Combine capacitors in series and parallel
Derive and use equivalent capacitance rules by tracking which quantity is common in each circuit arrangement.
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 19. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use charge, potential difference, current, resistance, graph area, energy and SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In a series capacitor combination, which quantity is common to each capacitor?
What you will learn
- Derive the series-capacitance rule from C = Q/V
- Derive the parallel-capacitance rule from C = Q/V
- Calculate equivalent capacitance for series and parallel combinations
Series capacitors share charge
For capacitors in series, each capacitor has the same charge magnitude Q. Their potential differences add: Vtotal = V₁ + V₂ + …. Since Vᵢ = Q/Cᵢ and Vtotal = Q/Ceq, dividing by the common Q gives 1/Ceq = 1/C₁ + 1/C₂ + ….
For two capacitors, Ceq = C₁C₂/(C₁ + C₂). The series equivalent is smaller than the smallest individual capacitance, a useful check on the calculation.
Parallel capacitors share potential difference
For capacitors in parallel, each has the same potential difference V. The total stored charge is the sum: Qtotal = Q₁ + Q₂ + …. With Qᵢ = CᵢV and Qtotal = CeqV, dividing by V gives Ceq = C₁ + C₂ + ….
The parallel equivalent exceeds the largest individual capacitance. These rules follow from C = Q/V; remember which circuit quantity stays common before choosing a formula.
Pause and explain
What is the equivalent capacitance of 2.0 μF and 5.0 μF capacitors in parallel?
Worked example
Find the equivalent capacitance of 6.0 μF and 3.0 μF capacitors (a) in series and (b) in parallel.
Show the worked solution
- For series, use 1/Ceq = 1/C₁ + 1/C₂.
- For parallel, add the capacitances.
- Check that series gives less than 3.0 μF and parallel gives more than 6.0 μF.
Answer Series: Ceq = 1/(1/6.0 + 1/3.0) = 2.0 μF. Parallel: Ceq = 6.0 + 3.0 = 9.0 μF.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Calculate the equivalent capacitance of 4.0 μF and 12 μF capacitors in series.
Give me a hint
Add reciprocals, then take the reciprocal of the result.
Compare my reasoning
- 1/Ceq = 1/4.0 + 1/12 = 1/3 μF⁻¹.
- Ceq = 3.0 μF.
- This is below the smaller capacitor, as a series check requires.
Ceq = 3.0 μF.
Look for these in your work
- Series rule selected
- Reciprocals combined
- Magnitude checked
Three capacitors of 2.2 μF, 4.7 μF and 10 μF are connected in parallel. Find Ceq.
Give me a hint
Add all capacitances in parallel.
Compare my reasoning
- Ceq = 2.2 + 4.7 + 10.
- Ceq = 16.9 μF.
- The result exceeds the largest individual capacitance.
Ceq = 16.9 μF.
Look for these in your work
- Parallel rule selected
- All three values included
- Magnitude checked
Two equal capacitors each have capacitance C. Compare their equivalent capacitance in series and in parallel.
Give me a hint
Apply the two combination rules with both values equal to C.
Compare my reasoning
- In series, 1/Ceq = 1/C + 1/C = 2/C.
- So Cseries = C/2.
- In parallel, Cparallel = C + C = 2C.
Series gives C/2; parallel gives 2C, four times the series equivalent.
Look for these in your work
- Series derivation shown
- Parallel derivation shown
- Ratio compared
Common mistakes
- Adding capacitances in series
- Adding reciprocals in parallel
- Using equal charge for parallel capacitors
- Forgetting the series equivalent must be below the smallest capacitor
What can you explain now?
In a series capacitor combination, which quantity is common to each capacitor?
Compare with the explanation
Charge magnitude
The series elements carry the same charge magnitude.
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Make it stick
Which quantity is shared in series and in parallel, and how does that produce the equivalent-capacitance rules?
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