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Capacitor networks

In practical circuits capacitors are often joined together. We will consider the cases of two capacitors, first in parallel and then in series.

(a) Capacitors in parallel

Consider two capacitors connected in parallel as shown in Figure 1

The potential across both capacitors is the same (V) and let the charges on the capacitors be Q1 and Q2 respectively.

Now Q = CV, and so Q = C1V and Q = C2V. But the total charge stored Q = Q1 + Q2, therefore

Q = Q1 + Q2 = V(C1 + C2)

Capacitors in parallel:    C = C1 + C2

where C is the capacitance of the combination.

(b) Capacitors in series

In this case the capacitors are connected as shown in Figure 2.

The charge stored by each capacitor is the same. If V1 and V2 are the potentials across C1 and C2 respectively then:
V1 = Q/C1 and V2 = Q/C2. Therefore: V = V1 + V2 = Q(1/C1 + 1/C2)

Capacitors in series:    1/C = 1/C1 + 1/C2

where C is the capacitance of the combination.

(Notice that they are the "reverse" of the formulae for two resistors in series and parallel)

The charge distribution on series and parallel capacitors

When two capacitors are joined tighter in a circuit and then connected to a voltage supply charge will move onto the plates. The actual distribution of charge for a series and parallel circuit is shown in Figure 3(a) and (b).

Joining two charged capacitors

If two capacitors are joined together as shown in Figure 4 then:
(a) there is no change in the total charge stored by the system;
(b) the potential across the two capacitors becomes equal
(c) the combined capacitance of the two capacitors in parallel becomes

C = C1 + C2

There is usually a loss in energy when the two capacitors are joined; this is because unless the potential differences across them are equal, charge will flow to equalise this difference.
The flow of charge results in heating in the connecting wires and a consequent loss of energy.

Breakdown potential for a capacitor

Every capacitor has a working voltage, this is the maximum potential that should be applied between the plates. Any more than this and the dielectric material between the plates will break down and become conducting and the capacitor will be destroyed, usually resulting in a small bang as the material breaks down.

Example problems
1. Calculate the capacitance of the capacitor formed by joining:
(a) two 100 mF capacitors in series
(b) two 100 mF capacitors in series

(a) 1/C = 1/C1 + 1/C2 = 2/100x10-6 = 20 000 Therefore C = 50 mF
(b) C = C1 + C2 = 100x10-6 + 100x10-6 = 200 mF

2. 250 mF capacitors are joined in series to a 12V supply.
(a) the potential difference across each capacitor
(b) the charge on each plate of each capacitor

(a) since the capacitors are equal the potential across each will be the same and equal to 6V
(b) for each capacitor C = Q/V therefore Q = CV = 250x10-6x12 = 3x10-3 C = 3 mC

Capacitance formulae

The formulae for capacitors of some other common shapes are given below, although the proofs of these formulae are not needed at this level.

Sphere (radius a) 4pea
Concentric spheres (radii a and b) 4peab/[b-a]
Concentric cylinders (radii a,b,length L) 2peL/[ln(b/a)]
Two long, parallel wires (separation d,radius a,length L (d >> a) eL/ln[d/a]
Parallel-plate capacitor of area A containing a thickness x of dielectric and thickness b of air eA/(erb + x)

© Keith Gibbs 2011