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Circuits overview

Topic 7 of 9

Capacitor combinations

Capacitor combinations follow charge and voltage constraints. Identify which plates share a node and whether charge can enter an intermediate node before choosing a rule.

For a capacitor with equal and opposite plate charges, C = Q/V, where Q is the magnitude on one plate and V is the p.d. magnitude. The net charge of the two-plate device can be zero while it stores energy. Capacitance has unit farad: 1 F = 1 C/V, with 1 µF = 10-6 F and 1 µC = 10-6 C. The capacitance explanation develops these quantities.

Parallel: the same p.d. across both capacitors

Parallel: the same p.d. across both capacitorsA twelve-volt source connects node A to the upper plates of separate three- and six-microfarad capacitors. Their lower plates both join C. The three-microfarad capacitor has plus and minus thirty-six microcoulomb on its plates; the six-microfarad capacitor has plus and minus seventy-two microcoulomb. The sum of positive plate charges is one hundred and eight microcoulomb. No conductor or current arrow crosses either dielectric gap.+-AC12 V3 µF6 µF+-+-|Q| = 36 µC|Q| = 72 µC

Both capacitors have 12 V across A/C. The equivalent is 9 microfarad; the source supplies 108 microcoulomb to the positive-plate side.

Series: the isolated middle conductor has zero net charge

Series: the isolated middle conductor has zero net chargeInitially uncharged three- and six-microfarad capacitors are in series between A at twelve volts and C at zero. The top capacitor plates carry plus twenty-four and minus twenty-four microcoulomb. The lower capacitor plates carry plus twenty-four and minus twenty-four microcoulomb. The two middle plates and their connecting wire form node B, whose negative and positive charges sum to zero, while its potential is four volts. The upper capacitor has eight volts across it and the lower four volts.+-A: 12 VB: 4 VC: 0 V12 V3 µF6 µF+24 µC-24 µC+24 µC-24 µCDashed boundary encloses node B.

All four plate-charge magnitudes are 24 microcoulomb. B has zero net charge under the stated initial conditions; zero net charge does not imply zero potential.

A third capacitor joins A to C

A third capacitor joins A to CThe series branch has three microfarads from A to B and six microfarads from B to C. A separate four-microfarad capacitor connects directly from A to C across the twelve-volt source. The series branch equivalent is two microfarads, giving a total of six microfarads. Source charge is twenty-four plus forty-eight, or seventy-two microcoulomb, not the sum of magnitudes on every plate.+-ABC12 V3 µF6 µF4 µF+-+-+-A/C branches share the same 12 V.

Reduce the 3/6 microfarad series branch first, then add the parallel 4 microfarad branch. Source charge is 72 microcoulomb.

The parallel pair shares both supply nodes. The series pair has an isolated, initially neutral middle node B, so its two plate charges cancel at B. The mixed circuit adds a separate branch across the same A/C supply.

Parallel: the same p.d.

Capacitors in parallel share both endpoints and therefore V. The total charge supplied to the positive plates is the sum:

Qtotal = C1V + C2V + …
Cparallel = C1 + C2 + …

For 3.0 µF and 6.0 µF in parallel across 12 V, Cequivalent = 9.0 µF. The plate-charge magnitudes are 36 and 72 µC, giving total supplied charge 108 µC. Do not double this by adding both positive and negative plate-charge magnitudes.

Series: the middle-node charge constraint

Start with uncharged capacitors and keep their intermediate node isolated from any other conducting connection. No net charge can enter that node, so its two connected plates acquire equal and opposite charges. The series capacitors therefore have equal plate-charge magnitudes Q, while their voltages add:

V = Q/C1 + Q/C2 + …
1/Cseries = 1/C1 + 1/C2 + …

For 3.0 µF on A/B and 6.0 µF on B/C across 12 V:

Cseries = (3.0 × 6.0)/(3.0 + 6.0) = 2.0 µF
Q = (2.0 µF)(12 V) = 24 µC
VAB = 24/3.0 = 8.0 V
VBC = 24/6.0 = 4.0 V

With C at 0 V and A at 12 V, B is at 4 V. Its connected plates carry -24 and +24 µC, so B has zero net charge, not zero potential. The smaller capacitance has the greater p.d. under this equal-charge condition.

Initially charged capacitors can leave a nonzero net charge on an isolated middle node. Then conserve that given node charge rather than assuming equal magnitudes automatically. The combination rules above use the stated initial and connection conditions.

A mixed arrangement and an energy check

Add a 4.0 µF capacitor directly across A/C, in parallel with the 2.0 µF series equivalent. The complete capacitance is 6.0 µF. At 12 V the added branch has 48 µC, while the series branch still has 24 µC, giving total supplied charge 72 µC.

For an ideal capacitor, stored energy is U = ½CV2. Add each component's energy using its own voltage:

Stored energy for the supplied 12 V arrangements
ArrangementComponent energies / µJTotal / µJ
3 and 6 µF parallel216 + 432648
3 and 6 µF series96 + 48144
Series pair plus 4 µF branch96 + 48 + 288432

Each total also equals ½CequivalentVsupply2. These are final stored energies, not automatically all the energy supplied during charging. An ideal capacitor has no steady conduction through its dielectric; the charging process determines the transient currents and other energy transfers.

Optional check Initially uncharged 3.0 and 6.0 microfarad capacitors are placed in series from A through isolated middle conductor B to C. A is at 12.0 V and C at 0 V; B retains zero net charge. Which final state is correct?
Initially uncharged 3.0 and 6.0 microfarad capacitors are placed in series from A through isolated middle conductor B to C. A is at 12.0 V and C at 0 V; B retains zero net charge. Which final state is correct?