Resistors in parallel add like conductances and in series like resistances — but capacitors do the exact opposite. Put capacitors in parallel and their capacitances simply add; put them in series and the total is smaller than the smallest one. Worse, in series the voltage does not split evenly: the smallest capacitor takes the biggest share of the voltage, which can destroy it. This post works the numbers for the same three capacitors in both arrangements.
Figure 1: The same three capacitors in parallel (a, top) and in series (a, bottom), and the voltage each one takes in series (b). In parallel every capacitor sees the full 12 V; in series the 10 µF capacitor takes 7.20 V of it. (Schematic, not to scale.)
In parallel — capacitances add
In parallel every capacitor connects across the same voltage, so each one stores its own charge Q = C × V and the charges add. The equivalent capacitance is just the sum:
Ceq = C1 + C2 + C3
Take 10 µF, 22 µF and 47 µF at 12 V. Then Ceq = 10 + 22 + 47 = 79 µF. The total charge is Q = 79 µF × 12 V = 948 µC — made up of 120 µC on the 10 µF capacitor, 264 µC on the 22 µF one and 564 µC on the 47 µF one (120 + 264 + 564 = 948 µC, as it must). The stored energy is
E = ½ Ceq V2 = 0.5 × 79 × 10−6 × 144 = 5.69 mJ
(0.72 mJ + 1.58 mJ + 3.38 mJ on the three capacitors — the energies add, as they must.) Paralleling capacitors is the everyday way to get a bigger capacitance: it is how power-supply filter banks and decoupling networks are built.
In series — the reciprocal rule, and uneven voltage
In series the same current flows into every capacitor, so every capacitor ends up with the same charge Q — but since V = Q / C, the voltage across each one is inversely proportional to its capacitance. The equivalent capacitance follows the reciprocal rule:
1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3
With the same three capacitors: 1 / Ceq = 1/10 + 1/22 + 1/47 = 0.100 + 0.0455 + 0.0213 = 0.1667 µF−1, so Ceq = 6.00 µF — smaller than the smallest capacitor in the chain. At 12 V the common charge is Q = 6.00 µF × 12 V = 72.0 µC, and the voltages are:
V1 = 72 / 10 = 7.20 V · V2 = 72 / 22 = 3.27 V · V3 = 72 / 47 = 1.53 V
Check: 7.20 + 3.27 + 1.53 = 12.0 V — the voltages add up to the supply, exactly as Kirchhoff's voltage law requires. The stored energy is now only ½ × 6.00 µF × 144 = 0.432 mJ (259 µJ + 118 µJ + 55 µJ) — about thirteen times less than the parallel arrangement at the same voltage.
The trap: voltage rating
Look at the series result again: the 10 µF capacitor — the smallest — takes 7.20 V, more than half the supply. If that capacitor were rated for only 6.3 V it would be overstressed and could fail, even though the supply is only 12 V. Always check the voltage on the smallest series capacitor, not just the total. Series connection is still useful: two 100 µF, 25 V capacitors in series make a 50 µF capacitor rated for roughly 50 V (in practice, add equal balancing resistors across each one, because leakage currents differ and the voltage will not split evenly on its own).
Key points to remember
- Parallel: Ceq = C1 + C2 + C3 — 10 + 22 + 47 = 79 µF at 12 V, holding 948 µC and 5.69 mJ, with every capacitor at the full 12 V.
- Series: 1/Ceq = 1/C1 + 1/C2 + 1/C3 — the same three give only 6.00 µF, smaller than the smallest one.
- In series every capacitor carries the same charge (72.0 µC here), so the voltages split in inverse proportion to capacitance: 7.20 V, 3.27 V and 1.53 V, adding to 12.0 V.
- Stored energy is far lower in series (0.432 mJ vs 5.69 mJ here) — series connection trades capacitance for voltage rating.
- Check the voltage rating of the smallest series capacitor: it takes the largest voltage (7.20 V here) and is the one that fails first.