The Combination of Capacitors has the same application as that of the combination of cells. How do we connect the cells when we need a \(6\,{\text{V}}\) D.C supply and have four Electric cells of \(1.5\,{\text{V}}\) D.C? We all would have done many science projects during our schooling when we connected many cells in a series combination to get a desired fixed D.C. voltage. Isn’t it?

Capacitors are fascinating devices. They allow us to store electrical energy inside a circuit by creating an electric field between two conductors. They are used in almost every type of electronic device and are vital to the function of many.

Capacitors are designed for providing fixed capacity. Then, how do we have the customized, effective capacitance that we need? We can achieve this goal by combining capacitors in different ways. Whenever a specific capacitance is needed, we can add a capacitor in series to reduce the effective capacitance of the circuit, or if we add a capacitor in parallel, it will increase the effective capacitance of the circuit.

The capacitance changes depending on the way capacitors are combined. This article will study two types of connections, simple and common, called series and parallel. We can easily calculate the total capacitance for them, and certainly, more complicated connections can also be coupled to combinations of series and parallel.

How are Capacitors Connected?

Capacitor combinations can be made in many ways. The combination is connected to a battery to apply a potential difference \(V\) and develop a charge \(Q\) on the plates. We can define the equivalent capacitance of the combination between two points to be:
\(C = \frac{Q}{V}\)
The frequently used methods of combination are:
1. Parallel combination
2. Series combination

The Parallel Combination of Capacitors

Capacitors are said to be in parallel when their two terminals are connected to each terminal of another capacitor. The voltage across all the capacitors that are connected in parallel is the same.
Consider three capacitors connected in parallel. Let the capacitance of each capacitor be \({C_1},\,{C_2}\) and \({C_3}\) their equivalent capacitance in parallel is \({C_P}.\)
A parallel combination of three capacitors, with one plate of each capacitor connected to one side of the circuit and the other plate connected to the other side, is illustrated in Figure (a).

Since the capacitors are connected in parallel, they all have the same voltage \(V\) across their plates. However, each capacitor may store a different charge in the parallel network. To find the equivalent capacitance \({C_P}\) of the parallel network as given in Figure (b), we note that the total charge \(Q\) stored by the network is the sum of all the individual charges:
\(Q = {Q_1} + {Q_2} + {Q_3}\)
We use the relation on the left-hand side of this equation \(Q = {C_P}V,\) which holds good for the entire network. On the right-hand side of the equation, for the three capacitors in the network, we use the relations:
\({Q_1} = {C_1}V,\,{Q_2} = {C_2}V\) and \({Q_3} = {C_3}V\)
In this way, we obtain \({C_P}V = {C_1}V + {C_2}V + {C_3}V.\)
This equation, when simplified, is the expression for the equivalent capacitance of the parallel network of three capacitors: \({C_P} = {C_1} + {C_2} + {C_3}\)
In general, for \(n\) capacitors,
\({C_P} = {C_1} + {C_2} + {C_3} + ….. + {C_n}\)
This expression is easily generalized to any number of capacitors connected in parallel in the network.

Series Combination of Capacitors

When one terminal of a capacitor is connected to the terminal of another capacitor end to end (like train coaches connected one after another), it is called the series combination of capacitors. In series, the same charge is developed on each capacitor.

Consider three capacitors connected in series, as shown in Figure (a). Let each capacitance be \({C_1},\,{C_2}\) and \({C_3}\) their equivalent capacitance be \({C_S}\) as shown in Figure (b).
As these capacitors are connected in series, the charge across each capacitor is the same \(Q.\) The potential difference of the battery \(V\) gets divided across each component as \({V_1},\,{V_2}\) and \({V_3}.\)

It can be noted that \(V = {V_1} + {V_2} + {V_3}\)
Using \(V = \frac{Q}{C},\)
\(\frac{Q}{{{C_S}}} = \frac{Q}{{{C_1}}} + \frac{Q}{{{C_2}}} + \frac{Q}{{{C_3}}}\)
Therefore Equivalent capacitance for series combination
\(\frac{1}{{{C_S}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}\)
In general, for \(n\) capacitors,
\(\frac{1}{{{C_S}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}} + …. + \frac{1}{{{C_n}}}\)

