AQA A-Level Physics: Electricity (7408) – 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 Electricity of the AQA AS and A-level Physics specification (7407/7408, version 1.4, AS and A-level exams June 2016 onwards): 3.5.1.1 to 3.5.1.6, including Required practicals 5 and 6. Electricity is AS and A-level content, assessed at A-level in Paper 1 and assumed in Paper 2; it is not one of the options (sections 3.9 to 3.13), one of which you study for Paper 3 Section B.

Learn the content first with the study guide. Take the charge on an electron as 1.60 × 10⁻¹⁹ C in magnitude. Treat meters as ideal and supplies as having negligible internal resistance unless told otherwise.

Questions

1. State what is meant by:

(a) electric current [1] (b) potential difference. [1]

2. A phone charger supplies a steady 0.75 A for 40 minutes at 5.0 V.

(a) Calculate the charge that flows. [2] (b) Calculate the number of electrons that flow. [1] (c) Calculate the energy transferred. [2]

3. Describe the I–V characteristic of a filament lamp, with I on the vertical axis, and explain its shape. [4]

4.

(a) Describe the I–V characteristic of a semiconductor diode in forward and reverse bias. [3] (b) In forward bias a diode has a pd of 0.70 V across it and a current of 14 mA. Calculate its resistance at this point. [1]

5. A wire is 1.20 m long with diameter 0.25 mm. The material has resistivity 1.1 × 10⁻⁶ Ω m.

(a) Calculate the resistance of the wire. [3] (b) The wire is connected across a 6.0 V supply. Calculate the power dissipated. [2]

6. Describe how to determine the resistivity of the metal in a wire using a micrometer, an ammeter, a voltmeter and a metre rule. Include how you would use a graph. [6]

7. A 12 V supply is connected to two parallel branches. Branch X is a 2.0 Ω resistor in series with a 4.0 Ω resistor. Branch Y is a 9.0 Ω resistor in series with a 3.0 Ω resistor. The 2.0 Ω and 9.0 Ω resistors are connected to the positive terminal. Point P lies between the 2.0 Ω and 4.0 Ω resistors; point Q lies between the 9.0 Ω and 3.0 Ω resistors.

(a) Calculate the total resistance of the circuit. [2] (b) Calculate the current from the supply. [1] (c) Calculate the current in each branch. [2] (d) Calculate the pd between P and Q. [3]

8. State the conservation law that explains each of the following:

(a) the total current into a junction equals the total current out [1] (b) round a closed loop, the sum of the emfs equals the sum of the pds. [1]

9. A 6.0 V supply is connected across a 10 kΩ fixed resistor in series with an LDR. V_out is measured across the fixed resistor. The LDR’s resistance is 90 kΩ in the dark and 2.0 kΩ in bright light.

(a) Calculate V_out in the dark and in bright light. [3] (b) Suggest how to change the circuit so that V_out rises when it gets dark. [1] (c) In bright light, a 10 kΩ load is connected across the output. Calculate the new V_out. [3]

10. A student investigates the emf and internal resistance of a battery (Required practical 6). The results are:

I / A	0.20	0.40	0.60	0.80	1.00
V / V	4.20	3.90	3.60	3.30	3.00

(a) Describe the circuit used and how the readings are varied. [3] (b) Use the data to determine the emf ε and internal resistance r. [4] (c) Explain why the student should switch off between readings. [1] (d) The battery is connected to a 6.0 Ω resistor. Calculate the terminal pd. [2]

11.

(a) Compare how the resistance of a metal conductor and of an ntc thermistor change as temperature rises, and explain each. [3] (b) State what is meant by a superconductor. [2] (c) Give two applications of superconductors. [2]

12. Four identical cells each have emf 1.5 V and internal resistance 0.40 Ω.

(a) The cells are connected in series to a 4.4 Ω resistor. Calculate the current and the terminal pd. [3] (b) The cells are instead connected in parallel to the same resistor. Calculate the current. [2] (c) For the series arrangement, calculate the energy transferred in the 4.4 Ω resistor in 10 minutes. [2]

Answers

1. (a) The rate of flow of charge (I = ΔQ/Δt) [1]. (b) The work done (energy transferred) per unit charge (V = W/Q) [1]. Examiner insight: “Energy per charge” without saying work done or energy transferred may not be accepted; give the full wording.

2. (a) Δt = 40 × 60 = 2400 s [1]; Q = IΔt = 0.75 × 2400 = 1800 C [1]. (b) N = 1800 / 1.60 × 10⁻¹⁹ = 1.1 × 10²² electrons [1]. (c) W = QV [1] = 1800 × 5.0 = 9000 J [1]. Examiner insight: Using 40 instead of 2400 s loses the conversion mark; write Q on its own line so that later steps can still be followed and credited.

3. Straight line through the origin at low current [1]; gradient decreases as V increases, and the graph is symmetrical for reversed pd [1]. Larger current raises the filament temperature [1], so the metal ions vibrate more, the charge carriers collide more and the resistance increases [1]. Examiner insight: The explanation needs the whole chain from current to temperature to ion vibration to resistance; “it gets hot” alone is not enough.

4. (a) Forward bias: almost no current below a threshold pd [1], then current rises steeply [1]. Reverse bias: very high resistance, current about zero [1]. (b) R = V/I = 0.70 / 0.014 = 50 Ω [1]. Examiner insight: Convert 14 mA to 0.014 A; a bare “0.05” with no unit or power of ten earns nothing.

