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CR · Combining capacitors

Capacitors in Series and Parallel

If the rules for combining resistors in series and parallel feel familiar but capacitors confuse you, hold on to one idea: a capacitor is the exact opposite of a resistor. In parallel the capacitances simply add, and in series their reciprocals add; you learn why by reasoning with plate area and shared charge.

Parallel widens the plates and raises capacitance

To C1 = 4 μF a second capacitor C2 is connected in parallel, and moving C2 changes its value. The top bar is the parallel capacitance C_p = C1 + C2, with the darker segment for the first capacitor and the lighter one for the second. The green bar below is the capacitance of the same two capacitors in series, and however large the second one gets, it never grows past the darker segment of the top bar. Set the parallel capacitance to the target of 10 μF on the orange dashed line by adjusting C2.

Second capacitance C2C2 = 2.0 μF
Parallel and series capacitance
C_p = 6.0 μF · C_s = 1.3 μF
Far from 10 μF

In parallel, plate areas add

A capacitor's capacitance is proportional to the plate area and inversely proportional to the plate spacing. In C = εA/d, doubling the area A doubles the capacitance. Connect two capacitors in parallel and both pairs of plates take the same voltage together, which is no different from widening one plate sideways to add area. So the parallel capacitance is C_p = C1 + C2, the plain sum of the two. At the same voltage a larger area holds more charge, so the combination is larger than either one, and that addition is also why the top bar looks like the darker and lighter segments joined end to end.

In series, reciprocals add

In series, the charge coming in from outside fills the two capacitors in turn, so the same amount of charge builds up on both. The charge Q is shared and only the voltage splits in two, V = V1 + V2, and since V = Q/C, Q/C_s = Q/C1 + Q/C2, that is, 1/C_s = 1/C1 + 1/C2. Rearranged, C_s = C1·C2/(C1+C2), so the series capacitance is always smaller than even the smaller of the two. With equal areas it is as if the two gaps were joined end to end and the plates moved farther apart, so a drop in capacitance is natural.

The exact opposite of resistors

Resistors add directly in series, and their reciprocals add in parallel. Capacitors swap the two: they add in parallel and their reciprocals add in series. Inductors are on the resistors' side and add in series, leaving capacitors alone on the other side. When the rules blur, think of area: parallel means wider plates, so capacitance grows, and series means plates moved apart, so it shrinks.

ObserveCp = C1 + C2
In parallel, capacitances add.
Choose1Cs = ?
In series, reciprocals add.
Fill inCs = ?
Series capacitance is the product over the sum.
On your ownC ∝ ? (C = εAd)
Capacitance is proportional to plate area.

Back to the first screen

As you raised C2, the parallel bar on top grew by exactly that much, and it reached the target of 10 μF when C2 was 6 μF (C1 = 4). Even then the series bar below was shorter than the darker segment of the top bar, showing plainly that a series combination is smaller than even the smaller capacitor. Parallel grows because widening the plates adds area, and series shrinks because the charge is shared, so the voltage splits and the reciprocals add. Turn the resistor rules inside out and you get the capacitor rules.

Connect capacitors in parallel and the plate areas add, raising the capacitance to Cp = C1 + C2; connect them in series and the same charge builds up on both while the voltage splits, giving 1/Cs = 1/C1 + 1/C2, that is, Cs = C1·C2/(C1+C2), smaller than either. The reason lies in C = εA/d, where capacitance is proportional to area. This is the mirror image of the resistor rules for series and parallel, and inductors fall on the same side as resistors.