Parallel Capacitor Calculator — Total Capacitance of a Bank

Background.

Capacitors in parallel simply add. Wire a 1 µF, a 5 µF and an 8 µF across the same two nodes and you have 14 µF, and that is the whole rule. It is the opposite of what capacitors do in series and the same as what resistors do in series, which is why the two get confused so persistently. This page adds up to twenty values in picofarads, nanofarads, microfarads, millifarads or farads, then tells you the charge the bank holds at your working voltage and how much of that charge sits on the largest capacitor.

The reason the rule is a plain sum is that every capacitor in a parallel bank has the same voltage across it — they share both terminals, so they have no choice. Each one stores Q = CV at that shared voltage, and the total charge is the sum of the individual charges. Factor out the common V and what remains is the sum of the capacitances. OpenStax University Physics states both halves of that: the equivalent capacitance is the sum of all individual capacitances, and in a parallel network the equivalent is always larger than any single member.

The practical consequence people miss is about voltage, and it goes in the dangerous direction. Paralleling capacitors does absolutely nothing for voltage withstand. A bank of ten 16 V capacitors is still a 16 V bank; put 25 V across it and every part in it is over-stressed at once. Series strings raise the voltage rating and lower the capacitance, parallel banks raise the capacitance and leave the rating alone, and mixing the two rules up is how supplies get destroyed. The page repeats this beside the result rather than burying it, because it is the one thing a total-capacitance number cannot tell you on its own.

Charge division is the other half of the answer. Since all members share the voltage, charge divides in proportion to capacitance — the exact mirror of a series string, where charge is shared and voltage divides inversely. In the worked example the 8 µF holds 96 µC of the bank's 168 µC, 57 % of the total, and it is therefore the part that sources most of the current when the bank dumps into a load. That matters for ripple-current rating: in a bulk bank the largest capacitor usually also has to be the one specified for the highest ripple current, not the smallest.

Two scope limits before you use the number. First, everything here is ideal — no equivalent series resistance, no series inductance, no leakage, no tolerance, no DC-bias derating. Second, and more subtly, the ideal sum cannot explain why real designs parallel a big electrolytic with a small ceramic. That trick is not about adding 100 nF to 1000 µF, which changes essentially nothing; it is about the ceramic's much lower ESR and ESL giving the combination a low impedance at frequencies where the electrolytic has already gone inductive. An ideal capacitance sum is blind to that effect, so treat this page as the DC and low-frequency answer and reach for an impedance-versus-frequency plot when decoupling is the question.

What is parallel capacitor calculator?

A parallel connection ties every capacitor between the same pair of nodes, so all of them experience the identical voltage V. Each stores Qᵢ = CᵢV, and the bank's total stored charge is Q = ΣQᵢ = VΣCᵢ. Because the equivalent capacitor must satisfy Q = C_eq·V at that same voltage, C_eq = ΣCᵢ — the capacitances add directly, with no reciprocals anywhere. Three consequences follow. The total is always greater than the largest member, since every other term in the sum is positive. Charge divides in proportion to capacitance, so the largest capacitor holds the biggest share and delivers the biggest share of any discharge current. And the voltage rating of the bank is the lowest rating present, because every capacitor sees the full applied voltage — paralleling buys capacitance and energy storage, never withstand. Physically, putting capacitors side by side is equivalent to enlarging the plate area of a single capacitor, and capacitance is proportional to plate area, which is the intuition behind the sum. This is the arrangement used for bulk energy storage, supply decoupling, motor-run banks and power-factor correction, all of which need more charge at a fixed voltage rather than a higher voltage.

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 first if the bank mixes scales — 100 nF is 0.1 µF.
3. 03Enter the DC voltage the bank works at. Every capacitor sees this same voltage; it sets the stored charge and does not change the total capacitance.
4. 04Read the total capacitance, then check it against the largest-capacitor figure beside it. A parallel total must exceed that value.
5. 05Check the voltage rating separately, by hand: the bank withstands only what its lowest-rated member withstands, whatever the total capacitance says.
6. 06Use the charge on the largest capacitor to decide which part needs the highest ripple-current rating — it is the one that sources most of the discharge.

