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
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:
Equivalent capacitance between a and b is:
C = C1 + C2
The charges on capacitors is given as:
In case of more than two capacitors, C = C1 + C2 + C3 + C4 + C5 + …………
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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
The ratio Q/V is called as the equivalent capacitance C between point a and b.
The equivalent capacitance C is given by:
The potential difference across C1 and C2 is V1 and V2 respectively, given as follows:
In case of more than two capacitors, the relation is:
- 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
(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.
- As 1 and 3 are in parallel, their effective capacitance is 4μF
- 4μF and 2μF are in series, their effective capacitance is 4/3μF
- 4/3μF and 2 10/3μF are in parallel, their effective capacitance is 10/3μF
- 10/3μF and 2μF are in series, their effective capacitance is 5/4μF
- 5/4μF and 2μF are in parallel, their effective capacitance is 13/4μF
Therefore the equivalent capacitance of the given system is 13/4μF.