OxfordAQA International GCSE Physics: Resistance, I-V graphs and components (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 follow the OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards, section 3.5.1 Electrical circuits, statements h to o (resistance, I-V characteristics and electronic components). Read the study guide first if a topic is new. Course pages: the OxfordAQA Physics hub and the printable checklist. The free diagnostics show which topics need work.

Questions

1. A resistor at constant temperature has a potential difference of 9.0 V across it and a current of 0.060 A.

(a) Calculate its resistance. [2] (b) The p.d. is raised to 18 V. Calculate the new current. [2]

2. Describe the I-V graph for a fixed resistor at constant temperature and state what it shows about the resistance. [3]

3. The table shows readings for a component.

p.d. (V)	2.0	4.0	6.0	8.0
current (A)	0.10	0.16	0.20	0.23

(a) Calculate the resistance at each p.d. Give each answer to 2 significant figures. [2] (b) State whether the component is ohmic and explain your answer. [2] (c) Suggest what the component could be. [1]

4. Explain, in terms of ions and electrons, why the I-V graph of a filament lamp is a curve. [4]

5. A thermistor is connected across a 12 V supply. Its resistance is 1800 Ω at 20 °C and 150 Ω at 70 °C.

(a) Calculate the current at each temperature. Give your answers in mA to 2 significant figures. [3] (b) State how the resistance of a thermistor changes as its temperature rises. [1] (c) Explain how a thermistor can be used in a fire alarm. [2]

6. An LDR is connected across a 6.0 V supply. In the dark its resistance is 2.0 MΩ. In bright light it is 400 Ω.

(a) Calculate the current in the dark (in µA) and in bright light (in mA). [3] (b) Explain how an LDR can be used to switch a lamp on when it gets dark. [3]

7. A diode has 0.90 V across it in the forward direction and carries 45 mA. When 6.0 V is applied in the reverse direction, the current is 2.0 µA.

(a) Calculate the forward resistance. [2] (b) Calculate the reverse resistance. [2] (c) State how many times larger the reverse resistance is than the forward resistance. [1] (d) State what these results show about the direction of current through a diode. [1]

8. A filament lamp and an LED lamp are each connected to the same p.d. The filament lamp carries 0.45 A and the LED lamp carries 0.054 A.

(a) State what an LED does when current flows through it in the forward direction. [1] (b) Calculate how many times larger the current in the filament lamp is than in the LED lamp. Give your answer to 2 significant figures. [2] (c) State why LEDs are increasingly used for lighting. [1] (d) Explain why an LED connected the wrong way round does not light. [1]

9. Describe how you would investigate the V-I characteristics of a filament lamp. Your answer should include the circuit, the measurements and how you would present the results. [6]

10. A student measures a diode in the forward direction.

p.d. (V)	0.60	0.70	0.75
current (mA)	2.0	10	30

(a) Calculate the resistance at each p.d. [3] (b) Describe the trend in resistance and state what it shows about the diode. [1] (c) Explain why a protective resistor is placed in series with the diode in this investigation. [1]

11. Three components were tested. Component A gave 0.040 A at 2.0 V, 0.080 A at 4.0 V and 0.12 A at 6.0 V. Component B gave 0.20 A at 1.0 V, 0.30 A at 2.0 V, 0.40 A at 4.0 V and 0.45 A at 6.0 V. For component C, no current could be measured when 5.0 V was applied in one direction, but it conducted when the connections were reversed.

(a) Calculate the resistance of A at each p.d. and state whether A is ohmic. [2] (b) Calculate the resistance of B at 1.0 V and at 6.0 V (2 significant figures). [2] (c) Identify B and explain the change in resistance in terms of ions and electrons. [3] (d) Identify C and justify your answer. [2]

12. A student makes four statements. For each, say whether it is correct. If it is wrong, write a correct version.

(i) “The current through a thermistor falls as it warms up, if the p.d. stays the same.” [2] (ii) “An LDR has a higher resistance in bright light than in the dark.” [2] (iii) “A filament lamp is an ohmic conductor because its graph passes through the origin.” [2] (iv) “A diode has a very high resistance in the reverse direction.” [1]

Answers

1. (a) R = V ÷ I [1] = 9.0 ÷ 0.060 = 150 Ω [1] (b) The resistance stays constant, so I = V ÷ R = 18 ÷ 150 [1] = 0.12 A (120 mA) [1] [4] Examiner insight: Show the rearranged equation and the substitution; a bare wrong number is hard to credit.

2. The graph is a straight line [1] that passes through the origin [1]. This shows that current is directly proportional to p.d., so the resistance is constant [1]. [3] Examiner insight: “A straight line” alone is not enough; the origin and the link to constant resistance are separate points.

3. (a) R = V ÷ I [1]: 2.0 ÷ 0.10 = 20 Ω, 4.0 ÷ 0.16 = 25 Ω, 6.0 ÷ 0.20 = 30 Ω, 8.0 ÷ 0.23 = 35 Ω (all four correct) [1] (b) Not ohmic [1], because the resistance is not constant: it increases as the p.d. and current increase [1] (c) A filament lamp [1] [5] Examiner insight: Quote values from your table when you explain.

