OxfordAQA International AS and A-level Physics: DC circuits, potential dividers, emf and internal resistance (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 are aligned to the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, sections 3.4.4 to 3.4.6 (shared International AS and A-level content). Treat meters as ideal and sources as having negligible internal resistance unless told otherwise.

Read the study guide and revision notes first. See also the course page, the checklist and the free diagnostics.

Questions

1. Define the emf of a cell, and explain why the terminal pd is less than the emf when the cell supplies a current. [3]

2. A 24 Ω and an 8.0 Ω resistor are in parallel, and this pair is in series with a 9.0 Ω resistor across a 7.5 V supply. (a) Calculate the total resistance and the supply current. [2] (b) Calculate the current in the 8.0 Ω resistor. [2]

3. A small motor runs from a 6.0 V supply and draws 0.80 A for 2.0 minutes. (a) Calculate the electrical energy supplied. [2] (b) The motor is 25% efficient. Calculate the useful energy output. [2]

4. A 20 Ω and a 30 Ω resistor are connected across a 10 V supply. (a) They are in series. Calculate the power dissipated in each. [3] (b) They are now in parallel across the same supply. Calculate the power in each. [2]

5. A 12 V supply sends 2.4 A into a junction, which splits into two parallel branches. One branch carries 1.5 A. (a) Find the current in the other branch. [1] (b) Find the resistance of the other branch. [2] (c) Name the conservation principle used in part (a). [1] (d) Calculate the total power from the supply and state how it compares with the power in the two branches. [2]

6. A 9.0 V supply is across a 1.2 kΩ resistor in series with a 1.8 kΩ resistor. (a) Calculate the pd across the 1.8 kΩ resistor. [2] (b) A second 1.8 kΩ resistor is now connected across the first one. Calculate the new pd across them. [2] (c) Explain why the pd changed. [1]

7. A 6.0 V supply drives an ntc thermistor and a 4.0 kΩ resistor in series. A switching circuit monitors the pd across the 4.0 kΩ resistor. The thermistor has a resistance of 8.0 kΩ at 15 °C and 2.0 kΩ at 45 °C (data supplied). (a) Calculate the output pd at each temperature. [2] (b) Explain why the monitored pd rises as the temperature rises. [2] (c) A switching circuit operates when the output reaches 3.0 V. Calculate the thermistor resistance at that point. [2]

8. A 12 V supply is across a 2.0 kΩ fixed resistor in series with a variable resistor of range 0 to 6.0 kΩ. The output is across the fixed resistor. (a) State the largest and smallest output pd, with the variable resistor setting for each. [3] (b) Calculate the variable resistor setting that gives 8.0 V. [2]

9. A battery has emf 9.0 V. When it supplies 1.5 A to a resistor, the terminal pd is 8.4 V. (a) Calculate the internal resistance. [2] (b) Calculate the resistance of the external resistor. [1] (c) Calculate the current if the terminals were joined by a wire of negligible resistance. [1] (d) Calculate the power wasted inside the battery in the original circuit. [1]

10. Two identical cells, each of emf 1.6 V and internal resistance 0.30 Ω, supply a 1.9 Ω resistor. (a) They are connected in series. Calculate the current. [3] (b) They are connected in parallel. Calculate the total current. [3]

11. A student investigates a cell by varying the current and measuring the terminal pd. (a) Describe the circuit, including the position of each meter. [2] (b) Explain how the emf and internal resistance are found from a graph of V against I. [2] (c) The line of best fit passes through (0.20 A, 1.82 V) and (0.80 A, 1.40 V). Calculate the internal resistance and emf. [3] (d) Give one reason for opening the switch between readings. [1]

Answers

1. The emf is the energy the cell supplies per coulomb of charge it drives round the circuit [1]. When a current flows, some energy per coulomb is dissipated in the cell’s internal resistance (lost volts, Ir) [1]. Terminal pd = ε − Ir, which is less than ε [1]. [3] Examiner insight: Definitions of emf gain credit for “per unit charge” or “per coulomb”; “the voltage of a cell” is too vague.

2. (a) Parallel: 1/R = 1/24 + 1/8 = 4/24, so R = 6.0 Ω [1]. RT = 6.0 + 9.0 = 15 Ω, and I = 7.5/15 = 0.50 A [1] (b) Pd across the pair = 0.50 × 6.0 = 3.0 V [1]. I = 3.0/8.0 = 0.375 A [1] Examiner insight: Keep the pd across the parallel pair separate from the supply pd. Using 7.5 V across the 8.0 Ω resistor is the usual wrong route.

3. (a) t = 120 s [1]. E = IVt = 0.80 × 6.0 × 120 = 576 J [1] (b) Useful = 0.25 × 576 [1] = 144 J [1]. [4] Examiner insight: Converting minutes to seconds is part of the working. A final answer in the wrong units undermines otherwise correct working.

