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Capacitors in Series Calculator

Find the combined capacitance of series capacitors.

How it is calculated

The reciprocals of the capacitances are added and the sum is inverted: Ct = 1 ÷ (1/C1 + 1/C2 + 1/C3). Enter all values in the same unit (for example µF); the answer comes back in that unit with four decimals. C1 and C2 must be greater than zero, and the third capacitor is optional: if it is blank, zero or negative it is ignored. For exactly two capacitors the formula reduces to C1C2 ÷ (C1 + C2). The total is always below the smallest single value.

Worked example

Three capacitors of 10, 22 and 47 µF are in series. Reciprocals: 1/10 = 0.1, 1/22 = 0.045455 and 1/47 = 0.021277. Their sum is about 0.166732. Inverting gives 1 ÷ 0.166732 = 5.9977. The series total is 5.9977 µF, smaller than the 10 µF capacitor, as it must be.

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Common mistakes

  • Adding the values like parallel capacitors. Series capacitance is always lower than the smallest part, so a result larger than that value signals this error.
  • Mixing units, such as 100 nF with 10 µF, without converting first. The tool does not convert, so the answer would be meaningless.
  • Assuming the voltage splits equally. It divides in inverse proportion to capacitance, so the smallest capacitor sees the largest share; unequal values need a voltage-rating check.
  • Forgetting that a blank or zero third field is simply ignored, so a third capacitor you meant to include is silently left out.

Sources

About this calculator

When capacitors are connected in series, their combined (total) capacitance is always smaller than the smallest individual capacitor, following the reciprocal formula 1/Cₜ = 1/C₁ + 1/C₂ + 1/C₃ + ... — the mirror image of resistors: capacitors in series combine the way resistors in parallel do. Each capacitor in the chain ends up storing the same charge, but the applied voltage divides unevenly across them in inverse proportion to their capacitance.

Electronics hobbyists and engineers connect capacitors in series to achieve a higher effective voltage rating than any single capacitor could handle alone (since the voltage splits across the series chain), or simply to reach a capacitance value that isn't available as a single off-the-shelf component.

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Comments (1)

  • Editorial team

    Whatever unit you type in, the total comes out in that same unit, so keep all values in microfarads or all in picofarads. In a series string the voltage divides in inverse proportion to capacitance, which means the smallest capacitor sees the largest share; check its rating, not just the total. A zero or empty third field is simply ignored. For four or more, feed the result back in as the first value. How many capacitors are in your string?

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