OxfordAQA International GCSE Physics: Charge, current, potential difference and circuit symbols (9203) – 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 section 3.5.1 Electrical circuits, statements a to g, of the OxfordAQA International GCSE Physics (9203) specification, for exams from May/June 2018. The qualification is untiered. Use the study guide if a topic is new. See also the OxfordAQA International GCSE Physics hub, the printable checklist and the free diagnostics.

Questions

1. A copper wire is connected to a cell and lamp in a closed circuit.

(a) Explain why the copper conducts but a strip of plastic in its place would not. [2] (b) State the direction in which the electrons move relative to the arrows drawn for conventional current. [1]

2. Calculate each quantity.

(a) The charge transferred when 0.60 A flows for 25 minutes. [2] (b) The current when 54 C flows in 36 s. [2] (c) The time, in minutes, for a current of 150 mA to deliver 330 C. [3]

3. A student rubs a polythene rod with a dry cloth.

(a) Explain why the rod becomes charged. [2] (b) The student rubs a metal rod held in the hand in the same way, and it does not stay charged. Explain why. [2]

4. A storm cloud discharges to the ground through a lightning flash.

(a) Explain how a lightning flash happens, in terms of charge. [4] (b) The flash transfers 24 C in 0.60 ms. Calculate the current. [2] (c) The potential difference between cloud and ground is 1.5 × 10⁸ V. Calculate the energy transferred. [2]

5. Calculate each quantity.

(a) The energy supplied by a 9.0 V battery that drives 40 C round a circuit. [2] (b) The potential difference across a lamp that transfers 84 J from 24 C. [2] (c) The charge that passes through a motor that transfers 150 J with 6.0 V across it. [2]

6. A 12 V battery drives a steady current of 0.50 A for 3.0 minutes round a circuit containing a lamp and another component.

(a) State the difference between the voltage of the battery and the potential difference across the lamp. [2] (b) Calculate the charge that flows. [2] (c) Calculate the energy supplied by the battery. [1] (d) The potential difference across the lamp is 8.0 V. Calculate the energy transferred by the lamp, and the potential difference across the other component. [3]

7. Name the component shown by each description of a circuit symbol.

(a) A circle containing a cross. [1] (b) A rectangle with a sloping arrow through it. [1] (c) A rectangle inside a circle, with arrows pointing towards it. [1] (d) A triangle pointing at a bar, inside a circle, with two arrows pointing away. [1] (e) A rectangle with a sloping line through it, ending in a bar. [1]

8. Draw a circuit diagram for a cell, a closed switch and a lamp in series. Add an ammeter to measure the current through the lamp and a voltmeter to measure the potential difference across it. [4]

State which meter measures the current through the lamp. [1]

9. A thermistor is connected to a 6.0 V cell and an ammeter. The current is 20 mA at 10 °C, 45 mA at 30 °C, 90 mA at 50 °C and 160 mA at 70 °C.

(a) State the trend in the current as the temperature rises. [1] (b) Explain what this shows about the resistance of the thermistor. [2] (c) Calculate the charge that flows in 2.0 minutes at 50 °C. [2] (d) Describe how a thermistor can be used in a room thermostat. [2]

10. An LDR is connected in series with a cell and an ammeter. The current is 14 mA in bright light, 6.2 mA at dusk and 1.2 mA in the dark.

(a) Describe how the resistance of the LDR changes as the light level falls, and explain your answer. [2] (b) Calculate the charge that flows in 5.0 minutes in the dark. [2] (c) Describe how an LDR can be used so that a street lamp comes on when it gets dark. [3]

11. A torch has a 4.5 V battery. The lamp carries 0.28 A for 1.5 hours.

(a) Calculate the charge that flows. [2] (b) Calculate the energy supplied by the battery. [2] (c) The potential difference across the lamp is 4.2 V. Calculate the energy transferred by the lamp, and the energy transferred elsewhere in the circuit. [2] (d) Explain why the charge passing through the battery equals the charge passing through the lamp. [2]

12. Correct each statement.

(a) “The lamp uses up the current, so less current comes out of it.” [2] (b) “The voltage flows through the wire.” [2]

Answers

1. (a) Copper contains many electrons that are free to move through it [1]; plastic has almost no free charges, so nothing can flow through it [1]. (b) The electrons move in the opposite direction to the arrows [1]. Examiner insight: name the free electrons in the metal. “Plastic has no electrons” is wrong, because it has electrons that cannot move freely.

2. (a) t = 25 × 60 = 1500 s [1]; Q = 0.60 × 1500 = 900 C [1]. (b) I = Q / t = 54 / 36 [1] = 1.5 A [1]. (c) I = 0.150 A [1]; t = Q / I = 330 / 0.150 = 2200 s [1]; 2200 / 60 = 37 minutes (36.7 minutes) [1]. Examiner insight: the conversions are where marks go. Convert minutes to seconds and milliamperes to amperes before substituting, and write each step.

3. (a) Rubbing transfers electrons from the cloth to the rod [1]; the rod gains an excess of electrons and is negatively charged, and the cloth is left positive [1]. (b) The metal is a conductor, so charge flows along it and through the student’s body to the ground [1]; there is a conducting route, so the charge does not stay on the rod [1]. Examiner insight: say that electrons move. Writing that “charge is made by rubbing” does not explain the imbalance.