4. The current makes the filament hotter [1]. The positive ions in the metal lattice vibrate with a larger amplitude [1]. The moving electrons collide with the ions more often [1]. The resistance increases, so the current rises by less for each extra volt and the graph curves [1]. [4] Examiner insight: Keep the ions and electrons separate: the ions vibrate, the electrons collide with them.

5. (a) I = V ÷ R [1]. At 20 °C: 12 ÷ 1800 = 0.0067 A = 6.7 mA [1]. At 70 °C: 12 ÷ 150 = 0.080 A = 80 mA [1] (b) The resistance decreases [1] (c) As the temperature rises, the thermistor’s resistance falls and the current increases [1]. When the change reaches a set level, the circuit switches the alarm on [1] [6] Examiner insight: Link the thermistor to the circuit’s response: say what changes (current or p.d.) and what the circuit then does.

6. (a) I = V ÷ R [1]. Dark: 6.0 ÷ 2 000 000 = 0.000 003 A = 3.0 µA [1]. Bright: 6.0 ÷ 400 = 0.015 A = 15 mA [1] (b) In the dark the LDR has a high resistance [1]. The circuit responds to this change in current or p.d. and switches the lamp on [1]. In daylight the resistance falls and the lamp is switched off [1] [6] Examiner insight: Say what happens to the LDR’s resistance in the dark.

7. (a) R = 0.90 ÷ 0.045 [1] = 20 Ω [1] (b) R = 6.0 ÷ 0.000 002 [1] = 3 000 000 Ω (3.0 MΩ) [1] (c) 3 000 000 ÷ 20 = 150 000 times [1] (d) Current flows easily in the forward direction and hardly at all in the reverse direction: it flows in one direction only [1] [6] Examiner insight: Convert mA and µA to A before dividing.

8. (a) It emits light [1] (b) 0.45 ÷ 0.054 [1] = 8.3 times [1] (c) LEDs use a much smaller current than other forms of lighting [1] (d) An LED conducts in the forward direction only, so no current flows the wrong way round and no light is emitted [1] [5] Examiner insight: Compare the two currents with a ratio and a unit-free answer; do not state “uses less power” if the question gives currents.

9. Connect the lamp in series with a variable power supply and an ammeter [1]. Connect a voltmeter in parallel across the lamp only [1]. Increase the p.d. in equal steps from zero and record the current at each step, with at least six readings [1]. Reverse the connections and repeat to get readings in the opposite direction [1]. Plot current on the vertical axis against p.d. on the horizontal axis [1] and draw a smooth curve through the points [1]. [6] Examiner insight: A good answer names where each meter goes and gives a range of readings; “measure current and voltage” alone is too general.

10. (a) R = V ÷ I [1]. 0.60 ÷ 0.0020 = 300 Ω [1]. 0.70 ÷ 0.010 = 70 Ω and 0.75 ÷ 0.030 = 25 Ω [1] (b) The resistance falls as the p.d. and current increase, so the diode is not ohmic [1] (c) Once the diode conducts, the current rises very steeply, so the resistor limits it and prevents damage [1] [5] Examiner insight: Give the reason for protection (large current can damage the diode); “to protect it” alone often does not say from what.

11. (a) 2.0 ÷ 0.040 = 4.0 ÷ 0.080 = 6.0 ÷ 0.12 = 50 Ω [1]. The resistance is constant, so A is ohmic [1] (b) At 1.0 V: 1.0 ÷ 0.20 = 5.0 Ω [1]. At 6.0 V: 6.0 ÷ 0.45 = 13 Ω [1] (c) B is a filament lamp [1]. As the current rises the filament heats up, so the ions vibrate more [1]. The electrons collide with them more often, which raises the resistance [1] (d) C is a diode [1]. It conducts in one direction only: its reverse resistance is very high [1] [9] Examiner insight: In a multi-part question, use the figures you calculated in earlier parts in your explanation.

12. (i) Wrong [1]. The thermistor’s resistance decreases as it warms, so the current increases [1] (ii) Wrong [1]. The LDR’s resistance decreases as light intensity increases, so it is lower in bright light [1] (iii) Wrong [1]. Its graph is a curve because its resistance increases with temperature; passing through the origin is not enough [1] (iv) Correct [1] [7] Examiner insight: A correction must state the right relationship, not just negate the statement; “it is false” alone is unlikely to earn the second point.

Where marks are usually lost

• Using mA, µA, kΩ or MΩ values in V = IR without converting them first.
• Leaving out units, or giving units that do not match the quantity.
• Giving “resistance increases because current increases” without referring to temperature, ions and electrons.
• Saying the electrons vibrate, or that the ions collide with the electrons “less often” or “more slowly”.
• Mixing up the thermistor (temperature) and the LDR (light intensity).
• Describing the I-V graph of a lamp or diode as straight.
• Describing the practical with the voltmeter in series, or omitting reversed readings.

Next steps

• Recap with the revision notes or re-read the study guide
• Tick off what you know on the printable checklist
• Return to the OxfordAQA Physics hub
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Official syllabus

This page follows the OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards, Version 5.2, published by OxfordAQA, section 3.5.1 Electrical circuits, statements h to o.