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Prove. The equivalent capacitance C_{E} of n capacitors C_{1}, C_{2}, ... , C_{n} connected in parallel is

1) C_{n} = C_{1} + C_{2} + ... + C_{n}

where C_{1}, C_{2}, ... , C_{n} are the capacitances of the capacitors.

Proof. We will prove it for three capacitors C_{1}, C_{2}, and C_{3}.

Let a potential difference V_{ab} be applied across the terminals a
and b of the parallel network shown in Fig. 1. The potential
difference across each of the capacitors is V_{ab}, however the
charges on the different capacitors is different. The charges on
the capacitors are

2) Q_{1} = C_{1}V_{ab} Q_{2} = C_{2}V_{ab} Q_{3} = C_{3}V_{ab}

The total charge Q on the network is

3) Q = Q_{1} + Q_{2} + Q_{3}

The equivalent capacitance C_{E} is that of a single capacitor which
would acquire the same total charge Q with the same potential
difference. Thus

4) Q = C_{E}V_{ab}

Substituting 2) and 4) into 3) gives

5) C_{E}V_{ab} = C_{1}V_{ab} + C_{2}V_{ab} + C_{3}V_{ab}

or

6) C_{E} = C_{1} + C_{2} + C_{3}

We note that the energy of the single equivalent capacitor is equal to the sum of the energies of
capacitors C_{1}, C_{2}, and C_{3}. The energy of the equivalent is ½ QV_{ab} and the sum of the energies of

capacitors C_{1}, C_{2}, and C_{3} is ½ Q_{1}V_{ab }+ ½ Q_{2}V_{ab }+ ½ Q_{3}V_{ab } = ½ QV_{ab} .

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