Differences between Series and Parallel Grouping of Capacitors

Series grouping	Parallel grouping
(1) The charge on each capacitor remains the same and equals the main charge supplied by the battery and \(V = {V_1} + {V_2} + {V_3}\)	The potential difference across each capacitor remains the same and equal to the applied potential difference and \(Q = {Q_1} + {Q_2} + {Q_3}\)
(2) Equivalent capacitance \(\frac{1}{{{C_{{\text{eq}}}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}\) or \({C_{{\text{eq}}}} = {\left( {C_1^{ – 1} + C_2^{ – 1} + C_3^{ – 1}} \right)^{ – 1}}\)	Equivalent capacitance \({C_{{\text{eq}}}} = {C_1} + {C_2} + {C_3}\)
(3) In a series combination, potential difference and energy distribute in the reverse ratio of capacitance, i.e., \(V\alpha \frac{1}{C}\) and \(U\alpha \frac{1}{C}.\)	In parallel, combination charge and energy distributes in the ratio of capacitance i.e. \(Q\alpha C\) and \(U\alpha C\)
(4) If two capacitors having capacitances \({C_1}\) and \({C_2}\) are connected in series, then \({C_{{\text{eq}}}} = \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} = \frac{{{\text{Multiplication}}}}{{{\text{Addition}}}}\) \({V_1} = \left( {\frac{{{C_1}}}{{{C_1} + {C_2}}}} \right).V\) and \({V_2} = \left( {\frac{{{C_2}}}{{{C_1} + {C_2}}}} \right).V\)	If two capacitors having capacitance \({{C_1}}\) and \({{C_2}}\) respectively are connected in parallel, then \({C_{{\text{eq}}}} = {C_1} + {C_2}\) \({Q_1} = \left( {\frac{{{C_1}}}{{{C_1} + {C_2}}}} \right).Q\) and \({Q_2} = \left( {\frac{{{C_2}}}{{{C_1} + {C_2}}}} \right).Q\)
(5) If \(n\) identical capacitors, each having capacitances, \(C\)are connected in series with supply voltage \(V,\) Equivalent capacitance \({C_{{\text{eq}}}} = \frac{C}{n}\) and the potential difference across each capacitor \(V’ = \frac{V}{n}.\)	If \(n\) identical capacitors are connected in parallel, Equivalent capacitance \({C_{{\text{eq}}}} = nC\) and Charge on each capacitor \(Q’ = \frac{Q}{n}\)

Solving Networks Containing Capacitors

To solve capacitive network for equivalent capacitance following guidelines should be followed.
Guideline 1. Identify the two points across which the equivalent capacitance is to be calculated.
Guideline 2. Connect (Imagine) a battery between these points.
Guideline 3. Solve the network from the point (reference point), which is farthest from the points we have to calculate the equivalent capacitance. (The point is likely to be not a node)

Simple circuits: Suppose equivalent capacitance is to be determined in the following networks between points \(A\) and \(B.\) The calculation is done as shown in the Figure below.

Circuit 1

Circuit 2

Circuits with extra wire: If there is no capacitor in any branch of a network, then every point of this branch will be at the same potential. Step by step process to find equivalent capacitance between any two points \(A\) and \(B\) is as follows:-

Circuit 1:

Circuit 2

Wheatstone bridge-based circuit: If five capacitors are arranged in a network as shown in the following Figure, the network is called a Wheatstone bridge type circuit. If it is balanced, then \(\frac{{{C_1}}}{{{C_2}}} = \frac{{{C_3}}}{{{C_4}}}\) hence \({C_5}\) is removed, and equivalent capacitance between \(A\) and \(B\) is as follows:-

Circuit 1

An infinite chain of capacitors: Here, we will study how to solve an infinite circuit of capacitors.

Circuit 1

Suppose the effective capacitance between \(A\) and \(B\) is \({C_R}.\) Since the network is infinite, even if we remove one pair of capacitors from the chain, the remaining network would still have infinite pair of capacitors, i.e., the effective capacitance between \(X\) and \(Y\) would also be \({C_R}.\)

Hence equivalent capacitance between \(A\) and \(B\)

\({C_{AB}} = \frac{{{C_1}\left( {{C_2} + {C_R}} \right)}}{{{C_1} + {C_2} + {C_R}}} = {C_R}\, \Rightarrow {C_{AB}} = \frac{{{C_2}}}{2}\left[ {\sqrt {\left( {1 + 4\frac{{{C_1}}}{{{C_2}}}} \right)} – 1} \right]\)

Note: 1. If \(n\) identical plates are arranged as shown below, the circuit constitutes \(\left( {n – 1} \right)\) capacitors in series. If each capacitor have capacitance, \(\frac{{{\varepsilon _0}A}}{d}\) and \({C_{{\text{eq}}}} = \frac{{{\varepsilon _0}A}}{{\left( {n – 1} \right)d}}\)

In this situation, except for two extreme plates, each plate is common to adjacent capacitors.
2. Suppose \(n\) identical plates are arranged such that an even number of plates are connected together, and an odd number of plates are connected. In that case, \(\left( {n – 1} \right)\) capacitors will be formed, and they will be in a parallel grouping.

