OCR GCSE Physics A: Series and parallel circuit calculations and electrical power (J249) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover learning outcomes P3.2i to P3.2m of the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026): series and parallel circuit calculations, measurement and testing circuits, and electrical power and energy. All of these outcomes are assessed in both tiers, so no question is labelled Higher Tier only.

Related: Study guide | Revision notes | OCR GCSE Physics hub | Printable checklist | All free 10-minute diagnostics

Questions

1. Explain why connecting a second identical resistor in series with the first increases the net resistance of a circuit. [2]

2. A student connects a second identical lamp in parallel with the first across the same cell. Explain why the net resistance of the circuit decreases. [3]

3. A 6.0 V supply is connected across three resistors in series: 5.0 Ω, 7.0 Ω and 3.0 Ω. Calculate the current and the p.d. across the 7.0 Ω resistor. [4]

4. A 20 Ω resistor and a 30 Ω resistor are connected in parallel across a 12 V supply. Calculate the supply current and the net resistance of the pair. [4]

5. A 16 V supply is connected to a 5.0 Ω resistor in series with a parallel pair of resistors. One of the pair is 12 Ω. An ammeter in the supply lead reads 2.0 A.

(a) Calculate the p.d. across the 5.0 Ω resistor. [2] (b) Calculate the current in the 12 Ω resistor. [2] (c) Calculate the resistance of the other resistor in the pair. [2]

6. A student wants to find the resistance of a filament lamp.

(a) Draw or describe the circuit, stating where the ammeter and the voltmeter are placed. [3] (b) The voltmeter reads 4.5 V and the ammeter reads 0.30 A. Calculate the resistance. [2]

7. An NTC thermistor is in series with a 300 Ω fixed resistor on a 9.0 V supply. A voltmeter is connected across the fixed resistor and reads 3.0 V.

(a) Find the current flowing round the circuit. [2] (b) Calculate the resistance of the thermistor. [2] (c) The thermistor is heated until its resistance is 200 Ω. Calculate the new reading on the voltmeter. [2]

8. A lamp is run at 12 V and draws 2.5 A. Find its power, then the energy transferred during a 10 minute test. [4]

9. A heating element of resistance 8.0 Ω carries a current of 3.0 A. Calculate its power, and the energy transferred in 2.0 minutes. [4]

10. An electric heater has a power of 2.5 kW and runs for 4.0 hours. Calculate the energy transferred in kW h, and then in joules (1 kW h = 3 600 000 J). [4]

11. A motor connected to a 12 V supply has 45 C of charge pass through it in 9.0 s. Calculate the energy transferred, the current and the power. [4]

12. A 12 V supply is connected to a 10 Ω resistor and a 20 Ω resistor in series, for 90 s.

(a) State the equivalent resistance of the circuit. [1] (b) Calculate the current. [2] (c) Calculate the p.d. across each resistor. [2] (d) Calculate the power in each resistor and the total power. [3] (e) Calculate the total energy transferred in 90 s, and show that the charge passing is consistent with E = Q × V. [2] (f) The 20 Ω resistor is replaced by a 30 Ω resistor. State and explain what happens to the current. [2]

Answers

1. Charge has to pass through both resistors in turn, so there is more opposition to the current [1]. Their resistances add, so the net resistance is greater [1]. [2] Examiner insight: “More wire” or “resistance is used up” gains little. State the single path and the addition of opposition.

2. The second lamp provides an extra path for charge [1]. The p.d. across each lamp is the same as before, so the total current from the cell is larger [1]. With a larger current for the same p.d., R = V / I is smaller [1]. [3] Examiner insight: A full answer links the extra path, the increased total current and the change in net resistance. Saying only “there is less resistance” restates the question.

3. Net R = 5.0 + 7.0 + 3.0 = 15 Ω [1]. I = V / R = 6.0 / 15 [1] = 0.40 A [1]. p.d. across 7.0 Ω = 0.40 × 7.0 = 2.8 V [1]. [4] Examiner insight: Show the equation and substitution. Keep the unit with the answer, and check that the three p.d.s (2.0, 2.8 and 1.2 V) add to 6.0 V.

4. Currents: 12 / 20 = 0.60 A and 12 / 30 = 0.40 A [1]. Supply current = 0.60 + 0.40 = 1.0 A [1]. R = V / I = 12 / 1.0 [1] = 12 Ω [1]. [4] Examiner insight: Both branches have the full 12 V, which is what makes the method work. A net resistance bigger than 20 Ω would be a red flag.

