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Capacitors in Series and Parallel Calculator

Physics

How are they connected?

Reciprocals add, so the result is always smaller than the smallest member. Every capacitor carries the same charge, and the smallest one takes the largest share of the voltage.

Optional. Leave blank for the equivalent capacitance alone; enter it to get charge, energy and the per-capacitor split.

Capacitor values

Paste a list instead

Values carrying their own suffix (100nF, 4.7uF, 2200pF) keep it.

Display and AC options

Optional. X_C = 1/(2πfC). Undefined at DC, so 0 Hz is rejected.
0 to 10. Applies to the results, the copied text and the exported files alike.
On screen the results auto-scale; CSV columns use this fixed unit so a spreadsheet can sort them.
Equivalent capacitance · Series
6.8750 µF

Smaller than the smallest capacitor (10.0000 µF) — series always reduces capacitance. The same parts in parallel would give 32.0000 µF.

C110.0000 µF8.2500 VC222.0000 µF3.7500 V12.0000 V

Network totals

QuantityValueWhat it means
Equivalent capacitance6.8750 µFWhat the whole network looks like to the rest of the circuit
Same parts in parallel32.0000 µFThe comparison that shows how far the two rules diverge
Total charge Q82.5000 µCQ = C_eq × V drawn from the supply
Stored energy E495.0000 µJE = ½ C_eq V², shared across the members
Nearest E12 standard value6.8000 µF-1.09 % from the computed value
Nearest E24 standard value6.8000 µF-1.09 % from the computed value

Per-capacitor breakdown

CapacitorCapacitanceVoltageChargeEnergy% of supplyRating
C110.0000 µF8.2500 V82.5000 µC340.3125 µJ68.8 %

C222.0000 µF3.7500 V82.5000 µC154.6875 µJ31.3 %

C18.2500 V · 68.8 %C23.7500 V · 31.3 %

Step by step

StepWorking
Formula1/C_eq = 1/C₁ + 1/C₂ + … + 1/Cₙ
Substitute1/C_eq = 1/(10.0000 µF) + 1/(22.0000 µF)
Add the reciprocals1/C_eq = 100000.0000 + 45454.5455 = 145454.5455 F⁻¹
InvertC_eq = 1 / 145454.5455 = 6.8750 µF
ResultC_eq = 6.8750 µF
Total chargeQ = C_eq × V = 6.8750 µF × 12.0000 V = 82.5000 µC
Stored energyE = ½ C_eq V² = ½ × 6.8750 µF × (12.0000 V)² = 495.0000 µJ
Product over sumC_eq = (C₁ × C₂) / (C₁ + C₂) = 10.0000 µF × 22.0000 µF / (32.0000 µF) = 6.8750 µF
Product over sum is the same identity
For exactly two capacitors in series the reciprocal sum collapses into C₁C₂/(C₁+C₂). Both routes are shown above so the algebra can be checked against the arithmetic.

About This Tool

Capacitors in Series and Parallel – Equivalent Capacitance, Charge and Voltage

Combining capacitors is one of the first things anyone learns in circuit theory, and one of the easiest to get backwards. Capacitors combine opposite to resistors: in parallel they add directly, C_eq = C₁ + C₂ + … + Cₙ, while in series their reciprocals add, 1/C_eq = 1/C₁ + 1/C₂ + … + 1/Cₙ. This capacitors in series and parallel calculator evaluates either rule for any number of capacitors, mixes units freely, and then works out how charge, voltage and stored energy divide across the network.

Why the two rules point in opposite directions

Both follow from the plate picture. Capacitance rises with plate area and falls with plate separation, C = ε·A/d. Wiring two capacitors in parallel is electrically the same as widening the plates, so the values add and the result is always larger than the largest member. Wiring them in series stacks the gaps instead, increasing the effective separation, so the result is always smaller than the smallest member. For exactly two in series the reciprocal identity collapses into the familiar product over sum shortcut, C_eq = C₁·C₂ / (C₁ + C₂), and for N identical parts it reduces further to C/N in series and N·C in parallel.

How charge and voltage divide

This is where the two topologies genuinely differ in behaviour. In a parallel bank every capacitor sees the full supply, so the voltages are equal and the charge splits in proportion to capacitance: Qᵢ = Cᵢ·V. In a series string the charge is common to every member — the charge pushed onto one plate is pulled straight off the plate facing it — so Q = C_eq·V and the voltage divides inversely with capacitance, Vᵢ = Q/Cᵢ. The stored energy follows from E = ½CV² for the network and Eᵢ = ½CᵢVᵢ² for each member, and the parts must always sum back to the whole.

The smallest capacitor fails first
Because Vᵢ = Q/Cᵢ, the smallest capacitor in a series string carries the largest voltage. Two parts rated 25 V do not make a 50 V string unless they are closely matched: a 1 µF and a 10 µF in series across 30 V put 27.27 V on the small one and only 2.73 V on the large one. The safe supply limit here is V_max = min(Cᵢ × V_ratingᵢ) / C_eq, which works out at 27.5 V — well under the 50 V the ratings alone suggest. Real high-voltage strings add balancing resistors across each capacitor for exactly this reason.

