Add potential differences across series components
Balance the total p.d. and find missing shares without assuming equal voltages across unequal resistors.
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028. Use the stated source, operating conditions and meter ranges. Written practice is self-checked. Human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Define p.d. as work per charge.
- Use the same current in series.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does 3 V across a component mean in its energy account?
What you will learn
- Use the sum of series p.d.s.
- Check voltage shares against the stated circuit model.
Follow the same charge through each component
Across a series arrangement, the total p.d. is the sum of individual component p.d.s. In a simple passive path with the same traversal, drops of 3 V and 6 V total 9 V. A 9 V ideal source feeding only those components supplies the matching total.
The same charge passes through each component, transferring different energy per coulomb if their p.d.s differ. For 2 C, the component transfers are 6 J and 12 J, adding to 18 J. Charge equality does not require equal work per charge.
Calculate a missing share
Subtract known p.d.s from the total to find a missing component p.d. For a 12 V arrangement with 2 V and 5 V known drops, the remaining drop is 5 V. Equal voltage division requires an appropriate equal-resistance or matching-device model, not just series connection.
For ordinary resistors with equal series current, V = IR explains why a larger resistance has a larger p.d. share. A proposed negative passive drop in the chosen current direction contradicts the stated simple positive-resistor model. Additional sources or non-ideal source drops need their own signed accounts.
Pause and explain
A 12 V series total has 2 V and 5 V known drops. What is the remaining drop?
Worked example
A 12 V total series p.d. contains drops of 2 V, 5 V and an unknown. Find the unknown.
Show the worked solution
- Add known drops: 2 + 5 = 7 V.
- Subtract from the 12 V total.
- The remaining drop is 5 V.
Answer 5 V.
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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.
Find the total p.d. across components with 3 V and 6 V series drops.
Give me a hint
Add the drops in the same traversal.
Compare my reasoning
- Both are passive drops in the stated direction.
- Vtotal = 3 + 6.
- Vtotal = 9 V.
9 V.
Look for these in your work
- Use a common direction.
- State volts.
A 12 V total has known series drops of 2 V and 5 V. Find the third.
Give me a hint
Remove the known drops from the total.
Compare my reasoning
- Known drops sum to 7 V.
- Vthird = 12 − 7.
- Vthird = 5 V.
5 V.
Look for these in your work
- Include both known drops.
- Check that all drops total 12 V.
Two resistors of 2 Ω and 4 Ω share a 6 V ideal source in series. Find current and each p.d., and assess an equal-division claim.
Give me a hint
Add resistance first, then apply the common current to each resistor.
Compare my reasoning
- Rtotal = 6 Ω, so I = 6/6 = 1 A.
- Drops are 1 × 2 = 2 V and 1 × 4 = 4 V.
- They total 6 V but are not equal.
1 A; drops 2 V and 4 V, so equal division is incorrect.
Look for these in your work
- Use common series current.
- Check the summed drops.
Common mistakes
- Series current equality is not equal voltage division.
- Add component drops, not the source twice.
- State the source and passive-path assumptions.
What can you explain now?
What is the total p.d. across series components?
Compare with the explanation
The sum of their individual p.d.s.
The same charge transfers the corresponding summed energy through the arrangement.
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Make it stick
Add series p.d.s and use common current to explain unequal resistor voltage shares.
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