Capacitors in Series and Parallel Calculator

Capacitors in series and parallel: the two rules

Capacitors in parallel add, C = C₁ + C₂ + C₃, and capacitors in series add as reciprocals, 1/C = 1/C₁ + 1/C₂ + 1/C₃. Those are the rules for resistors the other way round. Capacitors in parallel have both plates joined to the same two points, so they share one voltage and their charges add. Capacitors in series carry one and the same charge, because the plates joined between two neighbours form an isolated conductor that starts with no net charge, and their voltages add, which is why the reciprocals add.

That makes a series combination smaller than its smallest capacitor and a parallel one larger than its largest. In series the smallest capacitor takes the largest share of the voltage, V = Q/C; in parallel the largest capacitor holds the most charge, Q = CV.

How to use it

Choose how the groups connect. As Stages in series, each group is one link in a chain, and two or more capacitors inside a group are in parallel with each other, the same layout as the resistor network calculator. As Branches in parallel, each group is one branch across the supply, and two or more capacitors inside a group are in series. Type the capacitances in the unit chosen in the menu, and a supply voltage if you want the charges.

The headline is the equivalent capacitance, the single capacitor that would draw the same charge from the supply. Below it the table gives every capacitor’s own voltage, charge and stored energy, and the working shows each step with your numbers. The equivalent capacitance can be carried to the capacitor energy calculator or the RC time constant calculator, which each take one capacitance.

Worked example: 12 µF in series with 2 µF and 4 µF

This is the network loaded above, and it is Example 8.7 in OpenStax University Physics Volume 2 (Ling, Sanny and Moebs, 2016), section 8.2. The 2 µF and 4 µF are in parallel, so they add: 2 + 4 = 6 µF. That pair is in series with the 12 µF, so 1/C = 1/12 + 1/6 = 1/4, and C = 4 µF, below the 6 µF of the smaller stage.

On a 12 V supply the charge drawn is Q = 4 µF × 12 V = 48 µC, and both stages carry that same charge. The 12 µF takes 48 µC / 12 µF = 4 V and the pair takes 48 µC / 6 µF = 8 V, which add back to the 12 V supply. Inside the pair both capacitors have 8 V, so the 2 µF holds 2 µF × 8 V = 16 µC and the 4 µF holds 4 µF × 8 V = 32 µC, which add back to 48 µC.

The energy stored is ½ × 4 µF × (12 V)² = 288 µJ, shared as 96 µJ in the 12 µF, 64 µJ in the 2 µF and 128 µJ in the 4 µF. Example 8.8 in the book works the same energies to two figures, giving 130 µJ for the 4 µF and 0.29 mJ in all. Switch the arrangement to Branches in parallel and the same three numbers describe the 12 µF across a series pair of 2 µF and 4 µF, and the network becomes 13.333 µF.

Worked example: two in series with a third across them

Example 8.6 in the same section puts 1 µF and 5 µF in series and connects 8 µF across the pair. Choose Branches in parallel, with 1 and 5 in the first branch and 8 in the second. The series pair gives 1/C = 1/1 + 1/5 = 1.2, so C = 0.83333 µF, and the branches add: 0.83333 + 8 = 8.8333 µF, which the book rounds to 8.833 µF.

With 12 V applied, the pair holds 0.83333 µF × 12 V = 10 µC, which is the charge on each of its two capacitors, so the 1 µF takes 10 µC / 1 µF = 10 V and the 5 µF takes 10 µC / 5 µF = 2 V. The 8 µF has the full supply across it and holds 8 µF × 12 V = 96 µC. The supply delivers 106 µC in all, which is 8.8333 µF × 12 V.

Shortcuts worth knowing

Two capacitors in series combine as the product over the sum, C = C₁C₂/(C₁ + C₂), and n equal capacitors in series give C/n. Two in series across a supply form a capacitive divider: the voltage on the first is V₁ = V × C₂/(C₁ + C₂), with the other capacitor’s value on top, the reverse of a resistive divider. A very small capacitor in series with a large one sets the total almost on its own: 5 pF in series with 1 µF is still just under 5 pF.

What this does not cover

The capacitors are ideal: Q = CV exactly, with no tolerance, leakage or voltage rating. Real aluminium electrolytics are commonly ±20 percent, so five figures here are arithmetic rather than accuracy, and because leakage currents differ, a real series stack does not share its voltage by capacitance alone, which is why high-voltage stacks carry balancing resistors across each part. Every capacitor is taken to start uncharged, and the results are the steady state after charging has finished. The charging itself, and how long it takes, is what the RC charge and discharge simulator shows.

The network has two levels: groups in series with capacitors in parallel inside each, or groups in parallel with capacitors in series inside each. A bridge, or a network nested more deeply, is out of scope; reduce the innermost part first and type its equivalent as one capacitor.

Common mistakes

• Using the resistor rules. Adding 12 µF and 6 µF in series gives 18 µF, when the answer is 4 µF. Series capacitance is always below the smallest capacitor.
• Forgetting the final reciprocal. 1/12 + 1/6 is 0.25, which is not the answer. The answer is one over that, 4 µF.
• Sharing the voltage equally in series. Capacitors in series share the charge, and the smallest one takes the largest voltage, so it is the one whose rating matters most.
• Adding the charges along a chain. Each stage holds the same charge, and the supply delivers it once: the supply delivers only 48 µC in the example above, not 96 µC.
• Mixing units. A capacitance in microfarads times a voltage in volts is a charge in microcoulombs. The calculator works in farads throughout and adds prefixes only for reading.

Converting units first? Use the capacitance conversion table.