Mixed networks and design work

Ladder networks reduce in stages: collapse each group with its own rule, then combine the collapsed values with the opposite rule. Two parallel pairs of 10 + 10 µF and 22 + 47 µF become 20 µF and 69 µF, and putting those in series gives 15.506 µF. The calculator records every intermediate value so the reduction can be followed step by step. Running the problem backwards is just as common: given a target and the parts already in hand, C_missing = C_target − ΣC_known in parallel and 1/C_missing = 1/C_target − Σ(1/C_known) in series, with the nearest E12 and E24 preferred values suggested so the answer maps onto a part you can actually buy.

AC behaviour and tolerance

Once the equivalent capacitance is known, its capacitive reactance at any frequency follows from X_C = 1/(2πfC) — 6.875 µF presents about 231.5 Ω at 100 Hz, and less as frequency rises. That single number links the combination result to filter corners, coupling impedance and supply decoupling. Component tolerance matters just as much: recomputing the network with every value at its low and high extremes gives the true envelope, since both combination rules increase monotonically with each capacitance.

Nameplate values are only the starting point

The combination arithmetic is exact, but real parts are not. Class-2 ceramics lose a large fraction of their capacitance under DC bias and with temperature, electrolytics routinely carry −20/+80 % tolerances and drift as they age, and every capacitor has equivalent series resistance and inductance that dominate its behaviour at high frequency.

Use the calculated value as the nominal design point, and measure the parts when the tolerance actually matters.

Where this comes up in practice

Beyond coursework, the same arithmetic covers a lot of bench work: building an odd capacitance out of stock parts, sizing a decoupling bank where several ceramics sit in parallel with a bulk electrolytic, stacking capacitors in series to raise the working voltage across a DC bus, trimming an RC or 555 timing network, and tuning an RF circuit where two series capacitors make a value no single part offers. In every case the questions are the same ones the calculator answers: what is the equivalent value, how much charge and energy does it hold, and which component is closest to its limit.

Frequently Asked Questions

Is the Capacitors in Series and Parallel Calculator free?

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How does this capacitors in series and parallel calculator work?

Enter your capacitance values with whatever units they came in — pF, nF, µF, mF or F can be mixed freely in one network — and pick series or parallel. Every value is converted to farads before any arithmetic runs, capacitances are added directly for parallel or reciprocally for series, and the single unrounded equivalent capacitance then drives the charge, energy, per-capacitor voltage and reactance outputs. Because nothing is rebuilt from a rounded intermediate, the displayed voltage shares still sum to the supply and the per-capacitor energies still sum to the total.

Why do capacitors in series behave the opposite way to resistors?

Because the geometry that sets capacitance works the other way round. Wiring capacitors in parallel is effectively enlarging the plate area, and since C is proportional to area the values add. Wiring them in series is effectively increasing the plate separation, and since C is inversely proportional to separation the reciprocals add — which always leaves the result below the smallest member. A useful sanity check: parallel can only increase capacitance, series can only decrease it.

Why does the smallest capacitor in a series string take the most voltage?

In a series string the same charge sits on every capacitor, because the charge on one plate is drawn straight off the plate facing it. With Q common and V = Q/C, the voltage across a member is inversely proportional to its capacitance, so the smallest part carries the largest share and reaches its rating first. Two capacitors rated 25 V each do not give a 50 V string unless they are closely matched — which is exactly why real high-voltage strings use balancing resistors across each part.

How is the maximum safe supply voltage calculated?

For a series string the tool evaluates V_max = min(Cᵢ × V_ratingᵢ) / C_eq. The product Cᵢ × V_ratingᵢ is the charge each part can hold before it reaches its rating, and dividing the smallest of those by the equivalent capacitance converts it back into a supply voltage. For 1 µF and 10 µF parts both rated 25 V, that gives 27.5 V, not the 50 V a naive sum of ratings would suggest. In a parallel bank every capacitor simply sees the supply, so the weakest single rating sets the limit.

What does the tolerance band tell me?

It recomputes the equivalent capacitance with every component at its low extreme and again at its high extreme. Both combination rules increase monotonically with each individual capacitance, so those two corners are genuinely the widest the network can be — no interior combination can fall outside them. It matters most with electrolytics, where −20/+80 % tolerances are normal and the resulting band explains why they are never used to set a time constant or a filter corner.

How accurate is this for real circuits?

The combination arithmetic is exact, but real capacitors are not ideal. Equivalent series resistance and inductance, dielectric absorption, leakage, and the substantial capacitance drift that class-2 ceramics show with DC bias and temperature all move the true value away from the nameplate figure. Treat the equivalent capacitance as accurate for the values you entered, and treat the entered values themselves as nominal unless you have measured the parts.