Add potential differences across series components
Balance the total p.d. and find missing shares without assuming equal voltages across unequal resistors.
Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.
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.
Sources for this lesson
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.
Try a topic test
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Test this topic →From guided practice to a new situation
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.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Add series p.d.s and use common current to explain unequal resistor voltage shares.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.