 # Combination of Capacitors

## How Capacitors are connected?

Capacitors combination can be made in many ways. The combination is connected to a battery to apply a potential difference (V) and charge the plates (Q). We can define the equivalent capacitance of the combination between two points to be

$C=\frac{Q}{V}$

Two frequently used methods of combination are: Parallel combination and Series combination

## Parallel Combination of Capacitors

When capacitors are connected in parallel, the potential difference V across each is the same and the charge on C1, C2 is different i.e., Q1 and Q2. The total charge is Q given as:

$Q={{Q}_{1}}+{{Q}_{2}}$ $Q={{C}_{1}}V+{{C}_{2}}V$ $\frac{Q}{V}={{C}_{1}}+{{C}_{2}}$

Equivalent capacitance between a and b is:

C = C1 + C2

The charges on capacitors is given as:

• $Q1=\frac{{{C}_{1}}}{{{C}_{1}}+{{C}_{2}}}Q$
• $Q2=\frac{{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}Q$

In case of more than two capacitors, C = C1 + C2 + C3 + C4 + C5 + …………

### Watch this Video for More Reference ## Series Combination of Capacitors

When capacitors are connected in series, the magnitude of charge Q on each capacitor is same. 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

$V=\frac{Q}{{{C}_{1}}}+\frac{Q}{{{C}_{2}}}$

$\frac{V}{Q}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}$

The ratio Q/V is called as the equivalent capacitance C between point a and b.

The equivalent capacitance C is given by: $\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}$

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

${{V}_{1}}=\frac{{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}};\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{V}_{2}}=\frac{{{C}_{1}}}{{{C}_{1}}+{{C}_{2}}}V$

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

$\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}+\frac{1}{{{C}_{4}}}+……$

Important Points:

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

### Problems on Combination of Capacitors

Problem 1: Two capacitors of capacitance C1 = 6 μ F and C2 = 3 μ F are connected in series across a cell of emf 18 V. Calculate:

• The equivalent capacitance
• The potential difference across each capacitor
• The charge on each capacitor

Sol:

(a) $\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}$ $\Rightarrow \frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\frac{6\times 3}{6+3}=2\mu F$

(b) ${{V}_{1}}=\frac{C{}_{2}}{{{C}_{1}}+{{C}_{2}}}V=\frac{3}{6+3}\times 18=6\,volts$

${{V}_{2}}=\frac{C{}_{1}}{{{C}_{1}}+{{C}_{2}}}V=\frac{6}{6+3}\times 18=12\,volts$

(c) Q1 = Q2 = C1 V1 = C2 V2 = CV

Charge on each capacitor = Ceq V = 2μF x 18 volts = 36μC

In the above problem, note that the smallest capacitor has the largest potential difference across it.

Example 2: 3 capacitors are arranged as shown in the figure. Find the equivalent capacitance between A and B. Sol: All the 3 capacitors are in parallel C4 + C1 + C2 + C3 = 3C

Example 3: Find the equivalent capacitance between points A and B capacitance of each capacitor is 2 μF. Sol: In the system given, 1 and 3 are in parallel. 5 is connected between A and B. So, they can also be represented as follows.

1. As 1 and 3 are in parallel, their effective capacitance is 4μF
2. 4μF and 2μF are in series, their effective capacitance is 4/3μF
3. 4/3μF and 2 10/3μF are in parallel, their effective capacitance is 10/3μF
4. 10/3μF and 2μF are in series, their effective capacitance is 5/4μF
5. 5/4μF and 2μF are in parallel, their effective capacitance is 13/4μF

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