Capacitors in Series Calculator — Equivalent Capacitance and Voltage Split

Background.

Wiring capacitors in series lowers the capacitance and raises the voltage the chain can stand. That is the opposite of what resistors do, and it is the reason series capacitor strings exist at all: nobody puts capacitors in series to get less capacitance, they do it because a 450 V electrolytic in a 700 V bus has to share the stress with a partner. This page takes up to twenty values in picofarads, nanofarads, microfarads, millifarads or farads and returns the equivalent capacitance, the charge the string holds at your supply voltage, and — the part most calculators leave out — which capacitor ends up carrying the most voltage.

The rule is a reciprocal sum. Add up 1/C for every capacitor, then invert the total. OpenStax University Physics states it directly: for capacitors in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances. A useful sanity check falls straight out of that: the answer is always smaller than the smallest capacitor in the chain. Two 10 µF capacitors in series make 5 µF. A 1 µF in series with a 470 µF makes 0.998 µF — the big one barely participates, because in a series string the smallest value dominates the result.

The reason the arithmetic looks like that is charge. The plates between two adjacent capacitors are connected only to each other, so whatever charge is pushed onto one is pulled off the other, and every capacitor in a series chain therefore carries exactly the same charge. Since voltage is charge divided by capacitance, the capacitor with the least capacitance ends up with the most voltage across it. With 1, 5 and 8 µF on a 12 V supply, the 1 µF sees 9.06 V — more than three quarters of the supply — while the 8 µF sees 1.13 V. Anyone who assumed the 12 V would split evenly three ways and specified 6 V parts has just built a failure.

That asymmetry is why the voltage output on this page sits next to the capacitance rather than in a footnote. It is also why the classic use of series capacitors comes with balancing resistors. The split this page computes is the ideal capacitive split, and it is the right answer for the instant the string is energised and for AC operation. Under settled DC conditions in a real high-voltage bank the division is governed by leakage resistance instead, which varies part to part and with temperature, and it can drift far from the capacitive ratio — so a designer fits a resistor across each capacitor to force the split rather than trusting it.

A few more limits worth knowing before you use the number. Everything here is ideal: no equivalent series resistance, no leakage current, no dielectric absorption, no self-heating. Nothing here derates a Class 2 ceramic, which can lose the majority of its marked capacitance under DC bias, and nothing here accounts for tolerance — electrolytics are routinely specified at −20 %/+80 %, and an equivalent computed from two such parts is no more precise than they are, however many digits appear on screen. Use the result as the design centre, then check the corners against the tolerances on the datasheets in front of you.

What is capacitors in series calculator?

A series connection puts capacitors end to end, so the same current path runs through all of them and there is no junction where charge can leave the chain. Because of that, every capacitor in the string holds the same charge Q, while the supply voltage divides among them. Applying V = Q/C to each capacitor and summing gives V = Q(1/C₁ + 1/C₂ + … + 1/Cₙ), and since the equivalent capacitor by definition satisfies V = Q/C_eq, the two expressions can only agree if 1/C_eq = Σ(1/Cᵢ). Inverting gives the equivalent capacitance. Three consequences follow and all three are practical rather than academic. First, the equivalent is always below the smallest member — adding capacitors in series can only reduce capacitance. Second, the smallest capacitor takes the largest share of the voltage, because voltage is inversely proportional to capacitance when charge is fixed. Third, identical capacitors split the voltage evenly, which is exactly why series banks are built from matched parts. For two capacitors the reciprocal sum can be rewritten as the product over the sum, C_eq = C₁C₂/(C₁+C₂), which is faster by hand and is the form most circuits textbooks reach for.

How to use this calculator.

1. 01Type the capacitor values into the list box, separated by commas, spaces or new lines. Two to twenty values are accepted.
2. 02Choose the unit that applies to every value in the list. Convert before entering if your string mixes scales — 100 nF is 0.1 µF.
3. 03Enter the DC voltage applied across the whole chain, not across one capacitor. It does not affect the equivalent capacitance, only the charge and the voltage split.
4. 04Read the equivalent capacitance, then check it against the smallest-capacitor figure beside it. If the result is not smaller than that, something was entered wrong.
5. 05Read the voltage across the smallest capacitor. That, plus a margin, is the voltage rating that part needs — not the supply divided by the number of capacitors.
6. 06For a real high-voltage string, treat the split as the AC and transient answer and fit balancing resistors to control the settled DC split.

