Capacitor Calculator
Combine capacitors in series or parallel, with total capacitance, stored energy, charge and the voltage each series capacitor carries.
How to use this calculator
- 1Choose series or parallel, and pick the unit all your values are in.
- 2Enter up to four capacitances, leaving unused slots at zero.
- 3Enter the applied voltage to see stored energy, charge, and how voltage divides across a series chain.
How the calculation works
parallel: C = C₁ + C₂ + … series: 1/C = 1/C₁ + 1/C₂ + … E = ½CV² Q = CV- C
- Total capacitance of the combination, in farads
- C₁, C₂ …
- The individual capacitances
- E
- Energy stored, in joules
- Q
- Charge stored, in coulombs
- V
- Voltage applied across the combination
These rules are the reverse of the resistor rules. Resistors in series add and in parallel combine reciprocally; capacitors do the opposite. Applying the resistor rule to capacitors is the single most common error in this area.
The reason is geometric. Capacitance rises with plate area and falls with plate separation. Parallel capacitors effectively pool their area, so values add. Series capacitors effectively stack their dielectrics into a thicker gap, so the total falls.
Series combinations are always smaller than the smallest member, and parallel combinations always larger than the largest. That is a useful sanity check on any answer.
In a series chain every capacitor carries identical charge, so voltage divides inversely with capacitance. The smallest capacitor sees the highest voltage, which matters for voltage ratings.
Worked example
100 µF and 220 µF in parallel
- 1.Parallel capacitors add directly: 100 + 220 = 320 µF.
- 2.The total is larger than either capacitor — parallel connection always increases capacitance.
- 3.Both capacitors see the full 12 V, since they are connected across the same two nodes.
- 4.Energy stored: ½ × 0.00032 F × 12² = ½ × 0.00032 × 144 = 0.02304 J, or 23.04 mJ.
- 5.Total charge: Q = CV = 0.00032 × 12 = 3,840 µC.
Result: 320 µF, storing 23.04 mJ
The same two in series
- 1.Series capacitors combine reciprocally: 1/C = 1/100 + 1/220 = 0.01 + 0.0045455 = 0.0145455.
- 2.C = 1 ÷ 0.0145455 = 68.75 µF — smaller than the 100 µF capacitor, let alone the 220 µF one.
- 3.Total charge: Q = 0.00006875 × 12 = 825 µC, and every capacitor in the chain carries that same charge.
- 4.Voltage across the 100 µF: 825 µC ÷ 100 µF = 8.25 V. Across the 220 µF: 825 ÷ 220 = 3.75 V.
- 5.They sum to 12 V as they must, but the smaller capacitor takes more than twice the voltage — check its rating.
Result: 68.75 µF, with 8.25 V across the smaller capacitor
Why capacitors combine backwards from resistors
Anyone who has learned resistor networks arrives at capacitors with the wrong instinct. Resistors in series add; capacitors in series do not. Resistors in parallel combine reciprocally; capacitors in parallel simply add. The rules are swapped, and no amount of memorising helps as much as understanding why.
Capacitance is set by geometry: it rises in proportion to the plate area and falls in proportion to the gap between the plates. Wire two capacitors in parallel and you have, electrically, one capacitor with the combined plate area — so the values add. Wire them in series and the effect is a single capacitor with a thicker dielectric gap, so capacitance falls.
That gives a reliable sanity check that needs no arithmetic. Parallel always produces more than the largest capacitor present; series always produces less than the smallest. If your answer breaks either rule, you have applied the resistor formula.
Series chains and the voltage-rating trap
Capacitors are sometimes put in series specifically to withstand a higher voltage than any one of them is rated for. The idea is sound but the execution catches people out, because the voltage does not divide evenly unless the capacitors are identical.
In a series chain the same charge flows onto every capacitor, so each carries an identical Q. Since V = Q/C, the capacitor with the *smallest* capacitance ends up with the *largest* voltage across it. Put a 100 µF in series with a 220 µF and the smaller one takes 8.25 V of a 12 V supply — more than twice its neighbour.
This is why series capacitor strings in real designs use identical parts, and often add balancing resistors across each one. Tolerance alone can skew the division enough to push a capacitor past its rating, and an over-volted electrolytic fails loudly.
What this assumes, and where it stops
Assumptions
- Ideal capacitors: no leakage, no equivalent series resistance, no dielectric absorption.
- Capacitors are at their nominal values, ignoring manufacturing tolerance.
- Values entered as zero are treated as absent, not as a short circuit.
- A steady applied voltage, with the network fully charged.
Limitations
- Up to four capacitors in a single series or parallel group. Mixed series-parallel networks must be reduced in stages.
- Real capacitors carry wide tolerances — ±20% is common for electrolytics — so the voltage division in a series chain will not match the ideal figures exactly.
- Does not check voltage ratings, temperature coefficients or ripple current limits, all of which govern whether a combination is safe in practice.
- No frequency-dependent behaviour: equivalent series resistance and inductance matter at high frequency and are not modelled.
Common questions
How do capacitors add in series and parallel?
In parallel they add directly: C = C₁ + C₂. In series the reciprocals add: 1/C = 1/C₁ + 1/C₂. This is the opposite of resistors. A parallel combination is always larger than the biggest capacitor, and a series combination always smaller than the smallest.
Why do capacitors in series have less capacitance?
Because connecting them in series is electrically equivalent to increasing the distance between the plates, and capacitance falls as that gap widens. Parallel connection instead pools the plate area, and capacitance rises with area — which is why parallel values add.
How is voltage shared between capacitors in series?
Inversely to capacitance. Every capacitor in a series chain holds the same charge, and since V = Q/C the smallest capacitance carries the largest voltage. Two unequal capacitors will not split the supply evenly, so check that the smallest one is rated for its share.
How much energy does a capacitor store?
E = ½CV² joules. The half arises because voltage climbs from zero to its final value as the capacitor charges, so the average voltage during charging is half the final one. Because the term is squared, doubling the voltage stores four times the energy.
Sources
Formula and content last reviewed on .
Results are estimates for information only, not professional advice.
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