Equivalent capacitance \(C’ = \left( {n – 1} \right)C,\) Where \(C = \)Capacitance of a capacitor\( = \frac{{{\varepsilon _0}A}}{d}\)

Combination of Capacitors – Solved Examples

Q.1. Three capacitors \(2\mu F,\,3\mu F\) and \(6\mu F\) are joined in series, and the combination is charged using a \(24\) volt battery. The potential difference between the plates of the \(6\mu F\) capacitor is
(a) \(4\) volts (b) \(6\) volts (c) \(8\) volts (d) \(10\) volts
Ans: (a)

The equivalent capacitance of the network is \(\frac{1}{{{C_{{\text{eq}}}}}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}\)
\({C_{{\text{eq}}}} = 1\mu F\)
Charge supplied by a battery \(Q = {C_{{\text{eq}}}}V \Rightarrow 1 \times 24 = 24\mu C\)
Hence potential difference across \(6\mu F\) \( = \frac{{24}}{6} = 4\) volt

Q.2. In the following network, the potential difference across capacitance of \(4.5\,\mu F\) is:

(a) \(8\,{\rm{V}}\) (b) \(4\,{\rm{V}}\) (c) \(2\,{\rm{V}}\) (d) \(6\,{\rm{V}}\)
Ans: (a)

Equivalent capacitance \({C_{{\text{eq}}}} = \frac{{9 \times 4.5}}{{9 + 4.5}} = 3\,\mu F\)
Charge supplied by battery \(Q = {C_{{\text{eq}}}}V = 3 \times 12 = 36\,\mu C\)
Hence potential difference across \(4.5\,\mu F = \frac{{36}}{{4.5}} = 8\,{\rm{V}}.\)

Q.3. Several capacitors of \(8\,\mu F,\,250\,{\rm{V}}\) rating are given. Find the minimum number of capacitors needed to get an arrangement equivalent rating \(16\,\mu F,\,1000\,{\rm{V}}.\)
Ans: (c) Let \(C = 8\,\mu F,\,C’ = 16\,\mu F\) and \(V = 250\,{\text{volt,}}\,V{\text{‘ = 1000}}\,{\text{volt}}\)
Suppose \(m\) rows of given capacitors are connected in parallel in which each row contains \(n\) capacitors then,
The potential difference across each capacitor is \(V = \frac{{V’}}{n}\)
Equivalent capacitance of the network is \(C’ = \frac{{mc}}{n}\)
On putting the values, we get \(n = 4\) and \(m = 8.\) Hence total capacitors\( = m \times n = 8 \times 4 = 32.\)
Short Trick: For such type of problem, number of capacitors \(n = \frac{{C’}}{C} \times {\left( {\frac{{V’}}{V}} \right)^2}.\) Here \(n = \frac{{16}}{8}{\left( {\frac{{1000}}{{250}}} \right)^2} = 32.\)

Q.4. A finite ladder is constructed by connecting several sections of \(2\,\mu F,\,4\,\mu F\) capacitor combinations, as shown in the Figure. A capacitor of capacitance \(C\) terminates it. What value \(C\) makes the ladder’s equivalent capacitance between the points \(A\) and \(B\) independent of the number of sections between them?

(a) \(4\,\mu F\) (b) \(2\,\mu F\) (c) \(18\,\mu F\) (d) \(6\,\mu F\)
Ans: (a) By using formula \(C = \frac{{{C_2}}}{C}\left( {\sqrt {1 + 4\left( {\frac{{{C_1}}}{{{C_2}}}} \right)} + 1} \right);\,{C_1} = 4\,\mu F,\,{C_2} = 2\,\mu F\)
We get \(C = 4\,\mu F.\)

Summary

1. Capacitors are components that are used to store and release electrical energy in electronics.
2. The effective capacitance of a system is controlled by the way the capacitors are combined.
3. Capacitors in series have the same charge on all of them.
4. Capacitors in parallel have the charge distributed across all of them.
5. We can control the effective capacitance by changing how capacitors are combined in a system.
6. Adding a capacitor in series will decrease the effective capacitance, and a capacitor in parallel will increase the effective capacitance.

FAQs on Combination of Capacitors

Q.1. What are the combinations of capacitors?
Ans: The frequently used methods of combination are Parallel combination and Series combination.

Q.2. Why is the capacitor connected in parallel?
Ans: Parallel combinations of capacitors can store more energy since the equivalent capacitance is the sum of individual capacitances of all capacitors involved.

Q.3. Is voltage the same in series capacitors?
Ans: When capacitors are connected in series, the voltages across each capacitor are generally not equal but depend on the capacitance values.

Q.4. Define a series combination of capacitors.
Ans: When one terminal of a capacitor is connected to the terminal of another capacitor end to end (like train coaches connected one after another), it is called a series combination of capacitors.

Q.5. Define parallel combination of capacitors.
Ans: Capacitors are connected in parallel when both of their terminals are connected to each terminal of another capacitor. 

Q.6. In a series combination of capacitors, which physical quantity remains constant?
Ans: In a series combination of capacitors, the charge \(Q\) developed on each capacitor remains the same.

Q.7. In a parallel combination of capacitors, which physical quantity remains constant?
Ans: In a parallel combination of capacitors, the potential difference \(V\) across each capacitor remains the same.