Capacitores em série (Cs)

Two capacitors in series

Put two capacitors in series and the equivalent capacitance follows 1/C = 1/C₁ + 1/C₂, which you can rewrite as the product-over-sum form C = (C₁ · C₂) / (C₁ + C₂). Whatever you get, it is always smaller than the smallest of the two. The same charge Q flows through both, and the voltage divides inversely with capacitance, so the smaller C ends up with the larger voltage across it. Example: two 220 µF / 16 V electrolytics in series give C = 110 µF and a theoretical rating of 32 V. On the bench, though, you should add 100 kΩ balancing resistors in parallel with each capacitor, otherwise a mismatch in leakage current can push one of them above its rated voltage.

Applications: DC link banks and capacitive dividers

You see two capacitors in series all the time in the high-voltage DC link banks of three-phase inverters and VFDs, where a pair of 450 V electrolytics handles an 800 V bus (with balancing resistors). The same trick builds capacitive voltage dividers for AC metering, lets you hit precise non-standard values in tuned LC and RF circuits, and gets you to voltage ratings that no single component on the shelf can match.

FAQ

Why is the equivalent smaller than the smallest? Stacking them in series spreads the outer plates farther apart without adding any plate area, and that drops the capacitance.

How does voltage split between them? Inversely with capacitance, following V₁ = V·C₂/(C₁+C₂). Take a 10 µF in series with a 100 µF at 110 V and the 10 µF sees 100 V of it.

Do I really need balancing resistors? For high-voltage banks, yes. Real capacitors never leak at exactly the same rate, and without parallel resistors (100 kΩ–1 MΩ) one of them charges past its rating and fails.

Polarity for electrolytics? Keep both facing the same way, or switch to bipolar (non-polarized) types when there is AC in the path.