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Capacitors in Series Calculator

Free capacitors in series calculator — total capacitance in series or parallel, plus how the supply voltage divides across each part.

Capacitance values (µF)

Supply voltage (V, optional)

In series the total is always smaller than the smallest one — the plates are further apart in effect.

Total capacitance

59.977 µF

3 capacitors

smaller than 100 µF

Calculation breakdown

1/C = 1/100 + 1/220 + 1/470

C = 59.977 µF

In series each capacitor holds the same charge, so the smallest value takes the largest share of the voltage — and is the one that fails first.

Instant insight

3 in series gives 59.977 µF
Series capacitors combine like parallel resistors — the total is always smaller than the smallest one. It is parallel capacitors that add.
Quick check: with equal capacitors, the total is one value divided by the count. If your answer is bigger than the smallest part, the arithmetic is inverted.
Voltage does not divide evenly in practice
It splits in inverse proportion to capacitance, and tolerance and leakage make the split drift over time — one capacitor can end up over its rating.
Series strings intended to share mains-level voltage use balancing resistors across each capacitor. Without them, the string is not safe to rely on.
Capacitors carry a tolerance, typically ±10–20%, worse for electrolytics
The printed value is nominal. Actual parts vary within their tolerance band, and drift further with temperature, age and applied voltage.
Design so the circuit still works across the whole tolerance band. If it only works at the nominal value, it does not work.

Worked out from your figures in your browser — nothing is sent anywhere. Idealised figures: real installations need real margins.

Compare results

Save this result, change your inputs, and recalculate to compare scenarios side by side.

Formula

In series, 1/C = 1/C₁ + 1/C₂ + … and the total is always smaller than the smallest part. In parallel the values simply add. This is the exact opposite of resistors, which is why it is worth checking rather than trusting memory.

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Everyday Uses

Reaching a higher voltage rating

Two 400 V parts in series survive 800 V. The standard trick when the voltage you need exceeds the parts you can buy.

Making a value you do not have

Two 100 µF in series give 50 µF. Useful when the parts drawer has the wrong values and the shop is shut.

Finding which part fails first

In series the smallest capacitance sees the most voltage. Checking that share before powering up prevents a bang.

Spreading ripple current in parallel

Several smaller capacitors share the heating that one large one would absorb alone, and typically last considerably longer for it.

Checking coursework both ways

Series and parallel rules are inverted relative to resistors, which is the single most common exam slip in this topic.

Substituting during a repair

When the exact part is unavailable, combining what you have gets a board working — provided the voltage split is checked.

Frequently Asked Questions

How do you calculate capacitors in series?

Add the reciprocals and invert: 1/C = 1/C₁ + 1/C₂ + … For two capacitors that simplifies to C₁C₂/(C₁+C₂). The total is always smaller than the smallest individual value — two 100 µF capacitors in series give 50 µF. Placing capacitors in series is effectively increasing the distance between the plates, which reduces capacitance, and that is exactly the opposite of what happens with resistors.

How do capacitors in parallel add up?

They simply add: C = C₁ + C₂ + … Three 100 µF capacitors in parallel give 300 µF. Connecting them side by side is equivalent to enlarging the total plate area, and capacitance rises in direct proportion to area. This is the usual way to reach a large capacitance from components you already have, and it also spreads the ripple current across several parts, which helps them run cooler and last longer.

Why are capacitors the opposite of resistors?

Because capacitance depends on plate geometry rather than on opposition to current. Series resistors add because each one adds more material for current to fight through. Series capacitors reduce the total because the arrangement behaves like one capacitor with a much larger plate separation. The safest approach is not to reason by analogy with resistors at all but to remember the two formulas independently, since the intuition transfers badly.

How does voltage divide across capacitors in series?

In inverse proportion to capacitance, because every capacitor in a series chain carries exactly the same charge. The smallest capacitance therefore takes the largest share of the voltage. Put a 10 µF and a 100 µF in series across 110 V and the small one sees 100 V while the large one sees only 10. This catches people out constantly: the smallest part is the one most likely to be over its rating and fail first.

Why put capacitors in series at all?

Almost always to survive a higher voltage than any single part is rated for. Two 400 V capacitors in series will tolerate 800 V, at the cost of halving the capacitance. In practice each one gets a balancing resistor across it, because real capacitors have slightly different leakage currents and would otherwise drift into an unequal voltage split, pushing one of them past its rating.