How to calculate capacitors in series
Capacitors are in series when they sit one after another in a single path. The same charge then lands on every one of them, the voltages across them add up to the source, and the total capacitance comes out smaller than the smallest: it is like one capacitor with a wider gap between its plates. This calculator works out the total of any number of series capacitors and, for the voltage you enter, the shared charge and the voltage and energy of each one.
How to use the series capacitor calculator
- Enter the source voltage. It is shared out across the capacitors. Leave it at 0 V if you only need the total capacitance.
- Choose how many capacitors with the + and − buttons or by typing the number, up to 500.
- Type each capacitance the way it is written on parts: 100n, 4.7u,
10uF, 22p. The panel shows the three-digit code for each value. - Read the results under the drawing: total capacitance, the charge on each capacitor, the stored energy and the largest voltage across any one. Click a capacitor in the drawing for its own calculation.
- Open the Calculation tab for the maths with fractions, or the Results table for every capacitor's voltage, share and energy.
The series capacitance formula
The charge Q is the same on every capacitor, and the voltages add: V = Q/C1 + Q/C2 + … Dividing by Q gives:
1/Ceq = 1/C1 + 1/C2 + 1/C3 + 1/C4
Your 4 capacitors 1/Ceq = 1/100 nF + 1/220 nF + 1/470 nF + 1/1 µF, so Ceq = 56.58 nF.
Two capacitors: product over sum
Ceq = (C1 × C2) / (C1 + C2)
Two 10 µF in series make 5 µF; equal capacitors give C ÷ n. The total is always below the smallest capacitor in the chain.
Charge and the voltage across each capacitor
Q = Ceq × V Vk = Q / Ck
Because every capacitor carries the same charge, the smallest capacitor takes the largest voltage, the opposite of resistors in series. This is the practical trap of series capacitors: a small part next to a large one can end up with almost the whole source across it and must be rated for that. Chains of equal electrolytics for high voltages get balancing resistors across each one for the same reason.
Your chain, step by step
- Add the reciprocals: 1/Ceq = 1/100 nF + 1/220 nF + 1/470 nF + 1/1 µF, so Ceq = 56.58 nF, below the smallest, 100 nF.
- The charge on each: Q = Ceq × V = 56.58 nF × 12 V = 679 nC; energy ½CeqV² = 4.074 µJ.
- Each capacitor has V = Q / C across it:
| Capacitor | Voltage Q / C | Energy ½CV² | Share of V |
|---|---|---|---|
| C1 = 100 nF | 6.79 V | 2.31 µJ | 57% |
| C2 = 220 nF | 3.09 V | 1.05 µJ | 26% |
| C3 = 470 nF | 1.44 V | 490 nJ | 12% |
| C4 = 1 µF | 679 mV | 231 nJ | 5.7% |
| Total: 56.58 nF | 12 V | 4.07 µJ | 100% |
Worked example: 100 nF, 220 nF and 1 µF at 12 V
Add the reciprocals: 1/100 nF + 1/220 nF + 1/1 µF, and flip: Ceq = 64.33 nF, below the 100 nF. The charge on each is Q = 64.33 nF × 12 V = 771.9 nC. The 100 nF then has 7.72 V across it (64% of the 12 V), the 220 nF 3.51 V, and the 1 µF only 772 mV: the voltages add up to 12 V, and the smallest capacitor takes the most.
The three-digit code on capacitors
Small capacitors are marked with a three-digit code in picofarads: the first two digits, then a power of ten. 104 is 10 × 10⁴ pF = 100 nF; 223 is 22 nF; 475 is 4.7 µF. Values under 100 pF are printed as they are. The drawing prints each capacitor's code on its body and the panel shows it beside the value.
Questions
Why is the total smaller than the smallest?
Series plates behave like one capacitor whose plates are further apart; a wider gap means less capacitance. In the formula, adding any positive 1/C term makes 1/Ceq bigger, so Ceq smaller.
Why would anyone put capacitors in series?
To stand a higher voltage than one part can (the voltage divides), to get a value you do not have, or in DC-blocking and tuning circuits. Rarely to make a smaller capacitor on purpose.
Which capacitor needs the highest voltage rating?
The smallest one: it takes the largest share of the voltage. The Results table shows the voltage across each.
What about capacitors in parallel?
Then every capacitor has the same voltage and the capacitances simply add; the parallel capacitor calculator works those out.