Capacitors in Series and Parallel

Capacitor in Series and Parallel Formula

Let's first look at the formula for capacitors in parallel, this will become clearer later on why we choose parallel first.

Capacitors in Parallel

In the figure below, we see two parallel plate capacitors connected in parallel.

Fig. 2 - Parallel plate capacitors in a parallel orientation, connected by two wires.

In a parallel circuit, current is supplied to two components independently of one another by use of a junction. In order to calculate the total capacitance of this setup, we can use the following equation

\[ C_{\text{p}} = \sum_{\text{i}} C_{\text{i}} ,\]

where \(C_{\text{p}}\) is the total parallel capacitance measured in farads \(\mathrm{F}\), \(C_{\text{i}}\) is the individual capacitance of the capacitors also measured in farads \(\mathrm{F}\), and the sum sign \(\sum_{\text{i}}\) indicates that we add together the individual capacitances. From this equation, we can see the similarities between the parallel capacitance equation and the series resistor equation. An important thing to note about this equation is that the capacitance of a group of capacitors in parallel will always be greater than any of the involved capacitors, even the one with the greatest capacitance.

Capacitors in Series

Let's now introduce the series capacitance rule. We will see that it has the same form as that for the total resistance of a set of resistors connected in parallel. In the figure below, we see the same two capacitors \(C_1\) and \(C_2,\) now connected in series.

Fig.3 - The capacitors are placed in series, connected by one wire.

In this case, the elements are connected one after the other, and there is no junction. The equation for capacitors in series is

\[ \frac{1}{C_{\text{s}}} = \sum_{\text{i}} \frac{1}{C_{\text{i}}} ,\]

where \(C_{\text{s}}\) is the series capacitance measured in farads \(\mathrm{F}\), \(C_{\text{i}}\) is the individual capacitances measured in \(\mathrm{F}\), and again \(\sum_{\text{i}}\) denotes the sum across all the capacitances. As we can see, it is similar to the equation for parallel capacitors, except now we are adding the reciprocals of the individual capacitances to obtain the inverse of the total capacitance. The capacitance of a group of capacitors in series is always less than the capacitance of any of the capacitors involved, even the one with the lowest capacitance.

Capacitor in Series and Parallel Derivation

Firstly, let's look at the derivation for capacitors in parallel.

Capacitors in Parallel Derivation

Using the capacitor equation, we find that the first capacitor obeys the equation

\[ C_1 = \frac{Q_1}{V_1} ,\]

where \(C_1\) is the capacitance of the first capacitor, \(Q_1\) is the magnitude of the charge on the plates of the first capacitor, and \(V_1\) is the voltage across the first capacitor. Similarly, with the second capacitor, we find

\[ C_2 = \frac{Q_2}{V_2} ,\]

where \(C_2\) is the second capacitance, \(Q_2\) is the magnitude of the charge on the plates of the second capacitor, and \(V_2\) is the voltage across the second capacitor.

When we have electrical components in parallel with one another, we know that the voltage across them is equal and equivalent to the total voltage across the parallel section. Thus we can write that

\[ V_1 = V_2 = V_{\text{T}} .\]

Furthermore, we can rewrite the total voltage across the parallel section as

\[ V_{\text{T}} = \frac{Q_{\text{T}}}{C_{\text{T}}} .\]

The total charge \(Q_{\text{T}}\) is given by the addition of the charges across both parallel plate capacitors, this is because the current across two legs of a parallel circuit is shared. Thus we find that

\[ \begin{align} Q_\text{T} &= Q_1 + Q_2 \\ C_{\text{T}} V_{\text{T}} &= \left(C_1 V_1 \right) + \left( C_2 V_2 \right) \\ C_{\text{T}} &= \frac{1}{V_{\text{T}}} ( \left(C_1 V_{\text{T}} \right) + \left( C_2 V_{\text{T}} \right)) \\ C_{\text{T}} &= \frac{\bcancel{V_{\text{T}}}}{\bcancel{V_{\text{T}}}} (C_1 + C_2) \\ C_{\text{T}} &= C_1 + C_2, \end{align} \]

thus giving us our equation for capacitors in parallel.

