Capacitors in Series Calculator

Formula

How this is calculated

When capacitors are connected in series, the same charge Q must pass through each one in turn. Because each capacitor must store the same charge, the total effect is as if the dielectric gaps are stacked in sequence — increasing the effective plate separation and decreasing the total capacitance. The relationship is analogous to resistors in parallel: the reciprocals of the individual capacitances add, giving 1/Cs = 1/C1 + 1/C2 + … + 1/Cn, so the series equivalent is Cs = 1 / (sum of reciprocals).

As a result, the series equivalent is always smaller than the smallest capacitor in the chain. Adding more capacitors in series reduces the equivalent further. For two equal capacitors C in series, the result is C/2. For capacitors of very different values, the smallest capacitor dominates — a 10 µF in series with a 1000 µF gives almost 10 µF, because the small capacitor is the bottleneck.

All values must be entered in the same unit — pick the unit from the dropdown and enter every capacitor consistently. Stray capacitance, tolerance (typically ±5–20%) and voltage distribution among series capacitors (unequal values mean unequal voltages) are not accounted for. In high-voltage applications, resistors are placed in parallel with each series capacitor to equalise the voltage sharing.