OxfordAQA International AS and A-level Physics: Current, p.d., resistance, I-V characteristics and resistivity (9630) – 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 sections 3.4.1 to 3.4.3 of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, which is shared International AS and A-level content. Take the elementary charge as 1.60 × 10⁻¹⁹ C. Treat meters as ideal. Learn the content from the study guide or the revision notes. The course hub and checklist list the rest of the course.

Questions

1. A current of 2.5 mA flows through a light-emitting component for 8.0 minutes. Calculate the charge that passes and the number of electrons this represents. [3]

2. A charge of 45 C passes in 30 s through a motor with 6.0 V across it.

(a) Calculate the current. [1] (b) Calculate the electrical energy transferred in the motor in this time. [2]

3. Using V = W/Q and R = V/I, show that the ohm is equivalent to kg m² s⁻³ A⁻². [3]

4. A resistor at constant temperature gives these readings.

V / V	1.0	2.0	3.0	4.0	5.0
I / mA	12.4	25.1	37.3	49.8	62.4

(a) Calculate R for each pair of readings and state, with a reason, whether the resistor is ohmic. [3] (b) A graph of I (vertical) against V (horizontal) is drawn. The best-fit line passes through (0, 0) and (4.0 V, 50 mA). Find its gradient in A V⁻¹ and use it to find R. [2]

5. A semiconductor diode gives these readings.

Condition	Forward 0.50 V	Forward 0.70 V	Reverse 4.0 V
Current	0.40 mA	12.5 mA	too small to measure

(a) Calculate the resistance of the diode at 0.50 V and at 0.70 V forward. [2] (b) Describe and compare the diode’s resistance in forward and reverse bias. [2] (c) A student says the diode obeys Ohm’s law above 0.6 V because the graph is steep. Evaluate this claim. [1]

6. The characteristic of a filament lamp is plotted with V on the vertical axis and I on the horizontal axis. Two points on the curve are (0.24 A, 3.0 V) and (0.50 A, 9.0 V).

(a) Calculate the resistance of the lamp at each point. [2] (b) Explain why the curve bends upwards on these axes. [2] (c) State why the lamp does not obey Ohm’s law. [1] (d) A student finds ΔV/ΔI between the two points and says the answer is the lamp’s resistance at 9.0 V. Calculate that value and explain why the student is wrong. [1]

7. Describe an experiment to obtain the I-V characteristic of a filament lamp for both directions of current. [6]

8. An aluminium wire has diameter 0.50 mm and length 1.2 m. Its resistivity is 2.7 × 10⁻⁸ Ω m.

(a) Calculate its resistance. [3] (b) A second aluminium wire is 0.60 m long with diameter 0.25 mm. Calculate its resistance and compare it with the first wire. [2]

9. Several lengths L of nichrome wire, diameter 0.30 mm, are tested. The graph of R against L is a straight line from the origin through (0.80 m, 12.5 Ω).

(a) Calculate the gradient of the line. [1] (b) Calculate the cross-sectional area of the wire and hence its resistivity. [2] (c) Explain why the line should pass through the origin. [1] (d) State one way to reduce the uncertainty in the resistivity value. [1]

10. An ntc thermistor has resistance 8.0 kΩ at 15 °C and 1.2 kΩ at 75 °C. It is connected to a 3.0 V supply with an ideal ammeter in series.

(a) Calculate the ammeter reading at each temperature. [2] (b) Explain, in terms of charge carriers, why the resistance of the thermistor falls as it warms. [2] (c) Explain why a metal wire shows the opposite trend. [2] (d) State one use of an ntc thermistor. [1]

11. (a) State what is meant by the critical temperature of a superconductor. [2] (b) Describe the resistance-temperature graph for a superconductor. [1] (c) Give two applications of superconductors and, for each, state the advantage that zero resistivity provides. [2]

12. A tungsten filament wire has diameter 0.030 mm. At room temperature its resistance is 1.8 Ω and its resistivity is 5.6 × 10⁻⁸ Ω m. In use it carries 0.50 A with 12 V across it.

(a) Calculate the length of the wire. [3] (b) Calculate the resistance in use and the ratio of this to the room-temperature resistance. [2] (c) Explain why the resistance changes. [2] (d) Assuming the dimensions do not change, calculate the resistivity in use. [1]

Answers

1. t = 8.0 × 60 = 480 s [1]. Q = It = 2.5 × 10⁻³ × 480 = 1.2 C [1]. N = Q/e = 1.2 / 1.60 × 10⁻¹⁹ = 7.5 × 10¹⁸ electrons [1] [3] Examiner insight: Convert minutes and milliamps before substituting, and show the conversion.

2. (a) I = Q/t = 45/30 = 1.5 A [1] (b) W = VQ [1] = 6.0 × 45 = 270 J [1] [3] Examiner insight: Use the equation that contains the quantities you were given. Here V and Q give W directly.

3. Energy: J = N m = kg m s⁻² × m = kg m² s⁻² [1]. Charge: C = A s, so V = J/C = kg m² s⁻³ A⁻¹ [1]. Ω = V/A = kg m² s⁻³ A⁻² [1] [3] Examiner insight: On a “show that”, write each unit replacement on its own line.

4. (a) R = V/I gives 80.6, 79.7, 80.4, 80.3 and 80.1 Ω [1]. The values are constant within reading variation [1], so the resistor is ohmic, because I is proportional to V at constant temperature [1] (b) Gradient = 0.050 / 4.0 = 1.25 × 10⁻² A V⁻¹ [1]. R = 1/gradient = 80 Ω [1] [5] Examiner insight: Say what the constant R is evidence for and quote the condition (constant temperature). On an I-V graph, R is the reciprocal of the gradient.

