What will be the total capacitance if capacitors are connected in parallel combination

Greater than the individual capacitance values

Less than the individual capacitance values

Equal to the individual capacitance values

What will happen when three different capacitors are connected in series

All will consist of the same potential

Combination of Capacitors - Parallel and Series Combination, Examples

How Are Capacitors 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

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

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

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Parallel Combination of Capacitors

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

Parallel combination of Capacitors

The total charge in Q is given as:

\(\begin{array}{l}Q={{Q}_{1}}+{{Q}_{2}}\end{array} \)

\(\begin{array}{l}Q={{C}_{1}}V+{{C}_{2}}V\end{array} \)

\(\begin{array}{l}\frac{Q}{V}={{C}_{1}}+{{C}_{2}}\end{array} \)

The equivalent capacitance between a and b is:

C = C₁ + C₂

The charges on capacitors are given as:

\(\begin{array}{l}Q1=\frac{{{C}_{1}}}{{{C}_{1}}+{{C}_{2}}}Q\end{array} \)
\(\begin{array}{l}Q2=\frac{{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}Q\end{array} \)

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

Series Combination of Capacitors

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

\(\begin{array}{l}V=\frac{Q}{{{C}_{1}}}+\frac{Q}{{{C}_{2}}}\end{array} \)

\(\begin{array}{l}\frac{V}{Q}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}\end{array} \)

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

The equivalent capacitance C is given by:

\(\begin{array}{l}\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}\end{array} \)

The potential difference across C₁ and C₂ is V₁ and V₂ respectively, is given 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 the case of more than two capacitors, the relation is:

\(\begin{array}{l}\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}+\frac{1}{{{C}_{4}}}+……\end{array} \)

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

Watch These Videos for Reference

Parallel Combination of Capacitor

Capacitors in Series

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:

(a) The equivalent capacitance

(b) The potential difference across each capacitor

Sol:

(a)

\(\begin{array}{l}\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}\end{array} \)

\(\begin{array}{l}\Rightarrow C = \frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\frac{6\times 3}{6+3}=2\mu F\end{array} \)

(b)

Q = C_eqV

Substituting the values, we get

Q = 2 μF × 18 V = 36 μ C

V₁ = Q/C₁ = 36 μ C/ 6 μ F = 6 V

V₂ = Q/C₂ = 36 μ C/ 3 μ F = 12 V

(c) When capacitors are connected in series, the magnitude of charge Q on each capacitor is the same. The charge on each capacitor will equal the charge supplied by the battery. Thus, each capacitor will have a charge of 36 μC.

Example 2: Find the equivalent capacitance between points A and B. The capacitance of each capacitor is 2 μF.

Problem 2 on Combination of Capacitors