Practise 56 real questions on D.C. circuits from 2021 to 2026. See the answer and explanation as soon as you answer.
D.C. circuits is one of the 13 main topics in our A Level Physics (Cambridge (CAIE) 9702) question bank, and about 8% of its real questions come from it. Tayyari Karo has 56 real questions on this topic from 2021 to 2026, from Paper 1, Paper 2, Paper 4; 45 are multiple choice and marked instantly, and the rest are structured questions with a model answer. By difficulty: 26 hard, 22 medium, 8 easy.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
These are real past-paper questions. Think of your answer first, then open the explanation.
Two resistors of resistances R₁ and R₂ are connected in parallel between X and Y.
What is the combined resistance between X and Y?
Answer: B. R₁R₂/(R₁ + R₂)
For resistors in parallel, 1/R = 1/R₁ + 1/R₂, so R = R₁R₂/(R₁+R₂).
Kirchhoff's second law is a consequence of a basic principle.
What is this principle?
Answer: B. The energy in an electric circuit is conserved.
Kirchhoff's second law (sum of e.m.f.s = sum of p.d.s around a loop) follows from conservation of energy.
A cell with internal resistance is connected to a light-dependent resistor (LDR), a fixed resistor and a voltmeter (measuring the p.d. across the fixed resistor). The voltmeter reading increases.
Which quantity decreases as the voltmeter reading increases?
Answer: D. the terminal potential difference across the cell
If the p.d. across the fixed resistor increases, the current has increased (LDR resistance decreased, e.g. more light), so more p.d. is dropped internally, meaning the terminal p.d. across the cell decreases.
Kirchhoff's first and second laws are a consequence of the conservation of which quantities?
Answer: A. charge and energy
Kirchhoff's first law follows from conservation of charge (current in = current out at a junction) and the second law from conservation of energy (sum of e.m.f.s = sum of p.d.s around a loop).
A cell of internal resistance r and e.m.f. E is connected in series with an ammeter and a variable resistor R. When R is 10 Ω the ammeter reads 0.3 A. When R is 5 Ω the ammeter reads 0.4 A. What is the value of E?
Answer: D. 6 V
E = I(R+r). From the two readings: E = 0.3(10+r) = 0.4(5+r). Solving: 3+0.3r = 2+0.4r → 1 = 0.1r → r = 10 Ω. Then E = 0.3(10+10) = 6 V.
Three resistors of 120 Ω, 180 Ω and 150 Ω are connected in series with a 6.0 V battery of negligible internal resistance. What is the potential difference across the 180 Ω resistor?
Answer: B. 2.4 V
Total resistance = 120+180+150 = 450 Ω. I = 6.0/450 = 0.0133 A. V₁₈₀ = 0.0133 × 180 ≈ 2.4 V.
| Session | Paper 1 | Paper 2 | Paper 4 |
|---|---|---|---|
| May/June 2026 | 4 | 1 | – |
| May/June 2025 | 4 | – | – |
| October/November 2025 | 4 | 1 | – |
| May/June 2024 | 4 | 1 | – |
| October/November 2024 | 7 | 1 | – |
| May/June 2023 | 2 | 1 | – |
| October/November 2023 | 5 | 1 | – |
| May/June 2022 | 3 | 1 | – |
| October/November 2022 | 2 | 1 | – |
| May/June 2021 | 5 | 1 | 1 |
| October/November 2021 | 5 | 1 | – |
56 real A Level Physics (9702) questions on D.C. circuits from 2021 to 2026, from Paper 1, Paper 2, Paper 4. Each one shows the year, session and paper it came from.
Yes. Every question shows the correct answer and a short explanation the moment you answer.
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