5. (a) A = π(0.25 × 10⁻³)²/4 = 4.91 × 10⁻⁸ m² [1]. R = ρL/A [1] = 1.1 × 10⁻⁶ × 1.20 / 4.91 × 10⁻⁸ = 27 Ω [1]. (b) P = V²/R [1] = 6.0² / 26.9 = 1.3 W [1]. Examiner insight: Using the diameter as the radius in πr² gives four times the area; show A on its own line so the method mark is visible.

6. Measure the diameter with the micrometer at several points along the wire and in different orientations, then take the mean [1]. Calculate A = πd²/4 [1]. Connect the wire in series with the ammeter and a supply, with the voltmeter across a measured length L, using the metre rule [1]. Record V and I and find R = V/I for a range of lengths [1]. Plot R against L; the gradient is ρ/A [1], so ρ = gradient × A [1]. Examiner insight: A six-mark method needs measurements, the calculation and the graph analysis; a list of equipment with no gradient step leaves marks behind.

7. (a) Branch X = 6.0 Ω, branch Y = 12 Ω [1]. 1/R = 1/6.0 + 1/12, so R = 4.0 Ω [1]. (b) I = 12 / 4.0 = 3.0 A [1]. (c) I_X = 12 / 6.0 = 2.0 A [1]; I_Y = 12 / 12 = 1.0 A [1]. (d) Pd across 4.0 Ω = 2.0 × 4.0 = 8.0 V, so P is 8.0 V above the negative terminal [1]. Pd across 3.0 Ω = 1.0 × 3.0 = 3.0 V, so Q is 3.0 V above it [1]. Pd between P and Q = 8.0 − 3.0 = 5.0 V [1]. Examiner insight: Measure both potentials from the same reference point; subtracting a pd taken from the positive end from one taken from the negative end gives a wrong answer.

8. (a) Conservation of charge [1]. (b) Conservation of energy [1]. Examiner insight: “Conservation of current” or “of voltage” is not a physical conservation law, so it does not answer the question.

9. (a) Dark: V_out = 6.0 × 10/(10 + 90) [1] = 0.60 V [1]. Light: V_out = 6.0 × 10/12 = 5.0 V [1]. (b) Take V_out across the LDR instead (or swap the two components) [1]. (c) 10 kΩ in parallel with 10 kΩ = 5.0 kΩ [1]. V_out = 6.0 × 5.0/(5.0 + 2.0) [1] = 4.3 V [1]. Examiner insight: For a loaded divider, the parallel combination must replace the fixed resistor in both the numerator and the total.

10. (a) Battery in series with an ammeter and a variable resistor [1]; voltmeter connected across the battery terminals [1]; change the variable resistor to vary I and record V each time [1]. (b) Gradient = (3.00 − 4.20)/(1.00 − 0.20) = −1.5 V A⁻¹ [1], so r = 1.5 Ω [1]. Intercept: ε = V + Ir = 4.20 + 0.20 × 1.5 [1] = 4.5 V [1]. (c) To stop the battery warming up, which would change its internal resistance [1]. (d) I = 4.5/(6.0 + 1.5) = 0.60 A [1]; V = IR = 0.60 × 6.0 = 3.6 V [1]. Examiner insight: Read the gradient over a wide interval of the line, not from two adjacent points, and give r as a positive value.

11. (a) Metal: resistance increases, because the ions vibrate more and impede charge carriers [1]. ntc thermistor: resistance decreases [1], because more charge carriers are released as temperature rises [1]. (b) A material with zero resistivity [1] at and below a critical temperature, which depends on the material [1]. (c) Producing strong magnetic fields [1]; reducing energy loss in the transmission of electric power [1]. Examiner insight: “Very low resistance” is wrong for a superconductor; state zero resistivity and include the critical temperature condition.

12. (a) ε = 4 × 1.5 = 6.0 V and r = 4 × 0.40 = 1.6 Ω [1]. I = 6.0/(4.4 + 1.6) = 1.0 A [1]. V = IR = 4.4 V [1]. (b) ε = 1.5 V, r = 0.40/4 = 0.10 Ω [1]. I = 1.5/4.5 = 0.33 A [1]. (c) E = I²Rt = 1.0² × 4.4 × 600 [1] = 2.6 × 10³ J [1]. Examiner insight: Identical cells in parallel keep the emf of one cell; multiplying the emf by four is a common slip that loses both marks.

Where marks are usually lost

• Leaving minutes or mA unconverted before using I = ΔQ/Δt or R = V/I.
• Using the diameter as the radius when finding a wire’s cross-sectional area.
• Taking resistance from the gradient of a curved I–V graph.
• Forgetting that the parallel branches share the supply pd only when nothing is in series with them.
• Missing the temperature step when explaining the filament lamp.
• Ignoring the loading effect of a resistor placed across a potential divider output.
• Quoting r as negative because the gradient of V against I is negative.
• Writing “very low resistance” for a superconductor.

Next steps

• Recap with the Electricity revision notes.
• Go back to the Electricity study guide for worked examples.
• Uncertainties for Required practical 5: measurements practice.
• See the AQA A-Level Physics hub and the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA AS and A-level Physics specification (7407/7408), version 1.4, July 2026, for AS and A-level exams June 2016 onwards, published by AQA. Section 3.5 Electricity.