Physics 9702 · syllabus 10
Cambridge A Level Physics 9702 — D.C. circuits topical past papers
Analysing direct-current circuits with e.m.f. and internal resistance, Kirchhoff's laws, series and parallel resistors, potential dividers and potentiometers.
- Questions
- 405
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 102
Anyone can browse these pages. Opening the questions and their mark schemes takes a free Momentum account.
What D.C. circuits covers
The electromotive force of a source is the energy it transfers to each unit of charge driven round a complete circuit, while potential difference is the energy each unit of charge transfers to a component. Real sources have internal resistance, so part of the e.m.f. is used inside the source and the terminal p.d. falls as the current rises: V = E − Ir.
Kirchhoff's first law says the total current into a junction equals the total current out, a consequence of conservation of charge. His second law says that round any closed loop the sum of the e.m.f.s equals the sum of the p.d.s, a consequence of conservation of energy. You derive the series and parallel resistance formulae from these laws, then use them to solve circuits with several branches or sources.
A potential divider shares a supply p.d. between resistors in proportion to their resistances; with a thermistor or LDR in place of one resistor, the output depends on temperature or light. A potentiometer compares p.d.s by finding the balance point at which a galvanometer reads zero, a null method that draws no current from the source being measured. This is AS content, examined on Papers 1 and 2.
In the syllabus
The headings this topic has in the Cambridge International 2025–2027 syllabus, as the board prints them.
- 10 D.C. circuits
- 10.1 Practical circuits
- 10.2 Kirchhoff's laws
- 10.3 Potential dividers
What you should be able to do
- Distinguish e.m.f. from p.d. in terms of energy transferred per unit charge
- Use V = E − Ir and explain why terminal p.d. falls as the current increases
- Apply Kirchhoff's laws to circuits with several branches or sources
- Derive and use the formulae for resistors in series and in parallel
- Calculate the output of a potential divider, including one containing a thermistor or LDR
- Explain how a potentiometer and galvanometer compare p.d.s by a null method
Key ideas
- E = V + Ir, so the terminal p.d. is V = E − Ir
- Kirchhoff's first law: total current into a junction = total current out (conservation of charge)
- Kirchhoff's second law: sum of e.m.f.s = sum of IR round any closed loop (conservation of energy)
- Series: R = R₁ + R₂ + …; parallel: 1/R = 1/R₁ + 1/R₂ + …
- Potential divider: the output across R₂ is V(out) = V(in) × R₂/(R₁ + R₂)
How D.C. circuits has been examined
From 2022 to 2026, D.C. circuits was on Paper 2 · AS Level Structured Questions in 7 of 11 exam series, worth about 9% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 1 · Multiple Choice | 80 | 6% | 8 of 11 |
| Paper 2 · AS Level Structured Questions | 18 | 9% | 7 of 11 |
| Paper 4 · A Level Structured Questions | 4 | 1% | 2 of 11 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 1.
Filed in Momentum under: D.c. Circuits.
A worked example
A battery of e.m.f. 12.0 V and internal resistance 1.0 Ω is connected to a 3.0 Ω resistor in series with a parallel pair of 6.0 Ω and 3.0 Ω resistors. Find the terminal p.d. of the battery and the current in the 6.0 Ω resistor.
- Parallel pair: 1/R = 1/6.0 + 1/3.0 = 1/2.0, so R = 2.0 Ω
- External resistance = 3.0 + 2.0 = 5.0 Ω; total including internal resistance = 6.0 Ω
- Current from the battery: I = 12.0 / 6.0 = 2.0 A
- Terminal p.d. = E − Ir = 12.0 − 2.0 × 1.0 = 10.0 V
- P.d. across the parallel pair = 2.0 × 2.0 = 4.0 V, so the current in the 6.0 Ω resistor = 4.0 / 6.0 = 0.67 A
Answer: Terminal p.d. 10.0 V; current in the 6.0 Ω resistor 0.67 A
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Treating e.m.f. and terminal p.d. as equal when the source has internal resistance — the difference Ir is the p.d. lost inside the source, and it grows with current
- Adding parallel resistances directly, or forgetting to take the reciprocal at the end of 1/R = 1/R₁ + 1/R₂ — a parallel combination is always smaller than its smallest resistor
- Getting the direction of a potential divider's change wrong — as a thermistor warms its resistance falls, so the p.d. across it falls and the p.d. across the fixed resistor rises
- Going round a Kirchhoff loop without a fixed direction, so that an e.m.f. opposing the chosen direction is added instead of subtracted
Exam technique
- In potential-divider explanations, write the chain one step at a time: temperature rises, thermistor resistance falls, its share of the supply p.d. falls, so the output changes
- Mark currents with labelled arrows on the circuit diagram before writing Kirchhoff equations; a negative answer simply means that current flows the other way
- When defining e.m.f., say 'energy transferred per unit charge' and 'in driving charge round a complete circuit'; 'the voltage of the battery' earns nothing
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.