The formula.

Convert every entered value to farads first, because the sum only means anything 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 does that conversion once, correctly, for the whole list.

Then add. 1 + 5 + 8 = 14 µF, or 1.4×10⁻⁵ F. There is no reciprocal step and no product-over-sum shortcut needed, because the parallel rule is the simple one. The check that catches a mistyped entry is that the total must exceed the largest member: 14 µF is greater than 8 µF, as it must be.

Charge follows from Q = CV. The bank holds Q = 1.4×10⁻⁵ F × 12 V = 1.68×10⁻⁴ C, which is 168 µC. Unlike a series string — where that figure would sit on every capacitor — here it is a total that gets divided up. Capacitor by capacitor: the 8 µF holds 8×10⁻⁶ × 12 = 96 µC, the 5 µF holds 60 µC and the 1 µF holds 12 µC. Those add to 168 µC, which is the arithmetic check.

So charge divides in proportion to capacitance, which is worth stating alongside its opposite: in a series string, charge is common and voltage divides inversely with capacitance. Same two quantities, mirrored roles. If you remember only one of the pair you will get the other backwards, which is why this page and the series page each state both.

The share the largest capacitor takes — 96 of 168 µC, or 57.14 % — is what the verdict beside the result classifies, and it classifies the unrounded ratio rather than the displayed percentage, so a bank sitting exactly on the halfway point is described consistently with the number shown. Two equal capacitors put exactly 50 % on each, which is the edge case; it takes three or more members for any single one to hold less than half.

Rounding happens once, at the end. Intermediate steps run at full precision, so a twenty-capacitor bank does not accumulate error. And note which units the outputs use: microfarads and picofarads, not farads. That is deliberate. This page rounds outputs to ten decimal places, and a 200 pF total expressed in farads is 2×10⁻¹⁰ — right at the edge of that precision, where a correct calculation would print a visibly wrong number.

What this does not model: ESR, ESL, leakage, dielectric absorption, tolerance, temperature and DC-bias derating. The bias effect in particular can remove most of a Class 2 ceramic's marked value at its rated voltage, so a bank of ten 10 µF X5R parts at rated bias may be nothing like 100 µF. Enter derated figures from the manufacturer's curves if the exact value matters.

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 parallel combination, the equivalent (net) capacitance is the sum of all individual capacitances in the network: C_P = C_1 + C_2 + C_3 + ⋯' — for 'In a parallel network of capacitors, the equivalent capacitance is always larger than any of the individual capacitances in the network', and for 'Since the capacitors are connected in parallel, they all have the same voltage V across their plates.' Retrieved 2026-07-29.
2. [2]Engineering LibreTexts / James M. Fiore, Introduction to Circuit Analysis, §6.1.2 'Capacitance and Capacitors'. Independent second authority, consulted to check the rule rather than to derive it: 'Capacitors in parallel add directly: C_Total = C₁ + C₂ + C₃', together with C = Q/V. Agrees with the primary source on every value this page computes, including the 14 µF total and 168 µC of stored charge in the worked example. Retrieved 2026-07-29.
3. [3]OpenStax (Rice University), University Physics Volume 2, §8.1 'Capacitors and Capacitance'. Source of 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' — of the farad as 'one coulomb per one volt', and of the ~3.0 MV/m breakdown field of air that sets the physical basis for every capacitor voltage rating discussed on this page. Retrieved 2026-07-29.
4. [4]MIT OpenCourseWare, 6.002 Circuits and Electronics, Spring 2007 (Prof. Anant Agarwal). Course materials on the capacitor element law and the series/parallel reduction of energy-storage elements, used as a third independent treatment. Landing page verified to load and to list the course number, term and instructor; lecture materials are linked from it. Retrieved 2026-07-29.