5. (a) R = 0.50 / 0.40 × 10⁻³ = 1250 Ω (1.3 kΩ) [1]. R = 0.70 / 12.5 × 10⁻³ = 56 Ω [1] (b) Forward: resistance falls as p.d. rises (1250 Ω to 56 Ω) [1]. Reverse: current too small to measure, so resistance is very large [1] (c) The claim is wrong: R is not constant (it falls from 1250 Ω to 56 Ω), so I is not proportional to V [1] [5] Examiner insight: Support a comparison with figures from the table. A steep graph is not the same as a proportional one.

6. (a) R = 3.0 / 0.24 = 12.5 Ω [1]. R = 9.0 / 0.50 = 18 Ω [1] (b) A larger current makes the filament hotter, so R rises [1]. V = IR then increases faster than I, so the curve bends upwards [1] (c) Temperature changes, so R is not constant and I is not proportional to V [1] (d) ΔV/ΔI = 6.0 / 0.26 = 23 Ω. The resistance at 9.0 V is V/I = 18 Ω, because R is the ratio of V to I at the point, not a slope [1] [6] Examiner insight: On a curve, use both coordinates of the point, and check which quantity is on which axis.

7. Circuit: the lamp in series with a variable power supply and an ammeter [1]; a voltmeter across the lamp alone, in parallel [1]. Increase the p.d. in steps from near zero, recording V and I after the lamp settles [1], without exceeding the rated p.d. [1]. Reverse the supply connections and repeat for negative values [1]. Plot I against V: a curve through the origin, symmetrical about it [1] [6] Examiner insight: Say where each meter is connected and what is varied. State series or parallel explicitly.

8. (a) A = π(0.25 × 10⁻³)² = 1.96 × 10⁻⁷ m² [1]. R = ρL/A [1] = 2.7 × 10⁻⁸ × 1.2 / 1.96 × 10⁻⁷ = 0.17 Ω [1] (b) Half the length halves R. Half the diameter makes A a quarter, which multiplies R by four. The net factor is 2 [1]. R = 2.7 × 10⁻⁸ × 0.60 / π(0.125 × 10⁻³)² = 0.33 Ω, double the first wire [1] [5] Examiner insight: Use proportional reasoning as a check, and give the number too.

9. (a) Gradient = 12.5 / 0.80 = 15.6 Ω m⁻¹ [1] (b) A = π(0.15 × 10⁻³)² = 7.07 × 10⁻⁸ m² [1]. ρ = gradient × A = 15.6 × 7.07 × 10⁻⁸ = 1.1 × 10⁻⁶ Ω m [1] (c) R is directly proportional to L (R = ρL/A with ρ and A constant), so R is zero when L is zero [1] (d) Measure the diameter at several points and in perpendicular directions, then average. Squaring d makes it the dominant uncertainty [1] [5] Examiner insight: Link the improvement to the quantity that dominates the uncertainty. “Measure d at several places” is more specific than “take more readings”.

10. (a) At 15 °C: I = 3.0 / 8000 = 0.375 mA [1]. At 75 °C: I = 3.0 / 1200 = 2.5 mA [1] (b) Heating releases more charge carriers [1], so more current flows for the same p.d. and R falls [1] (c) In a metal the number of free electrons barely changes [1]. Ions vibrate more, so collisions increase and resistance rises [1] (d) For example, a temperature sensor in a thermostat [1] [7] Examiner insight: Name what changes in each material: the number of carriers in the thermistor, the collision rate in the metal.

11. (a) The temperature below which (and at which) the material has zero resistivity [1]. It depends on the material [1] (b) Resistance falls abruptly to zero at the critical temperature and stays zero below it [1] (c) Strong electromagnets (for example scanner or accelerator coils): large currents without heating [1]. Power transmission cables: no energy lost as heat [1] [5] Examiner insight: Pair each application with its reason. A bare application does not link to zero resistivity.

12. (a) A = π(0.015 × 10⁻³)² = 7.07 × 10⁻¹⁰ m² [1]. L = RA/ρ [1] = 1.8 × 7.07 × 10⁻¹⁰ / 5.6 × 10⁻⁸ = 0.023 m [1] (b) R = V/I = 12 / 0.50 = 24 Ω [1]. Ratio = 24 / 1.8 = 13 [1] (c) The filament gets very hot in use [1]. Its ions vibrate more, so there are more collisions with electrons and the resistance is larger [1] (d) ρ = 13.3 × 5.6 × 10⁻⁸ = 7.5 × 10⁻⁷ Ω m [1] [8] Examiner insight: Keep extra digits during the calculation and round only the final answers. Use metres for the diameter in the area.

Where marks are usually lost

• Substituting minutes, milliamps or millimetres without converting.
• Using the gradient of a curve as the resistance.
• Mixing up the gradient on V-I and I-V graphs.
• Using diameter for radius in the area, or not squaring.
• Describing a superconductor as having low resistance rather than zero resistivity.
• Giving a resistivity in ohms instead of ohm metres.

Next steps

• Review the revision notes and study guide.
• Tick off the topic on the printable checklist and visit the course hub.
• Take one of the free 10-minute diagnostics to check what to revise next.
• Book a free trial class if you want help with any of these questions.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Sections 3.4.1 to 3.4.3, shared International AS and A-level content.