How are Capacitors connected?

The equivalent capacitance of a combination of capacitors connected to a battery to apply a potential difference (V) and charge the plates (Q) can be defined as the capacitance between two points.

$\begin{array}{l}C=\frac{Q}{V}\end{array}$

Two frequently used methods of combination are:

• Parallel combination
• Series combination

Related Topics

• Capacitor, Types and Capacitance
• Energy Stored in a Capacitor

Parallel Combination of Capacitors

The potential difference V across each capacitor when they are connected in parallel is the same, however, the charge on C₁ and C₂ (Q₁ and Q₂) is different.

The total charge is Q:

$$\begin{array}{l}\frac{Q_1+Q_2}{V}=C_1+C_2\end{array}$$

Equivalent capacitance between a and b is:

C = C1 + C2

The charge on capacitors is given as:

$\begin{array}{l}Q1=\frac{C_1}{C_1+C_2}Q\end{array}$

$\begin{array}{l}Q2=\frac{C_2}{C_1+C_2}Q\end{array}$

$C=\sum _{i=1}^n{C}_i$

Series Combination of Capacitors

The magnitude of charge Q on each capacitor is the same when connected in series, however, the potential difference across C₁ and C₂ is different, i.e., V₁ and V₂.

Q = C1 V1 = C2 V2

The total potential difference across combination is:

V = V1 + V2

$\begin{array}{l}V=\frac{Q}{C_1}+\frac{Q}{C_2}\end{array}$

$\begin{array}{l}\frac{V}{Q}=\frac{1}{C_1}+\frac{1}{C_2}\end{array}$

The ratio of Q to V, denoted as C, is referred to as the equivalent capacitance between point a and b.

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}\Rightarrow C=\frac{C_1{C}_2}{C_1+C_2}$

The potential difference across C1 and C2 is V1 and V2 respectively, as follows:

$\begin{array}{l}{V}_1=\frac{C_2}{C_1+C_2};V_2=\frac{C_1}{C_1+C_2}V\end{array}$

In case of more than two capacitors, the relation is:

$\begin{array}{l}\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\frac{1}{C_4}+\dots \dots \end{array}$

Important Points:

If N identical capacitors of capacitance C are connected in series, then the effective capacitance = C/N

If N identical capacitors of capacitance C are connected in parallel, then the effective capacitance is equal to C multiplied by N

Problems on Combination of Capacitors

Problem 1: Calculate the potential difference across each capacitor when two capacitors of capacitance C1 = 6 μF and C2 = 3 μF are connected in series across a cell of emf 18 V.

• (a) The Equivalent Capacitance
• (b) The potential difference across each capacitor
• (c) The Charge on Each Capacitor

Sol:

(a)

$\begin{array}{r}C=\frac{C_1{C}_2}{C_1+C_2}=\frac{6\times 3}{6+3}=2\mu F\end{array}$

(b)

Q = CeqV

Substituting the values, we obtain

Q = 36 μC = 2 μF × 18 V

V1 = Q/C1 = 36 μC/ 6 μF = 6V

V2 = Q/C2 = 36 μC/ 3 μF = 12 V

The magnitude of charge Q on each capacitor when connected in series will be the same and will equal 36 μC.

Example 2: Find the equivalent capacitance between points A and B.

Answer: The equivalent capacitance between points A and B is 4 μF.

Sol: In the system given, 1 and 3 are in parallel and 5 is connected between A and B. They can be represented as follows:

• 1. As 1 and 3 are in parallel, their effective capacitance is 4μF
• 2. The effective capacitance of a series circuit with 2.4μF and 2μF is 4/3μF.
• 3. The effective capacitance of 3.4/3μF and 2μF in parallel is 10/3μF.
• 4. The effective capacitance of a series circuit of 10/3μF and 2μF is 5/4μF.
• 5. The combined capacitance of 5/4μF and 2μF in parallel is 13/4μF

Therefore, the equivalent capacitance of the given system is 13/4 μF.