Capacitors in Series Calculator

Enter up to four capacitor values connected in series (leave unused ones at 0) to get the equivalent capacitance, plus the shared charge and per-capacitor voltage split.

Quick Facts

Series formula
1/C_eq = 1/C1 + 1/C2 + ...
The equivalent capacitance is always smaller than the smallest capacitor in the chain.
Two-capacitor shortcut
C_eq = (C1 × C2) / (C1 + C2)
Product-over-sum form, valid only when exactly two capacitors are in series.
Shared charge
Q = C_eq × V
Every capacitor in the series chain stores this same charge Q.

Your Results

Calculated
Equivalent Capacitance
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1/C_eq = 1/C1 + 1/C2 + ...
Shared Charge
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Q = C_eq × V, same on every capacitor
Voltage per Capacitor
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Vi = Q / Ci for each capacitor
Number of Capacitors
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Non-zero values used in the chain

Ready

Enter at least two capacitor values and press Calculate.

About Capacitors in Series

A capacitor stores electric charge on two conductive plates separated by an insulator, and its capacitance C (in farads) relates the charge it holds to the voltage across it: C = Q/V. When two or more capacitors are wired end-to-end in a single chain — the output terminal of one connected directly to the input terminal of the next — they are in series. Series wiring behaves very differently from parallel wiring, and the equivalent capacitance always ends up smaller than the smallest individual capacitor.

Understanding the formula

In a series chain, the same current flows through every capacitor, so each one accumulates exactly the same charge Q. The total voltage across the chain is the sum of the individual voltages: V = V1 + V2 + V3 + ... Since each Vi = Q/Ci, dividing through by the shared Q gives the reciprocal (harmonic) addition rule: 1/C_eq = 1/C1 + 1/C2 + 1/C3 + ... + 1/Cn. For exactly two capacitors this reduces to the convenient product-over-sum shortcut C_eq = (C1 × C2) / (C1 + C2). Because reciprocals are being added, C_eq is always less than whichever individual capacitor is smallest — physically, stacking capacitors in series is equivalent to increasing the effective distance between the outermost plates, which lowers overall capacitance.

Working with units and the charge/voltage split

  • Capacitance is commonly specified in picofarads (pF), nanofarads (nF), microfarads (µF), or farads (F) — 1 F = 10³ mF = 10⁶ µF = 10⁹ nF = 10¹² pF. Convert all capacitors to the same unit before combining them, which is what the unit selector above does automatically.
  • Once you know C_eq and the applied voltage V, the shared charge is Q = C_eq × V (in coulombs when C is in farads and V is in volts).
  • Each capacitor's individual voltage is then Vi = Q / Ci — smaller capacitors pick up a larger share of the total voltage, and the individual voltages always sum back to the applied voltage V.

Practical notes and common mistakes

Series capacitor banks are used in real circuits to raise the effective working voltage rating (splitting voltage across several capacitors so no single one exceeds its rated voltage) and to fine-tune an odd capacitance value from standard parts. A frequent mistake is applying the resistor-in-series rule (simple addition) to capacitors — capacitors do the opposite of resistors: they add directly in parallel and combine by reciprocals in series. Another common error is forgetting that a "0" or blank value for an unused slot must be excluded from the reciprocal sum entirely, not treated as a capacitor of zero capacitance (which would make 1/C infinite and force C_eq to zero).

Frequently Asked Questions

What is the formula for capacitors in series?
For capacitors connected in series, the reciprocals of the individual capacitances add: 1/C_eq = 1/C1 + 1/C2 + 1/C3 + ... For just two capacitors this simplifies to the product-over-sum form C_eq = (C1 × C2) / (C1 + C2). The equivalent capacitance of a series combination is always smaller than the smallest individual capacitor.
Why is the equivalent capacitance smaller than any individual capacitor?
Series capacitors share the same charge Q but split the source voltage across each one (V = V1 + V2 + ...). Since capacitance C = Q/V, keeping Q fixed while increasing the total voltage needed to store it mathematically forces the equivalent C down — the opposite of resistors in series, which add directly.
How much charge and voltage does each capacitor get?
Every capacitor in a series chain stores the same charge Q = C_eq × V_total, because charge cannot accumulate on the internal nodes between them. Each capacitor's own voltage is then Vi = Q / Ci, so smaller capacitors end up holding a larger share of the total voltage.
How is series different from parallel for capacitors?
Parallel capacitors add directly (C_eq = C1 + C2 + ...) because they share the same voltage and their plate areas effectively combine, increasing total capacitance. Series capacitors combine by reciprocals because the effective plate separation increases, which is why the series equivalent is always less than the smallest capacitor in the chain.