University Physics II · Capacitance and Dielectrics · 5.2
Capacitors in Series & Parallel
Two capacitors across the same pair of nodes share a potential difference. Two joined end to end share a charge, forced by the isolated conductor between them. Which constraint holds decides how the capacitances combine.
Build the model
Connect the measurement to the mechanism.
Every capacitor obeys C = Q/ΔV, so combining them is a matter of working out which of Q and ΔV the wiring forces to be common. Wire two between the same pair of nodes and they share ΔV, so their charges add and their capacitances add. Join them end to end and the conductor between them is isolated: it began neutral and cannot change its total charge, so induction gives both the same magnitude of charge while their potential differences add, the reciprocals add, and the pair is weaker than either member.
The rest is bookkeeping. Collapse series and parallel groups until one equivalent capacitance faces the source, find the charge it draws, then walk back through the reduction applying the common-ΔV rule at every parallel join and the common-Q rule at every series join. The equal-charge rule is not a property of the word series; it is charge conservation on one floating node, which fails the moment anything else is wired to it.
- Simple definition
- The equivalent capacitance of a network is the single capacitance that would draw the same charge from the same source: Ceq = Q/ΔV. Parallel members share ΔV and their capacitances add; series members share Q and their reciprocals add.
- Example
- A 12.0 μF and a 4.0 μF in series give Cₛ = (12.0 × 4.0)/16.0 = 3.0 μF. Put 5.0 μF in parallel with that branch and the network presents 8.0 μF, which at 24.0 V draws 192 μC.
Q is the magnitude on one plate, not the sum: the pair as a whole stays neutral.
1 F = 1 C V⁻¹; 1 μF at 1 V holds 1 μC
Parallel plates in parallel is added area at fixed separation, so the largest member sets the scale.
Charges add: Q = Cₚ ΔV. Work in μF and V and the charges come out in μC.
Cₛ falls below the smallest member; n identical capacitors in series give C/n.
Two members only: Cₛ = C₁C₂/(C₁ + C₂). Never sum reciprocals in mixed units.
The smallest capacitance takes the largest share, so check it against its voltage rating.
Volts; the shares sum to the applied ΔV because potential drops accumulate
This, not the word series, is what forces equal charge along a chain of capacitors.
q₀ in C is the net charge trapped on the middle conductor before wiring
Charge is the conserved quantity when charged capacitors are joined, never voltage or energy.
Signed charges, summed over the plates joined at the node; parallel geometry only
Read the wiring, not the picture
A capacitor network is defined by its nodes: sets of conductor joined by wire, each sitting at one potential. Two capacitors are in parallel when one plate of each lands on the same node and the other plate of each lands on a second node, so both span the same potential difference no matter what the rest of the circuit does. They are in series when they are joined end to end and the conductor between them touches nothing else, which leaves that middle conductor isolated and its total charge fixed. Neither test says anything about how the diagram is drawn. Two capacitors sketched side by side can be in series, and a chain redrawn as a loop can be in parallel; count the nodes and the ambiguity disappears. Settle this first, because the constraint the wiring selects — common ΔV or common Q — is what every combination formula below is built from.
Parallel: the capacitances add
Put C₁ and C₂ between the same two nodes and both carry the applied ΔV. Each obeys its own defining relation, so Q₁ = C₁ΔV and Q₂ = C₂ΔV, and the charge drawn from the source is their sum, Q = (C₁ + C₂)ΔV. Divide by ΔV: Cₚ = C₁ + C₂ + …, for any number of members. The equivalent exceeds the largest member, and for parallel plates the reason is geometric — wiring plates in parallel is a way of adding area at fixed separation, and C = ε₀A/d is linear in A. Two 4.7 μF units in parallel present 9.4 μF. A 0.010 μF added in parallel with a 4.7 μF gives 4.710 μF, a shift of 0.2%, which is why a small stray capacitance across a large one is usually ignorable and a small one in series is not.