4. (a) Charge is separated inside the cloud [1]; the ground below becomes charged oppositely to the base of the cloud [1]; the potential difference between them becomes very large [1]; the air stops acting as an insulator and charge flows through it as a discharge [1]. (b) I = Q / t = 24 / 0.00060 [1] = 4.0 × 10⁴ A [1]. (c) E = Q V = 24 × 1.5 × 10⁸ [1] = 3.6 × 10⁹ J [1]. Examiner insight: in the explanation, a build-up of charge plus a conducting path is the core. Convert 0.60 ms to 0.00060 s before you divide.

5. (a) E = Q V = 40 × 9.0 [1] = 360 J [1]. (b) V = E / Q = 84 / 24 [1] = 3.5 V [1]. (c) Q = E / V = 150 / 6.0 [1] = 25 C [1]. Examiner insight: write the rearranged equation before the numbers. A correct substitution with no unit leaves the answer incomplete.

6. (a) The voltage of the battery is the energy supplied to each coulomb driven round the circuit [1]; the potential difference across the lamp is the energy each coulomb transfers in the lamp [1]. (b) t = 3.0 × 60 = 180 s [1]; Q = 0.50 × 180 = 90 C [1]. (c) E = 90 × 12 = 1080 J [1]. (d) Lamp: E = 90 × 8.0 = 720 J [1]; remaining energy = 1080 − 720 = 360 J [1]; V = 360 / 90 = 4.0 V [1]. Examiner insight: use your answer to part (b) in the later parts and show it. Clear working lets a reader see how each later figure was obtained.

7. (a) Lamp [1]. (b) Variable resistor [1]. (c) LDR (light-dependent resistor) [1]. (d) LED (light-emitting diode) [1]. (e) Thermistor [1]. Examiner insight: give the full name or the standard abbreviation. An LED and an LDR both have arrows, so check which way the arrows point.

8. Cell, closed switch and lamp drawn in a single loop [1]; ammeter drawn in the loop in series with the lamp [1]; voltmeter drawn with its two connections on either side of the lamp, outside the loop [1]; all symbols correct and the long line of the cell shown as the positive terminal, drawn with straight lines [1]. The ammeter measures the current through the lamp [1]. Examiner insight: a voltmeter placed in the loop, or an ammeter placed across the lamp, is a common reason a diagram does not gain full credit.

9. (a) The current increases as the temperature rises [1]. (b) The potential difference is constant at 6.0 V [1]; a larger current means the resistance has decreased as the temperature rises [1]. (c) I = 0.090 A and t = 120 s [1]; Q = 0.090 × 120 = 10.8 C [1]. (d) As the room warms, the thermistor’s resistance falls and the current in the sensing circuit rises [1]; at a set value the circuit switches the heater off, and it switches it on again as the room cools [1]. Examiner insight: link the current to the resistance explicitly, with the constant potential difference stated. “It conducts better” alone is vague.

10. (a) The resistance of the LDR increases as the light level falls [1]; the current falls while the potential difference stays the same [1]. (b) I = 1.2 mA = 0.0012 A and t = 5.0 × 60 = 300 s [1]; Q = 0.0012 × 300 = 0.36 C [1]. (c) As it gets dark, the resistance of the LDR increases [1]; the current in the sensing circuit falls [1]; this change operates a switch that turns the street lamp on [1]. Examiner insight: a bare “light dependent” is not an answer. State the direction of change and what the circuit then does.

11. (a) t = 1.5 × 3600 = 5400 s [1]; Q = 0.28 × 5400 = 1512 C (1500 C to 2 significant figures) [1]. (b) E = Q V = 1512 × 4.5 [1] = 6800 J (6804 J) [1]. (c) Lamp: 1512 × 4.2 = 6350 J (6350.4 J) [1]; elsewhere: 6804 − 6350.4 = 454 J (453.6 J) [1]. (d) The torch is a single loop with one path [1]; charge is not used up or stored, so the same charge passes each point in the same time [1]. Examiner insight: in a long multi-part calculation, keep unrounded values until the final line and round at the end.

12. (a) Current is the rate of flow of charge, and it is the same at the lamp’s input and output in a single loop [1]; the lamp transfers the energy that the charge carries, but the charge is not used up [1]. (b) Charge flows through the wire, and this flow is the current [1]; potential difference is measured across a component, as energy transferred per coulomb [1]. Examiner insight: “flows through” for current and “across” for potential difference is a simple way to show you have the ideas right.

Where marks are usually lost

• Putting a time in minutes or hours into Q = It.
• Using milliamperes without converting to amperes.
• Giving a discharge answer with no conducting route and no mention of charge building up.
• Saying charge is “made” by rubbing instead of transferred.
• Mixing up the energy supplied by a source and the energy transferred by a component.
• Quoting the thermistor or LDR rule the wrong way round.
• Drawing a voltmeter in the loop or an ammeter across a component.

Next steps

• Revise with the revision notes.
• Return to the OxfordAQA International GCSE Physics hub and the printable checklist.
• Take the free 10-minute diagnostics.
• Ask about a free trial class.

Official syllabus

OxfordAQA International GCSE Physics (9203), specification version 5.2, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.5.1 Electrical circuits, statements a to g.