OCR GCSE Physics A: Current, charge and circuit conventions (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 follow the OCR GCSE (9-1) Physics A (Gateway Science) (J249) specification, Version 5.0, July 2026, statements P3.1e, P3.1f, P3.1g, P3.2a and P3.2b. All of these statements are assessed in both Foundation Tier and Higher Tier papers, so none of the questions is marked Higher Tier only. Work in seconds and amperes unless a question says otherwise.

Related pages: study guide, revision notes, the OCR GCSE Physics hub and the printable checklist.

Questions

1. State what is meant by electric current, and name the particles that carry the current in a metal wire. [2]

2. A new cell, a switch and a lamp are joined into a loop with copper wires. The switch is open and the lamp is off.

(a) State the two conditions needed for charge to flow in a circuit. [2] (b) Explain why no charge flows in any part of this circuit. [2]

3. A steady current of 0.45 A flows through a bulb for 40 s. Calculate the charge that flows through the bulb. Give the unit. [3]

4. A heater element carries a steady current. 96 C of charge passes through it in 8.0 minutes. Calculate the current. [3]

5. A charger delivers a current of 750 mA to a battery. Calculate the time, in seconds, taken to deliver 45 C of charge. [3]

6. A cell, a lamp and a buzzer are connected in series in a single loop. Ammeter A1 is between the cell and the lamp and reads 0.24 A. Ammeter A2 is between the lamp and the buzzer.

(a) State the reading on A2. [1] (b) A student says, “The lamp uses up some of the current, so A2 reads less than A1.” Explain why the student is wrong. [2] (c) Calculate the charge that passes through the buzzer in 5.0 s. [2]

7. Name the circuit component that matches each description of its symbol.

(a) A circle with a cross inside. [1] (b) A triangle pointing at a vertical bar. [1] (c) A rectangle crossed by a sloping arrow. [1] (d) Two parallel lines side by side, one long and thin, one short and thick. [1] (e) State which of the two lines in (d) shows the positive terminal. [1]

8. Describe two differences between a series circuit and a parallel circuit. One must be about current and one about potential difference. [4]

9. A student investigates a filament lamp. Describe how to connect an ammeter and a voltmeter to measure the current through the lamp and the potential difference across it, and explain each connection. [4]

10. Two lamps, X and Y, are connected in parallel to a power supply. The current from the supply is 0.90 A. The current in lamp X is 0.55 A.

(a) Calculate the current in lamp Y. [2] (b) Calculate the charge that passes through lamp Y in 3.0 minutes. [3] (c) Calculate the charge that passes through the power supply in the same time. [2] (d) Use your answers to explain what happens to charge at the junction where the branches meet. [1]

11. A phone charger supplies a steady current of 1.2 A to a phone battery. The charger, lead and battery form a single loop.

(a) Calculate the charge delivered in 45 minutes. [3] (b) The battery needs 5400 C in total. Calculate the time needed, in minutes, at this current. [3] (c) State the current in the lead and explain how it compares with the current in the battery. [2] (d) The lead is unplugged from the phone while the charger stays switched on. Explain why charge stops flowing. [2]

12. A student has one cell, two lamps and two switches. The student wants each lamp to be switched on and off without affecting the other.

(a) Describe how the components should be connected, saying where each switch goes. [3] (b) The filament in one lamp breaks. State and explain what happens to the other lamp. [2] (c) The student now connects the same two lamps in series with the cell. State and explain what happens to the second lamp if one filament breaks. [2]

Answers

1. Current is how fast charge flows past a point [1]. In a metal wire the charge carriers are electrons [1]. [2] Examiner insight: “Flow of charge” is the idea the question is after; writing “flow of electricity” tends to be too vague. Add “rate” or “per second” when the question asks what current is.

2. (a) A cell, battery or power supply to provide potential difference [1]. A closed circuit, meaning a complete conducting loop [1]. (b) The open switch leaves a gap in the loop, so the circuit is not closed [1]. With no complete path, no charge can flow anywhere in the loop, even though the new cell is a source of potential difference [1]. [4] Examiner insight: When a question gives you a working cell, say clearly that the second condition, the closed loop, is the one that fails. Naming the condition and linking it to the open switch is stronger than “the switch is off”.

3. Q = I t [1]. Q = 0.45 × 40 [1]. Q = 18 C [1]. [3] Examiner insight: Write the equation first, then substitute. A clear equation and substitution show your method to the reader, and writing the unit C with the answer completes it.

4. Time in seconds: 8.0 × 60 = 480 s [1]. I = Q / t = 96 / 480 [1]. I = 0.20 A [1]. [3] Examiner insight: Using 8.0 instead of 480 gives a current 60 times too large and loses the conversion step. Writing the conversion on its own line makes it visible to the reader.

