Figure 1(a) shows a series connection of three capacitors with a voltage applied. As for any capacitor, the capacitance of the combination is related to charge and voltage by C=QVC=QV size 12{C= { {Q} over {V} } } {}.

Note in Figure 1 that opposite charges of magnitude QQ size 12{Q} {} flow to either side of the originally uncharged combination of capacitors when the voltage VV size 12{V} {} is applied. Conservation of charge requires that equal-magnitude charges be created on the plates of the individual capacitors, since charge is only being separated in these originally neutral devices. The end result is that the combination resembles a single capacitor with an effective plate separation greater than that of the individual capacitors alone. (See Figure 1(b).) Larger plate separation means smaller capacitance. It is a general feature of series connections of capacitors that the total capacitance is less than any of the individual capacitances.

Figure 1: (a) Capacitors connected in series. The magnitude of the charge on each plate is Q Q . (b) An equivalent capacitor has a larger plate separation dd size 12{d} {}. Series connections produce a total capacitance that is less than that of any of the individual capacitors.

We can find an expression for the total capacitance by considering the voltage across the individual capacitors shown in Figure 1. Solving C=QVC=QV size 12{C= { {Q} over {V} } } {} for VV size 12{V} {} gives V=QCV=QC size 12{V= { {Q} over {C} } } {}. The voltages across the individual capacitors are thus V1=QC1V1=QC1 size 12{ {V} rSub { size 8{1} } = { {Q} over { {C} rSub { size 8{1} } } } } {}, V2=QC2V2=QC2 size 12{ {V} rSub { size 8{2} } = { {Q} over { {C} rSub { size 8{2} } } } } {}, and V3=QC3V3=QC3 size 12{ {V} rSub { size 8{3} } = { {Q} over { {C} rSub { size 8{3} } } } } {}. The total voltage is the sum of the individual voltages:

Now, calling the total capacitance CSCS size 12{C rSub { size 8{S} } } {} for series capacitance, consider that

Entering the expressions for V1V1 size 12{V rSub { size 8{1} } } {}, V2V2 size 12{V rSub { size 8{2} } } {}, and V3V3 size 12{V rSub { size 8{3} } } {}, we get

Canceling the QQ size 12{Q} {}s, we obtain the equation for the total capacitance in series CSCS size 12{ {C} rSub { size 8{S} } } {} to be

where “...” indicates that the expression is valid for any number of capacitors connected in series. An expression of this form always results in a total capacitance CSCS size 12{ {C} rSub { size 8{S} } } {} that is less than any of the individual capacitances C1C1 size 12{ {C} rSub { size 8{1} } } {}, C2C2 size 12{ {C} rSub { size 8{2} } } {}, ..., as the next example illustrates.

Figure 2(a) shows a parallel connection of three capacitors with a voltage applied. Here the total capacitance is easier to find than in the series case. To find the equivalent total capacitance CpCp size 12{ {C} rSub { size 8{p} } } {}, we first note that the voltage across each capacitor is VV size 12{V} {}, the same as that of the source, since they are connected directly to it through a conductor. (Conductors are equipotentials, and so the voltage across the capacitors is the same as that across the voltage source.) Thus the capacitors have the same charges on them as they would have if connected individually to the voltage source. The total charge QQ size 12{Q} {} is the sum of the individual charges:

Figure 2: (a) Capacitors in parallel. Each is connected directly to the voltage source just as if it were all alone, and so the total capacitance in parallel is just the sum of the individual capacitances. (b) The equivalent capacitor has a larger plate area and can therefore hold more charge than the individual capacitors.

Using the relationship Q=CVQ=CV size 12{Q= ital "CV"} {}, we see that the total charge is Q=CpVQ=CpV size 12{Q= {C} rSub { size 8{p} } V} {}, and the individual charges are Q1=C1VQ1=C1V size 12{ {Q} rSub { size 8{1} } = {C} rSub { size 8{1} } V} {}, Q2=C2VQ2=C2V size 12{ {Q} rSub { size 8{2} } = {C} rSub { size 8{2} } V} {}, and Q3=C3VQ3=C3V size 12{ {Q} rSub { size 8{3} } = {C} rSub { size 8{3} } V} {}. Entering these into the previous equation gives

Canceling VV size 12{V} {} from the equation, we obtain the equation for the total capacitance in parallel CpCp size 12{C rSub { size 8{p} } } {}:

Total capacitance in parallel is simply the sum of the individual capacitances. (Again the “...” indicates the expression is valid for any number of capacitors connected in parallel.) So, for example, if the capacitors in the example above were connected in parallel, their capacitance would be

The equivalent capacitor for a parallel connection has an effectively larger plate area and, thus, a larger capacitance, as illustrated in Figure 2(b).

More complicated connections of capacitors can sometimes be combinations of series and parallel. (See Figure 3.) To find the total capacitance of such combinations, we identify series and parallel parts, compute their capacitances, and then find the total.

Figure 3: (a) This circuit contains both series and parallel connections of capacitors. See Example 2 for the calculation of the overall capacitance of the circuit. (b) C1C1 size 12{ {C} rSub { size 8{1} } } {} and C2C2 size 12{ {C} rSub { size 8{2} } } {} are in series; their equivalent capacitance CSCS size 12{ {C} rSub { size 8{S} } } {} is less than either of them. (c) Note that CSCS size 12{ {C} rSub { size 8{S} } } {} is in parallel with C3C3 size 12{ {C} rSub { size 8{3} } } {}. The total capacitance is, thus, the sum of CSCS size 12{ {C} rSub { size 8{S} } } {} and C3C3 size 12{ {C} rSub { size 8{3} } } {}.