Series: the reciprocals add
Join C₁ and C₂ end to end across a source. Each carries the same magnitude of charge Q — the next section says why — so their potential differences are ΔV₁ = Q/C₁ and ΔV₂ = Q/C₂. Those add, because the drops accumulate as you walk along the chain: ΔV = Q(1/C₁ + 1/C₂). Divide by Q and 1/Cₛ = 1/C₁ + 1/C₂ + …, which for two members rearranges to Cₛ = C₁C₂/(C₁ + C₂). With C₁ = 12.0 μF and C₂ = 4.0 μF, Cₛ = 48.0/16.0 = 3.0 μF, below either one. That is the pattern in series: the equivalent sits under the smallest member, and n identical capacitors give C/n. Geometry agrees again — stacking two identical parallel-plate capacitors end to end is effectively doubling d at fixed A, which halves C.
Equal charge comes from an isolated node
Nothing pumps charge onto the middle conductor of a series pair. A wire only moves charge between the plates it connects, and no wire reaches that conductor. Start with everything neutral and connect the chain to a source: it draws +Q onto the far plate of C₁ and leaves −Q on the far plate of C₂. The middle conductor still holds zero net charge, but it polarises, with −q₁ appearing on the plate facing C₁ and +q₂ on the plate facing C₂. Conservation then reads −q₁ + q₂ = 0, so q₁ = q₂ = Q, and every capacitor in the chain carries the same magnitude of charge whatever its capacitance. Stated that way, the failure conditions come free. Trap a net charge q₀ on that conductor before wiring and the balance becomes −q₁ + q₂ = q₀, so the two charges differ. Run a third wire to it and it is no longer isolated, the junction is not a series junction, and the equal-charge rule does not apply there at all.
Reduce forward, then walk back
Collapse the network until one capacitance faces the source, then reverse the steps to recover the individual charges and voltages. Take C₁ = 12.0 μF in series with C₂ = 4.0 μF, that branch in parallel with C₃ = 5.0 μF, all across 24.0 V. Forward: the series pair is 3.0 μF, and in parallel with 5.0 μF that is 8.0 μF, so the source delivers Q = 8.0 μF × 24.0 V = 192 μC. Back: at the parallel join both branches hold 24.0 V, so C₃ takes 5.0 × 24.0 = 120 μC and the series branch takes 3.0 × 24.0 = 72 μC. At the series join both members carry that same 72 μC, giving ΔV₁ = 72/12.0 = 6.0 V and ΔV₂ = 72/4.0 = 18.0 V. Two checks close the work: 6.0 + 18.0 = 24.0 V along the branch, and 120 + 72 = 192 μC at the source. Energy agrees as well, since 216 μJ + 648 μJ + 1440 μJ = 2304 μJ matches ½ × 8.0 μF × (24.0 V)².
Checks, limits, and networks that will not reduce
Three tests catch most errors: Cₚ must exceed every parallel member, Cₛ must fall below every series member, and the branch voltages must sum to the applied ΔV while the branch charges sum to the charge drawn. Limits sharpen the habit. A 1.0 pF in series with a 10 μF gives 1.0 pF to six figures, because the small member dominates a series chain; the same 1.0 pF in parallel with 10 μF is invisible. That dominance is a design constraint, not a curiosity: a 1.0 nF and a 10 nF in series across 400 V share 364 nC, leaving 364 V on the 1.0 nF and only 36 V on the 10 nF, so the small unit is the one that has to survive. And not every network reduces. A five-capacitor bridge has no pure series or parallel pair anywhere in it; you assign node potentials and impose charge conservation at each floating node, which is the capacitor form of the rules coming in Unit 7.
Change one variable at a time
Make the relationship visible.
Leave C₁ near 12 μF and drag C₂ down to 1 μF: the parallel bar gives up only the 3 μF you removed and still clears the upper dashed line, while the series bar collapses onto C₂ itself and the shaded block swells to fill nearly the whole column — the smallest member limits the chain and takes almost all of the voltage.
