My name is Alexander FufaeV and here I will explain the following topic:

# Capacitors: Series and Parallel Connection

## Important Formula

What do the formula symbols mean?

## Total capacitance

Unit

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.

## Individual capacitance

Unit
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.