In a similar way to the case of capacitors connected in parallel, we can consider two or more capacitors connected in series (Figure 23.12). The total capacitance ${C_{total}}$ of two capacitors of capacitances ${C_1}$ and ${C_2}$ is given by:

$\frac{1}{{{C_{total}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}$

Here, it is the reciprocals of the capacitances that must be added to give the reciprocal of the total capacitance. For three or more capacitors connected in series, the equation for their total capacitance is:

$\frac{1}{{{C_{total}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}} + ...$

Capacitors in series: deriving the formula

The same principles apply here as for the case of capacitors in parallel. Figure 23.13 shows the situation.
${C_1}$ and ${C_2}$ are connected in series, and there is a p.d. V across them. This p.d. is divided (it is shared between the two capacitors), so that the p.d. across ${C_1}$ is ${V_1}$ and the p.d. across ${C_2}$ is ${V_2}$. It follows that:

$V = {V_1} + {V_2}$

Now we must think about the charge stored by the combination of capacitors. In Figure 23.13, you will see that both capacitors are shown as storing the same charge Q. How does this come about? When the voltage is first applied, charge $−Q$ arrives on the left-hand plate of ${C_1}$. This repels charge $−Q$ off the right-hand plate, leaving it with charge $+Q$. Charge $−Q$ now arrives on the left-hand plate of ${C_2}$, and this in turn results in charge $+Q$ on the right-hand plate.

Note that charge is not arbitrarily created or destroyed in this process – the total amount of charge in the system is constant. This is an example of the conservation of charge.
Notice also that there is a central isolated section of the circuit between the two capacitors. Since this is initially uncharged, it must remain so at the end. This requirement is satisfied, because there is charge $−Q$ at one end and $+Q$ at the other. Hence, we conclude that capacitors connected in series store the same charge. This allows us to write equations for ${V_1}$ and ${V_2}$:

${V_2} = \frac{Q}{{{C_1}}}$ and ${V_2} = \frac{Q}{{{C_2}}}$

The combination of capacitors stores charge Q when charged to p.d. V, and so we can write:

$V = \frac{Q}{{{C_{tatal}}}}$

Substituting these in $V = {V_1} + {V_2}$ gives:

$\frac{Q}{{{C_{total}}}} + \frac{Q}{{{C_1}}} + \frac{Q}{{{C_2}}}$

Cancelling the common factor of Q gives the required equation:

$\frac{1}{{{C_{total}}}} + \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}$

Worked example 2 shows how to use this relationship.