4. (a) I = 10/50 = 0.20 A [1]. P = I²R: 0.20² × 20 = 0.80 W [1]; 0.20² × 30 = 1.2 W [1] (b) P = V²/R: 10²/20 = 5.0 W [1]; 10²/30 = 3.3 W [1] Examiner insight: Choose the power form that matches the quantity shared: I²R in series, V²/R in parallel. Showing the form used makes the working easy to credit.

5. (a) 2.4 − 1.5 = 0.9 A [1] (b) R = V/I = 12/0.9 [1] = 13 Ω (13.3 Ω) [1] (c) Conservation of charge [1] (d) P = IV = 2.4 × 12 = 29 W (28.8 W) [1]. Branches: 1.5 × 12 + 0.9 × 12 = 18 + 10.8 = 28.8 W, equal to the supply power, so energy is conserved [1]. [6] Examiner insight: If asked to name a principle, name it and say what it applies to, such as “charge is conserved at the junction”.

6. (a) Vout = 9.0 × 1.8/(1.2 + 1.8) [1] = 5.4 V [1] (b) Parallel pair: 1/R = 2/1.8, so R = 0.90 kΩ [1]. Vout = 9.0 × 0.90/(1.2 + 0.90) = 3.86 V [1] (c) The load draws extra current, lowering the resistance of the lower section, so it takes a smaller share of the supply pd [1]. [5] Examiner insight: “The resistance went down” alone is incomplete; link it to the share of the supply pd.

7. (a) 15 °C: 6.0 × 4.0/(4.0 + 8.0) = 2.0 V [1]. 45 °C: 6.0 × 4.0/(4.0 + 2.0) = 4.0 V [1] (b) Thermistor resistance falls as temperature rises [1], so the total resistance falls, the current rises and the 4.0 kΩ resistor’s pd rises [1] (c) 3.0 = 6.0 × 4.0/(4.0 + Rt) [1], so 4.0 + Rt = 8.0 and Rt = 4.0 kΩ [1]. [6] Examiner insight: State which component the output is across before describing a trend. Output direction depends on that choice.

8. (a) Largest: variable resistor at 0 Ω, output 12 V [1]. Smallest: variable resistor at 6.0 kΩ [1], output 12 × 2.0/8.0 = 3.0 V [1] (b) 8.0 = 12 × 2.0/(2.0 + R) [1], so 2.0 + R = 3.0 and R = 1.0 kΩ [1]. [5] Examiner insight: Pair each extreme output with the setting that produces it. A bare pair of values is incomplete.

9. (a) Lost volts = 9.0 − 8.4 = 0.6 V [1]. r = 0.6/1.5 = 0.40 Ω [1] (b) R = 8.4/1.5 = 5.6 Ω [1] (c) I = ε/r = 9.0/0.40 = 22.5 A (23 A) [1] (d) P = I²r = 1.5² × 0.40 = 0.90 W [1]. [5] Examiner insight: Calculating r from 9.0/1.5 (the emf over the current) gives the total resistance, not the internal resistance.

10. (a) Series: ε = 3.2 V, r = 0.60 Ω [1]. I = 3.2/(1.9 + 0.60) [1] = 1.28 A [1] (b) Parallel: ε = 1.6 V, r = 0.30/2 = 0.15 Ω [1]. I = 1.6/(1.9 + 0.15) [1] = 0.780 A [1]. [6] Examiner insight: Show the combined emf and internal resistance before substituting. That way your method is easy to follow.

11. (a) Cell, variable resistor, ammeter and switch in series [1]; voltmeter connected across the cell’s terminals only [1] (b) Plot V against I. V = ε − Ir [1], so the gradient is −r and the vertical intercept is ε [1] (c) Gradient = (1.40 − 1.82)/(0.80 − 0.20) = −0.70 V A⁻¹, so r = 0.70 Ω [1]. Using V = ε − Ir: ε = 1.82 + 0.70 × 0.20 [1] = 1.96 V [1] (d) Continuous current heats the cell and drains it, changing its internal resistance and emf [1]. [8] Examiner insight: A correct r is a positive number, so give the magnitude of the gradient. Reading ε from the last data point rather than the intercept is a common slip.

Where marks are usually lost

• Not inverting 1/R in a parallel calculation.
• Using the supply pd across a branch that only receives part of it.
• Leaving time in minutes in E = IVt.
• Choosing P = V²/R in a series circuit where the pd is not known across the resistor.
• Treating a loaded divider as unloaded.
• Describing a thermistor or LDR trend without stating which component the output is across.
• Dividing the emf by the current and calling the result the internal resistance.
• Not stating the combined emf and internal resistance for cells in series or parallel.
• Answering with the terminal pd where the emf is asked for.

Next steps

• Review the revision notes and the study guide.
• Open the course page and tick off the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

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 3.4.4, 3.4.5 and 3.4.6.