Some circuits have capacitors connected in series and parallel combinations. To determine, for instance, the total charge stored by the set of capacitors, we have to find the single equivalent capacitance of the set. This is done by identifying a pair of capacitors in the set that are either in series or in parallel with one another, replacing that pair with its equivalent capacitor (thereby reducing the number of capacitors by one) and iterating until we're left with one capacitor that is the equivalent of the set.

Take the situation above, for example. The four capacitors have the following values:

C_{1} = C_{2} = 90 pF.

C_{3} = 45 pF

C_{4} = 120 pF

What is the potential difference across each capacitor? How much charge is on each capacitor?

To solve this we need to find the equivalent capacitance of the set of capacitors. The first step is to re-draw the circuit so that C_{1} is drawn vertically - this makes it more obvious what's in parallel or series.

Now contract the circuit from 4 capacitors to 1.

Step 1 - C_{2} and C_{3} are in series. Replace this pair by a single capacitor C_{23}:

1/C_{23} = 1/C_{2} + 1/C_{3} = 1/90 + 1/45 = 3/90.

Therefore C_{23} = 90/3 = 30 pF.

Step 2 - C_{1} and C_{23} are in parallel. Replace that pair by a single capacitor C_{123} = 90 + 30 = 120 pF.

Step 3 - C_{4} and C_{123} are in series. Replace that pair by a single capacitor C_{eq}:

1/C_{eq} = 1/C_{4} + 1/C_{123} = 1/120 + 1/120 = 2/120

C_{eq} = 120/2 = 60 pF

Step 4 - Determine the charge on C_{eq}.

Q = C_{eq} DV = 60 pF * 12 V = 720 pC.

Now we need to expand the circuit back to the original four capacitors, and determine the charge and potential difference across each one as we go.

Step 1 - C_{eq} represents C_{4} and C_{123} in series. Capacitors in series have the same charge but split the potential difference.

Q_{4} = Q_{123} = 720 pC.

The capacitors are equal, so they each have 6 volts across them.

Step 2 - C_{123} represents C_{1} and C_{23} in parallel. Devices in parallel have the same potential difference (6 V in this case) across them.

Q_{1} = C_{1} * 6 = 540 pC.

Q_{23} = C_{23} * 6 = 180 pC.

These add to 720 pC, as they should.

Step 3 - C_{23} represents C_{2} and C_{3} in series.

Q_{2} = Q_{3} = 180 pC.

DV_{2} = Q_{2}/C_{2} = 180/90 = 2 volts.

DV_{3} = Q_{3}/C_{3} = 180/45 = 4 volts.

These add to 6 volts, as they should.

Step 4 - A good way to check for consistency is to label the potential at different points. Pick some point as a reference (say, 0 V at the negative terminal of the battery) and label other points relative to that. Check that the potential differences across the capacitors are consistent with these potential values.