Capacitors in Series Derivation

Similar to the derivation of capacitors in parallel, let's apply the capacitor equation to both the capacitors in series. This results in

\[ C_1 = \frac{Q_1}{V_1} ,\]

and

\[ C_2 = \frac{Q_2}{V_2} .\]

When electrical components are in series with one another, their total voltage is shared whilst the current across both components is equal. Since the current is equal, this also means that the charge across both the plates of the capacitors will be equal, allowing us to write

\[ Q_1 = Q_2 = Q_{\text{T}} .\]

We can then also write that the total voltage across the combination of capacitors as

\[ V_{\text{T}} = V_1 + V_2 .\]

Thus rearranging our individual capacitor equations and substituting this into the total voltage equation, we find that

\[ \begin{align}V_{\text{T}} &= V_1+V_2\\ V_{\text{T}} &= \frac{Q_1}{C_1} + \frac{Q_2}{C_2} \\ V_{\text{T}} &= \frac{Q_{\text{T}}}{C_1} + \frac{Q_{\text{T}}}{C_2} \\ \frac{V_{\text{T}}}{Q_{\text{T}}} &= \frac{1}{C_1} + \frac{1}{C_2} \\ \frac{1}{C_{\text{T}}} &= \frac{1}{C_1} + \frac{1}{C_2} , \end{align} \]

which gives us our series capacitor equation.

Capacitor in Series and Parallel Combination

Now that we have established the rules of capacitors in both series and parallel orientations, let's consider combining the two cases and determine how to calculate the total capacitance. Referring to the diagram below, we can see that capacitor \(C_1\) is connected in series whilst \(C_2\) and \(C_3\) are connected in parallel.

Fig. 4 - The capacitors are now combined in both series and parallel.

Energy Stored in Capacitors in Series and Parallel

Before we calculate the total energy stored in a configuration of capacitors, we first define the energy stored in a single capacitor as

\[ U_{\text{C}} = \frac{1}{2} \frac{Q^2}{C} ,\]

where \(U_{\text{C}}\) is the energy stored in a single capacitor measured in joules \(\mathrm{J}\), \(Q\) is the magnitude of the charge on the parallel plates measured in coulombs \(\mathrm{C}\), and \(C\) is the capacitance of a capacitor measured in farads \(\mathrm{F}\).

Energy Stored in Capacitors in Series

Similar to how we derived the total capacitance of capacitors in series, we derive the total energy of two capacitors in series. The first capacitor obeys the equation

\[ C_1 = \frac{Q_1}{V_1} ,\]

whilst the second capacitor obeys the equation

\[ C_2 = \frac{Q_2}{V_2} .\]

Again, since they are in series, the charge \(Q\) on both capacitors are equal, allowing us to write

\[ Q_1 = Q_2 = Q_{\text{T}} .\]

Thus the total energy is

\[ \begin{align} E_{\text{T}} &= E_1 + E_2 \\ E_{\text{T}} &= \frac{1}{2} \frac{Q_1^2}{C_1} + \frac{1}{2}\frac{Q_2^2}{C_2} \\ E_{\text{T}} &= \frac{1}{2} Q^2 \left( \frac{1}{C_1} + \frac{1}{C_2} \right) . \end{align} \]

Energy Stored in Capacitors in Parallel

In the case of capacitors in parallel, we have to define another energy capacitor equation that involves voltage. This is given by

\[ U_{\text{C}} = \frac{1}{2} CV^2 ,\]

where \(V\) is the voltage across the capacitor measured in volts \(\mathrm{V}\).

For this definition, our two capacitors are now in parallel, thus their individual voltages are equal, allowing us to write

\[ V_1 = V_2 = V_{\text{T}} .\]

Thus the total energy is

\[ \begin{align} E_{\text{T}} &= E_1 + E_2 \\ E_{\text{T}} &= \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2 \\ E_{\text{T}} &= \frac{1}{2} V^2 \left(C_1 + C_2 \right) . \end{align} \]

Properties of Capacitors in Series and Parallel

Let's recap some important properties of capacitors in series and parallel are the following.

1. The capacitance of a group of capacitors in series is always less than the capacitance of any of the capacitors involved, even the one with the lowest capacitance.

2. The capacitance of a group of capacitors in parallel will always be greater than any of the involved capacitors, even the one with the greatest capacitance.