The formula.

Start by converting every entered value into farads, because the reciprocal sum only works in consistent units. A list of 1, 5 and 8 in microfarads becomes 1×10⁻⁶, 5×10⁻⁶ and 8×10⁻⁶ F. The unit selector exists so this conversion happens once, correctly, instead of being done by hand three times.

Now sum the reciprocals. In microfarads that is 1/1 + 1/5 + 1/8 = 1 + 0.2 + 0.125 = 1.325 per microfarad, and the equivalent capacitance is 1 ÷ 1.325 = 0.7547 µF. OpenStax works this exact combination in its section on series and parallel capacitors and gives 0.755 µF, which is the same number to three significant figures. Note what happened: three capacitors, the largest of them 8 µF, and the chain behaves like three quarters of a microfarad. The reciprocal sum is dominated by its largest term, which comes from the smallest capacitor.

Charge comes next. The equivalent capacitor holds Q = C_eq × V = 7.547×10⁻⁷ F × 12 V = 9.057×10⁻⁶ C, or 9.0566 µC. That figure is not just the total — it is the charge on each and every capacitor in the string, because there is nowhere for charge to go except through the chain. This is the single fact that makes series behaviour intuitive once you have it.

The voltage split then follows from V = Q/C applied capacitor by capacitor. The 1 µF gets 9.0566 µC ÷ 1 µF = 9.0566 V. The 5 µF gets 1.8113 V. The 8 µF gets 1.1321 V. Those three add to exactly 12.0000 V, which is Kirchhoff's voltage law and a useful arithmetic check on any hand calculation. The smallest capacitor is taking 75.5 % of the supply on its own.

Rounding happens once, at the end. Every intermediate step runs at full precision and only the displayed outputs are rounded, so the reciprocal sum never accumulates rounding error. The verdict sentence beside the result classifies the unrounded voltage share, not the rounded display, so a value sitting exactly on the halfway point is described consistently with the number shown.

What this does not model: equivalent series resistance, leakage, dielectric absorption, self-heating, tolerance and DC-bias derating. All of them are real and some of them are large. The DC-bias effect on Class 2 ceramics in particular can remove most of a part's marked capacitance at its rated voltage, and no formula can recover that — only the manufacturer's bias curve can.

A worked example.

Frequently asked questions.

References& sources.

1. [1]OpenStax (Rice University), University Physics Volume 2, §8.2 'Capacitors in Series and in Parallel'. Primary source for the rule implemented here — 'For capacitors connected in a series combination, the reciprocal of the equivalent capacitance is the sum of reciprocals of individual capacitances: 1/C_S = 1/C_1 + 1/C_2 + 1/C_3 + ⋯' — for 'In a series network of capacitors, the equivalent capacitance is always less than the smallest individual capacitance in the network', and for 'All capacitors of a series combination have the same charge.' The same section works the 1.000 / 5.000 / 8.000 µF string used as this page's example and states 0.755 µF; this calculator returns 0.7547169811 µF. Retrieved 2026-07-29.
2. [2]OpenStax (Rice University), University Physics Volume 2, §8.1 'Capacitors and Capacitance'. Source of the definition C = Q/V — 'the ratio of the maximum charge Q that can be stored in a capacitor to the applied voltage V across its plates' — and of the farad as 'one coulomb per one volt'. Also the source of the ~3.0 MV/m breakdown field of air that underlies every capacitor voltage rating discussed on this page. Retrieved 2026-07-29.
3. [3]Engineering LibreTexts / James M. Fiore, Introduction to Circuit Analysis, §6.1.2 'Capacitance and Capacitors'. Independent second authority, consulted specifically to check the rule rather than to derive it. Gives the series form 'C_Total = 1/(1/C₁ + 1/C₂ + 1/C₃)', the two-capacitor product-sum form, and states that in a series string voltage 'divides inversely to capacitance, calculated as V = Q/C where Q is the total charge'. Agrees with the primary source on every value this page computes. Retrieved 2026-07-29.
4. [4]MIT OpenCourseWare, 6.002 Circuits and Electronics, Spring 2007 (Prof. Anant Agarwal). Course materials covering the capacitor element law and series/parallel reduction of energy-storage elements, used as a third independent treatment of the same reduction. Landing page verified to load and to list the course, term and instructor; lecture materials are linked from it. Retrieved 2026-07-29.