Combination of Capacitors

Table of Contents:

How Capacitors are Connected?

We are also interested to calculate the amount of charge stored in a system where two or more capacitors are connected. The 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.

Parallel combination of Capacitors

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.

Series Combination of Capacitors

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.

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