the two capacitors. Since the series circuit has no nodes at which the current could split, the same current $ \class{red}{I}(t) $ flows through the two capacitors. The current is defined as charge $\class{red}{Q(t)}$ per time $t$:

$$ \begin{align} \class{red}{I(t)} ~=~ \frac{\class{red}{Q(t)}}{t} \end{align} $$

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

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

$$ \begin{align} U(t) ~=~ U_1(t) ~+~ U_2(t) \end{align} $$

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:

$$ \begin{align} \class{red}{Q(t)} &~=~ C_1 \, U_1(t) \\\\
\class{red}{Q(t)} &~=~ C_2 \, U_2(t) \end{align} $$

The total capacitance $C$ of the series circuit is related to the total voltage $ U(t) $ in the same way as the individual capacitances in 3:

$$ \begin{align} \class{red}{Q(t)} ~=~ C \, U(t) \end{align} $$

These equations state that the charge $ \class{red}{Q(t)} $ 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:

$$ \begin{align} U_1(t) &~=~ \frac{\class{red}{Q(t)}}{C_1} \\\\
U_2(t) &~=~ \frac{\class{red}{Q(t)}}{C_2} \\\\
U(t) &~=~ \frac{\class{red}{Q(t)}}{C} \end{align} $$

Now you can replace the voltages in 3 and 4 with those in 5:

$$ \begin{align} U(t) &~=~ U_1(t) ~+~ U_2(t) \\\\
\frac{\class{red}{Q(t)}}{C} &~=~ \frac{\class{red}{Q(t)}}{C_1} ~+~ \frac{\class{red}{Q(t)}}{C_2} \end{align} $$

Divide both sides by the charge $\class{red}{Q(t)}$ to eliminate it:

$$ \begin{align} \frac{1}{C} ~=~ \frac{1}{C_1} ~+~ \frac{1}{C_2} \end{align} $$

This equation can be rearranged for $C$:

$$ \begin{align} C ~=~ \frac{C_1\, C_2}{C_1 ~+~ C_2} \end{align} $$

If you have more than two capacitors, you can repeat all the steps above to derive the following formula:

Total capacitance of a series circuit

$$ \begin{align} \frac{1}{C} ~=~ \frac{1}{C_1} ~+~ \frac{1}{C_2} ~+~... ~+~ \frac{1}{C_n} \end{align} $$

Here $C_n$ denotes the capacitance of the $n$th capacitor.

Total capacitance of a parallel connection of capacitors

Parallel connection of two capacitors.

Let's consider a slightly different circuit. Again we take two capacitors with the capacitances $C_1$ and $C_2$. We apply an AC voltage $ U(t) $ to these two capacitors, but this time as shown in Illustration 2. In this way, we have constructed a parallel circuit of capacitors.

Current through capacitor:
In a parallel circuit, the total current $ \class{red}{I(t)} $ 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(t)} $ and through the second capacitor as $ \class{red}{I_2(t)} $. The total current must of course be the sum of the two individual currents because of charge conservation:

$$ \begin{align} \class{red}{I(t)} ~=~ \class{red}{I_1(t)} ~+~ \class{red}{I_2(t)} \end{align} $$

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 $\class{red}{Q(t)}$:

$$ \begin{align} \frac{\class{red}{Q(t)}}{t} ~=~ \frac{\class{red}{Q_1(t)}}{t} ~+~ \frac{\class{red}{Q_2(t)}}{t} \end{align} $$

Divide both sides by the time $t$:

$$ \begin{align} \class{red}{Q(t)} ~=~ \class{red}{Q_1(t)} ~+~ \class{red}{Q_2(t)} \end{align} $$

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:

$$ \begin{align} \class{red}{Q_1(t)} &~=~ C_1 \, U(t) \\\\
\class{red}{Q_2(t)} &~=~ C_2 \, U(t) \\\\
\class{red}{Q(t)} &~=~ C \, U(t) \end{align} $$

Replace the charges in 12 with those from 13:

$$ \begin{align} C \, U(t) ~=~ C_1 \, U(t) ~+~ C_2 \, U(t) \end{align} $$

Divide both sides by the voltage $U(t)$ to eliminate it:

$$ \begin{align} C ~=~ C_1 ~+~ C_2 \end{align} $$

If you have more than two capacitors, you can follow the same steps above to derive the following formula:

Total capacitance of a parallel circuit

$$ \begin{align} C ~=~ C_1 ~+~ C_2 ~+~ ...~+~ C_n \end{align} $$

Here $C_n$ denotes the capacitance of the $n$th capacitor.

Next, let's look at how to derive the total inductance of coils connected in parallel and in series.