Capacitors in Series and in Parallel

Series: one charge, shared voltage

Join two capacitors end to end, the second plate of \(C_1\) wired to the first plate of \(C_2\). Then put a battery across the pair. This is a series combination. The battery pulls charge \(-Q\) onto the far plate of \(C_2\) and pushes \(+Q\) onto the far plate of \(C_1\). What happens in the middle?

The two inner plates and the wire between them form one conductor that no battery touches. It started with no charge, and it cannot gain any. The \(+Q\) on the first outer plate pulls \(-Q\) onto the inner plate of \(C_1\). That leaves \(+Q\) on the inner plate of \(C_2\), and it pulls \(-Q\) onto the far plate. So every capacitor in series holds the same charge \(Q\).

The potential differences add along the chain, \(V = V_1 + V_2\), with \(V_1 = Q/C_1\) and \(V_2 = Q/C_2\). The smaller capacitor takes the larger share of the voltage. One capacitor \(C\) that holds the same \(Q\) at the same total \(V\) is the equivalent of the pair.

With three capacitors the rule simply grows. Take \(2\), \(3\) and \(6\,\mu\text{F}\) in series across \(60\,\text{V}\). Then \(1/C = 1/2 + 1/3 + 1/6\) \(= 1\), so \(C = 1\,\mu\text{F}\) and each holds \(60\,\mu\text{C}\). The voltages are \(30\), \(20\) and \(10\,\text{V}\), again in the inverse ratio of the capacitances. A series combination is always less than its smallest member.

The safe voltage of a series pair

Each capacitor has a largest safe voltage. A \(2\,\mu\text{F}\) capacitor rated \(300\,\text{V}\) can safely hold \(2 \times 300\) \(= 600\,\mu\text{C}\). A \(3\,\mu\text{F}\) rated \(150\,\text{V}\) can hold \(3 \times 150\) \(= 450\,\mu\text{C}\). In series both carry one charge, so it may not exceed the smaller limit, \(450\,\mu\text{C}\).

At \(450\,\mu\text{C}\) the first capacitor has \(450/2\) \(= 225\,\text{V}\), inside its rating, and the second has \(150\,\text{V}\), at its limit. So the largest safe voltage for the pair is \(225 + 150\) \(= 375\,\text{V}\). Adding the two ratings, \(450\,\text{V}\), would overload the \(3\,\mu\text{F}\) capacitor.

Parallel: one voltage, shared charge

Now join the capacitors side by side: every first plate to one wire and every second plate to another. This is a parallel combination. Each capacitor has its plates on the same two wires, so each has the same potential difference \(V\). Each takes its own charge, \(Q_1 = C_1V\) and \(Q_2 = C_2V\), and the battery supplies the total.

So the total charge is \(Q = (C_1 + C_2)V\), and the equivalent capacitance is \(C = C_1 + C_2\). Parallel capacitors act like one capacitor with a larger plate area. A parallel combination is always more than its largest member.

The same \(2\) and \(3\,\mu\text{F}\) in parallel across \(100\,\text{V}\) make \(5\,\mu\text{F}\). The first holds \(200\,\mu\text{C}\), the second \(300\,\mu\text{C}\), and the battery supplies \(500\,\mu\text{C}\). Compare the energies: in series the pair stores \(\tfrac12 \times 1.2 \times 10^{-6} \times 100^2\) \(= 6 \times 10^{-3}\,\text{J}\). In parallel it stores \(\tfrac12 \times 5 \times 10^{-6} \times 100^2\) \(= 0.025\,\text{J}\), because it holds more charge at the same voltage.

Take \(n\) equal capacitors \(C\). In series they give \(C/n\), and in parallel \(nC\). The ratio of the two is \(n^2\), so four equal capacitors give sixteen times as much in parallel as in series. Three \(6\,\mu\text{F}\) capacitors give \(2\,\mu\text{F}\) in series and \(18\,\mu\text{F}\) in parallel. Two of them in parallel, in series with the third, give \(12\) with \(6\), which is \(4\,\mu\text{F}\).

Solving a network

A network of capacitors is solved in two passes. First reduce it: replace any series pair or parallel group by its equivalent, again and again, until one capacitor is left. Then work back out: the charge on the last one is the battery's charge. Use the series rule (same charge) and the parallel rule (same voltage) to share it out.

Turn it round: a \(6\,\mu\text{F}\) capacitor beside a series pair of \(2\) and \(3\,\mu\text{F}\), all across \(20\,\text{V}\). The pair gives \(1.2\,\mu\text{F}\), so the whole is \(6 + 1.2\) \(= 7.2\,\mu\text{F}\). The \(6\,\mu\text{F}\) has the full \(20\,\text{V}\) and holds \(120\,\mu\text{C}\). The pair also has \(20\,\text{V}\) and holds \(1.2 \times 20\) \(= 24\,\mu\text{C}\) on each, with \(12\,\text{V}\) across the \(2\,\mu\text{F}\) and \(8\,\text{V}\) across the \(3\,\mu\text{F}\).

A wire across a capacitor empties it. Put \(4\) and \(6\,\mu\text{F}\) in series across \(100\,\text{V}\): the pair is \(2.4\,\mu\text{F}\) and each holds \(240\,\mu\text{C}\). Now join a wire across the \(6\,\mu\text{F}\). Its plates are at one potential, so it holds nothing. The whole \(100\,\text{V}\) falls across the \(4\,\mu\text{F}\), which now holds \(400\,\mu\text{C}\).

The balanced bridge

Some networks cannot be reduced into series and parallel groups. The bridge of Fig 3.2 is one. An upper arm holds \(2\) and \(4\,\mu\text{F}\) in series, with the point \(P\) between them. A lower arm holds \(3\) and \(6\,\mu\text{F}\), with the point \(Q\) between them. A fifth capacitor of \(5\,\mu\text{F}\) joins \(P\) and \(Q\).

Imagine the \(5\,\mu\text{F}\) removed and a voltage \(V\) put across \(A\) and \(B\). In a series pair the voltage divides inversely as the capacitances. So the \(2\,\mu\text{F}\) takes \(\tfrac46V\) and the \(3\,\mu\text{F}\) takes \(\tfrac69V\), both \(\tfrac23V\). The points \(P\) and \(Q\) are therefore at the same potential. Put the \(5\,\mu\text{F}\) back: with no voltage across it, it takes no charge and changes nothing.

With the middle capacitor gone, the arms are two series pairs in parallel. The upper arm is \(2 \times 4/6\) \(= \tfrac43\,\mu\text{F}\) and the lower arm \(3 \times 6/9\) \(= 2\,\mu\text{F}\). So \(C_{AB}\) \(= \tfrac43 + 2\) \(= \tfrac{10}{3}\) \(\approx 3.33\,\mu\text{F}\). Five equal capacitors \(C\) make a balanced bridge too. Each arm is then \(C/2\), and the two arms give \(C\).