5. Convert the current: 750 mA = 0.750 A [1]. t = Q / I = 45 / 0.750 [1]. t = 60 s [1]. [3] Examiner insight: Dividing 45 by 750 gives 0.06, which is a sign that the milliampere conversion was missed. Check that your answer is sensible for the situation before moving on.

6. (a) 0.24 A [1]. (b) There is a single loop with only one path [1], so every point in the loop carries an equal current; the lamp transfers energy, but charge is not used up [1]. (c) Q = I t = 0.24 × 5.0 [1]. Q = 1.2 C [1]. [5] Examiner insight: A strong explanation separates energy from charge: the lamp transfers energy, but the charge flowing per second is unchanged. Writing “because it is a series circuit” alone repeats the question without giving a reason.

7. (a) Filament lamp [1]. (b) Diode [1]. (c) Variable resistor [1]. (d) Cell [1]. (e) The long thin line is the positive terminal [1]. [5] Examiner insight: Choosing the short thick line as positive is an easy slip. If you are drawing a cell, label the plus and minus signs as well, so the reader does not have to judge the line lengths.

8. Current: in a series circuit every point carries an equal current because there is only one path [1], whereas in a parallel circuit the current splits at a junction and the branch currents add to give the supply current [1]. Potential difference: in a series circuit it is shared between the components in the loop [1], whereas in a parallel circuit each branch has the same potential difference as the supply [1]. [4] Examiner insight: Mention both circuit types in each comparison. A sentence about only one of them, such as “the current splits”, leaves the contrast incomplete.

9. Connect the ammeter in series with the lamp [1], so that the same current passes through the ammeter as through the lamp [1]. Connect the voltmeter in parallel with the lamp, with one terminal on each side of it [1], because it measures the potential difference between the two ends of the lamp [1]. [4] Examiner insight: Each connection needs its reason to earn full credit. “In series” or “in parallel” on its own gives the position but not the explanation the question asks for.

10. (a) I(Y) = 0.90 - 0.55 [1] = 0.35 A [1]. (b) t = 3.0 × 60 = 180 s [1]. Q = I t = 0.35 × 180 [1]. Q = 63 C [1]. (c) Q = 0.90 × 180 [1] = 162 C [1]. (d) The charge through X is 0.55 × 180 = 99 C, and 99 + 63 = 162 C, so the charge through the supply equals the total through the branches; charge is not lost at the junction [1]. [8] Examiner insight: Reusing your answer from (a) in (b) is sensible, but write out the substitution each time so the reader can follow your method.

11. (a) t = 45 × 60 = 2700 s [1]. Q = I t = 1.2 × 2700 [1]. Q = 3240 C [1]. (b) t = Q / I = 5400 / 1.2 [1]. t = 4500 s [1]. In minutes, 4500 / 60 = 75 minutes [1]. (c) The current in the lead is 1.2 A [1]. It is the same as the current in the battery because they are in a single closed loop [1]. (d) Unplugging the lead makes a gap, so the circuit is no longer closed [1]. Although the charger is still a source of potential difference, a closed circuit is also needed, so charge cannot flow [1]. [10] Examiner insight: In a two-stage question, state the unit conversion each time it happens. In (b), converting to minutes at the end is part of the question, so reread the unit asked for before you box the answer.

12. (a) Connect the two lamps in parallel, on separate branches across the cell [1]. Put one switch in series with lamp 1 in its own branch [1] and the other switch in series with lamp 2 in its branch [1]. (b) The other lamp stays lit [1], because its branch still forms a complete circuit with the cell and has the full supply potential difference [1]. (c) The second lamp goes out [1], because in a series circuit there is one loop and a broken filament opens it, so no current flows anywhere [1]. [7] Examiner insight: In design questions, say where each switch goes by naming the component it is in series with. “A switch on each lamp” is less clear than “a switch in series with each lamp, in that lamp’s own branch”.

Where marks are usually lost

• Giving only one of the two conditions for charge to flow, or writing “a battery” without mentioning the closed loop.
• Writing that current is “used up” in a component instead of saying that energy is transferred.
• Leaving time in minutes when using Q = I t.
• Using milliamperes without converting to amperes.
• Giving an answer with no unit, or with the wrong unit (A for charge, or C for current).
• Explaining a series circuit without saying that there is only one path.
• Describing parallel branches as sharing the potential difference, when each branch has the supply value.
• Wiring a voltmeter into the loop, or an ammeter on a branch by itself.
• Swapping the long and short lines of a cell, or drawing a diode the wrong way round.

Next steps

• Revise the key points in the revision notes.
• Re-read the explanations in the study guide.
• Tick off statements on the printable checklist and browse the OCR GCSE Physics hub.
• Try the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, GCSE (9-1) Physics A (Gateway Science) J249, Version 5.0, July 2026, for first assessment in 2018, Topic P3: Electricity, statements P3.1e to P3.1g and P3.2a to P3.2b.