5. (a) p.d. = I × R = 2.0 × 5.0 [1] = 10 V [1]. (b) p.d. across the pair = 16 − 10 = 6.0 V [1]. I = 6.0 / 12 = 0.50 A [1]. (c) Current in the other resistor = 2.0 − 0.50 = 1.5 A [1]. R = 6.0 / 1.5 = 4.0 Ω [1]. [6] Examiner insight: Part (b) needs the p.d. across the pair, not the supply p.d. Writing the intermediate value 6.0 V on the page shows your method.

6. (a) Lamp, ammeter and variable resistor (or cell) in series [1]. Ammeter in series with the lamp [1]. Voltmeter connected in parallel across the lamp [1]. (b) R = V / I = 4.5 / 0.30 [1] = 15 Ω [1]. [5] Examiner insight: A diagram with the voltmeter in the wrong position fails to show the correct placement, even if the equation is correct.

7. (a) I = V / R = 3.0 / 300 [1] = 0.010 A [1]. (b) p.d. across thermistor = 9.0 − 3.0 = 6.0 V [1]. R = 6.0 / 0.010 = 600 Ω [1]. (c) Net R = 300 + 200 = 500 Ω, so I = 9.0 / 500 = 0.018 A [1]. Voltmeter reading = 0.018 × 300 = 5.4 V [1]. [6] Examiner insight: A common slip is using the 9.0 V supply across the thermistor. State that the supply p.d. is shared between the two.

8. P = V × I = 12 × 2.5 [1] = 30 W [1]. t = 10 × 60 = 600 s [1]. E = P × t = 30 × 600 = 18 000 J [1]. [4] Examiner insight: Leaving time as 10 gives 300 J, which is wrong by a factor of 60. Convert minutes to seconds first.

9. P = I² × R = 3.0² × 8.0 [1] = 72 W [1]. t = 120 s [1]. E = 72 × 120 = 8640 J [1]. [4] Examiner insight: Square the current before multiplying by the resistance. Squaring the product I × R gives a very different value.

10. E = P × t = 2.5 × 4.0 [1] = 10 kW h [1]. Energy in joules = 10 × 3 600 000 [1] = 36 000 000 J (3.6 × 10⁷ J) [1]. [4] Examiner insight: Keep kW with hours for the first part, and convert to joules only at the end. Give each answer in the unit the question asks for.

11. E = Q × V = 45 × 12 [1] = 540 J [1]. I = Q / t = 45 / 9.0 = 5.0 A [1]. P = V × I = 12 × 5.0 = 60 W [1]. [4] Examiner insight: You can check the power with E / t = 540 / 9.0 = 60 W. A second route that agrees is a good way to catch an error.

12. (a) 10 + 20 = 30 Ω [1]. (b) I = 12 / 30 [1] = 0.40 A [1]. (c) 10 Ω resistor: 0.40 × 10 = 4.0 V [1]. 20 Ω resistor: 0.40 × 20 = 8.0 V [1]. (d) 10 Ω: 0.40² × 10 = 1.6 W [1]. 20 Ω: 0.40² × 20 = 3.2 W [1]. Total = 4.8 W, which equals 12 × 0.40 [1]. (e) E = 4.8 × 90 = 432 J [1]. Q = I × t = 0.40 × 90 = 36 C, and E = Q × V = 36 × 12 = 432 J, which agrees [1]. (f) The current decreases (to 12 / 40 = 0.30 A) [1] because the equivalent resistance increases for the same supply p.d. [1]. [12] Examiner insight: In a multi-part question, set out each part separately. Visible working for each part lets the reader follow your method, even if an earlier value was wrong.

Where marks are usually lost

• Using the supply p.d. across one resistor in a series chain instead of its share.
• Adding branch resistances in parallel instead of working from branch currents.
• Placing a voltmeter in series, or an ammeter across a component.
• Forgetting to convert minutes to seconds in E = P × t, or hours to seconds in joule calculations.
• Squaring the wrong quantity in P = I² × R.
• Giving power in joules or energy in watts.
• Missing the unit, or writing a unit with the wrong case (for example mA for A without converting).
• Writing “current is used up” or “resistance is used up” in an explanation.
• Not showing substitution into the equation.

Next steps

• Review the revision notes and the study guide.
• Open the course hub and record progress on the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR GCSE (9-1) Gateway Science Physics A (J249), Version 5.0 (July 2026), first assessment in 2018, published by OCR. Learning outcomes P3.2i to P3.2m.