Capacitors in series and parallel
Key idea: H2 Physics lessons on circuit representation, material behaviour, networks and RC transients.
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The core idea
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Big question: Why do capacitor combination rules look opposite to resistor rules?
Parallel capacitors share potential difference and their stored charges add, giving Ctotal = ΣC. Series capacitors carry equal charge magnitude while potential differences add, giving 1/Ctotal = Σ(1/C). Derive the rule from Q = CV instead of relying on memory.
Add parallel capacitors by common voltage
Capacitors connected across the same two nodes share potential difference V. Their charges are Q₁ = C₁V and Q₂ = C₂V, while the source supplies Q = Q₁ + Q₂. Therefore C_eq = C₁ + C₂ + ….
Parallel combination increases plate area in effect, so equivalent capacitance exceeds every individual value. This physical check catches accidental use of the reciprocal rule.
Check your understanding: Two capacitors 3 µF and 5 µF are in parallel at 12 V. Find total charge.
C_eq = 8 µF, so Q = CV = 96 µC.
Add series voltages at common charge
Initially uncharged capacitors in series acquire equal charge magnitude because the connecting conductor cannot gain net charge. Their voltages Q/C add to the supply, giving 1/C_eq = 1/C₁ + 1/C₂ + ….
The smaller capacitor has the larger voltage for the same Q. Equivalent series capacitance is smaller than the smallest individual capacitance; unequal series capacitors do not share voltage equally.
Check your understanding: A 2 µF and 6 µF capacitor are in series. Which has greater p.d.?
The 2 µF capacitor. With equal Q, V = Q/C, so its p.d. is three times larger.
Key ideas to keep
- Series capacitance is smaller than the smallest individual capacitance.
- Parallel capacitance is larger than any individual capacitance.
- Equal series charge does not imply equal voltage unless capacitances are equal.
See the reasoning
Worked example
Derive both combination rules instead of swapping a memory rule
Question: Explain why capacitor-combination rules are opposite in form to resistor rules.
Step 1: Use common voltage in parallel
Why: Every parallel capacitor joins the same two nodes.
Working: Qtotal = ΣCᵢV = C_eqV, so C_eq = ΣCᵢ.
Step 2: Use common charge in series
Why: The isolated connecting plates acquire equal and opposite charge.
Working: Vtotal = Σ(Q/Cᵢ) = Q/C_eq.
Step 3: Read off the reciprocal rule
Why: Cancel the shared Q from the series voltage equation.
Working: 1/C_eq = Σ(1/Cᵢ).
Answer: Parallel capacitors share p.d. and their plate charges add, so Ceq = ΣC. Series capacitors carry equal charge while their p.d.s add, so 1/Ceq = Σ(1/C). The result follows from Q = CV, not from memorising an analogy.
Check: Parallel C exceeds every branch value; series C is below the smallest value.
Use a hint if needed
Practise with support
Try this
Three 12 μF capacitors are in series. Find Ceq.
Hint: Add reciprocals, not capacitances.
Check your answer
For equal capacitors in series, Ceq = C/3 = 4.0 μF.
Now work without the hint
Practise independently
Your turn
Derive both capacitor rules from shared charge or shared p.d.
Check your answer
Parallel capacitors share V, so Qtotal = ΣCiV = CeqV and Ceq = ΣCi. Series capacitors have equal Q and Vtotal = Σ(Q/Ci) = Q/Ceq, giving 1/Ceq = Σ(1/Ci).
Avoid these traps
Common mistakes
Common mistake
Capacitances combine exactly like resistances.
What is wrong with this reasoning?
Show better thinking
Capacitances add in parallel; reciprocal capacitances add in series.
Write for the examiner
Exam guidance
Annotate which quantity is common—Q in series or V in parallel—before calculating.
Exam-style practice [5 marks]
Find Ceq for 4.0 μF in parallel with a series pair of 6.0 μF and 3.0 μF.
Plan before you answer
- Reduce the series pair first.
- Then add the parallel branch.
- Check the limiting size.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
The series pair is 2.0 μF. In parallel with 4.0 μF, total capacitance is 6.0 μF.
Come back in three days
Check what stayed with you
Recall question
Two equal capacitors C are connected first in series and then in parallel. State both equivalent capacitances.
Check the answer
Series gives C/2; parallel gives 2C.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Topic 16 states no explicit exclusions. I–V graph gradient language always names the plotted axes; resistance is V/I at an operating point rather than automatically a graph gradient. Metal temperature dependence is explained through drift velocity and NTC behaviour through carrier number density. Terminal p.d. is E − Ir while a source supplies current. Capacitor rules are not copied from resistor rules. The exponential forms assume constant R and C, an ideal source or isolated discharge loop, and τ = RC.
- GCE A-Level H2 PhysicsTopic 16(k) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027