SERIES Cₛ3.00 μF
PARALLEL Cₚ16.0 μF
CHARGE ON EACH, SERIES72.0 μC
CHARGE DRAWN, PARALLEL384.0 μC
Live interpretationSERIES Cₛ: 3.00 μF. PARALLEL Cₚ: 16.0 μF. CHARGE ON EACH, SERIES: 72.0 μC. CHARGE DRAWN, PARALLEL: 384.0 μC
Catch the common trap
Explain before calculating.
A 3.0 μF and a 6.0 μF capacitor, both initially uncharged, are connected in series across a 9.0 V battery. What charge does each hold, and how does the 9.0 V divide?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA 6.0 μF and a 3.0 μF capacitor. Find the equivalent capacitance when they are wired in parallel, and when they are wired in series.
- Parallel means one plate of each lands on the same node and the other plate of each on a second node, so ΔV is common and the capacitances add: Cₚ = 6.0 + 3.0 = 9.0 μF.
- Series means the conductor between them touches nothing else, so Q is common and the reciprocals add: 1/Cₛ = 1/6.0 + 1/3.0 = 0.1667 + 0.3333 = 0.5000 μF⁻¹.
- Invert the sum, never the terms: Cₛ = 1/0.5000 = 2.0 μF. The two-member shortcut agrees, Cₛ = (6.0 × 3.0)/9.0 = 18/9.0 = 2.0 μF.
- Rank check: 9.0 μF sits above the larger member and 2.0 μF sits below the smaller one, as parallel and series always must.
AnswerCₚ = 9.0 μF; Cₛ = 2.0 μF
MediumA 1.0 nF and a 10 nF capacitor, both initially uncharged, are wired in series across 400 V. Find the charge on each and the potential difference across each, and say which capacitor the voltage rating has to protect.
- Series equivalent: Cₛ = C₁C₂/(C₁ + C₂) = (1.0 × 10)/(1.0 + 10) = 10/11 = 0.909 nF, below the smaller member as expected.
- Charge drawn: Q = Cₛ ΔV = 0.909 nF × 400 V = 363.6 nC ≈ 364 nC. The middle conductor is isolated and started neutral, so that same 364 nC sits on both capacitors — it is not shared out.
- Divide the voltage with ΔVᵢ = Q/Cᵢ: ΔV₁ = 363.6 nC ÷ 1.0 nF = 364 V, and ΔV₂ = 363.6 nC ÷ 10 nF = 36 V.
- Check and read it: 364 + 36 = 400 V along the chain. The 1.0 nF, the smaller capacitance, takes 363.6/400 = 91% of the supply, so it is the unit that must be rated for 400 V.
Answer364 nC on each; 364 V across the 1.0 nF and 36 V across the 10 nF — the smaller capacitor carries 91% of the supply and sets the rating
HardA 2.0 μF and a 4.0 μF capacitor are wired in parallel, and that combination is in series with a 12 μF capacitor across a 15 V supply. Find the equivalent capacitance, then the charge and potential difference for every capacitor.
- Reduce the parallel pair first: both span the same two nodes, so Cₚ = 2.0 + 4.0 = 6.0 μF.
- That block is in series with the 12 μF: Ceq = (6.0 × 12)/(6.0 + 12) = 72/18 = 4.0 μF, which is what the supply actually faces.
- Charge drawn: Q = Ceq ΔV = 4.0 μF × 15 V = 60 μC. The series join is one isolated node, so the 12 μF and the whole parallel block each carry that same 60 μC.
- Walk back through the series join with ΔV = Q/C: 60 μC ÷ 12 μF = 5.0 V across the 12 μF, and 60 μC ÷ 6.0 μF = 10.0 V across the block. They sum to 15 V, as the loop demands.
- Walk back through the parallel join with Q = CΔV, both members at 10.0 V: 2.0 μF × 10.0 V = 20 μC and 4.0 μF × 10.0 V = 40 μC. Those add to 60 μC, back to the charge drawn.
AnswerCeq = 4.0 μF; 60 μC through the series join, giving 5.0 V on the 12 μF and 10.0 V across the parallel block, which holds 20 μC on the 2.0 μF and 40 μC on the 4.0 μF