Electronics & Circuits
Intermediate

Capacitors in Series

Reciprocals of series capacitances add.

Formula

1Ceq=1C1+1C2+⋯\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots

Variables

C_iIndividual capacitances

How it works

When capacitors are connected end-to-end in series, the same charge must pass onto each capacitor, so each one stores the same magnitude of charge. The total voltage across the chain is the sum of the individual voltage drops, and because V = Q/C, the voltages add in a way that makes the reciprocals of the capacitances add. This means the series combination always has a capacitance smaller than the smallest single capacitor in the set.

When to use it

Use this formula when you need the equivalent capacitance of capacitors connected in one continuous path, such as in circuit analysis, filter design, or when replacing several capacitors with a single equivalent component. It is especially useful whenever the capacitors are not connected across the same two nodes.

Worked example, step by step

Suppose three capacitors are connected in series: C1 = 4.0 μF, C2 = 6.0 μF, and C3 = 12.0 μF. 1/Ceq = 1/C1 + 1/C2 + 1/C3 1/Ceq = 1/4.0 + 1/6.0 + 1/12.0 1/Ceq = 0.25 + 0.1667 + 0.0833 1/Ceq = 0.5000 μF^-1 Ceq = 1 / 0.5000 = 2.0 μF Final answer: The equivalent capacitance is 2.0 μF.

Common mistakes to avoid

  • Adding series capacitances directly instead of adding reciprocals
  • Forgetting that the equivalent capacitance in series must be smaller than the smallest capacitor
  • Mixing up series and parallel rules for capacitors and using the wrong formula

Did You Know?

Capacitors in series combine like resistors in parallel, always giving less than the smallest one.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026