AQA A-Level Physics Paper 1, November 2020: Question 29

1 mark · Medium difficulty · Multiple Choice

Calculate the internal resistance of a cell given the change in potential difference across its terminals when a switch is closed in a circuit containing a resistor.

Practise this question

Question

Multiple choice question 29. A circuit diagram shows a cell connected in series with a switch and a 2.0 ohm resistor, with a voltmeter connected directly across the cell terminals. The text states that initially the voltmeter reading is V, and when the switch is closed, the reading becomes V/3. Four multiple-choice options for the internal resistance of the cell are given: A 0.33 ohms, B 0.67 ohms, C 4.0 ohms, and D 6.0 ohms.
Question text

29 In the circuit, the reading of the voltmeter is V.

V

When the switch is closed the reading becomes .

What is the internal resistance of the cell?

[1 mark]

A 0.33 Ω

B 0.67 Ω

C 4.0 Ω

D 6.0 Ω

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 29 is C.

29 C

How to answer it

Determining Internal Resistance from Terminal PD

📌 What this question tests

This question assesses your understanding of electromotive force (emf), terminal potential difference, and internal resistance in DC circuits. Specifically, it tests how open-circuit and closed-circuit conditions affect voltmeter readings across a cell with internal resistance.

Question 29: Multiple Choice Analysis

AQA A-Level Physics

✅ Correct Answer: C (4.0 Ω)

Option C is the correct value for the internal resistance of the cell based on the drop in potential difference when the circuit is completed.

💡 Key Knowledge

  • When the switch is open, no current flows. The voltmeter measures the open-circuit emf ( E ) of the cell, so V = E .
  • When the switch is closed, current I flows through the external resistor R and internal resistance r .
  • The terminal p.d. drops to V / 3 because of the "lost volts" across the internal resistance.

🧠 Exam Technique

For quick multiple-choice solutions, use proportional reasoning or potential divider equations rather than full simultaneous equations to save valuable time in the exam.

❌ Common Errors

  • Assuming the voltmeter reads terminal p.d. when the switch is open (it actually reads emf as current is zero).
  • Forgetting that internal resistance acts in series with the load resistor, meaning the total resistance of the circuit is R + r .

📐 Step-by-Step Calculation

  1. Identify open-circuit state: With the switch open, I = 0 , so the voltmeter reads the emf of the cell: emf = V .
  2. Identify closed-circuit state: With the switch closed, the voltmeter reads the terminal p.d., which is given as V / 3 .
  3. Apply the potential divider rule or current formula: The total resistance of the circuit is R + r (where R = 2.0 Ω ). Using the potential divider relationship for the external resistor:
    Terminal p.d. = emf × [ R / (R + r) ]
    V / 3 = V × [ 2.0 / (2.0 + r) ]
  4. Simplify and solve: Cancel V from both sides:
    1 / 3 = 2.0 / (2.0 + r)
    2.0 + r = 3 × 2.0
    2.0 + r = 6.0
    r = 6.0 - 2.0 = 4.0 Ω
Examiner Note: Top-scoring students instantly recognise that if the terminal p.d. drops to one-third of the emf, the potential drop across the internal resistance must be two-thirds of the emf. Since resistance is directly proportional to voltage in a series circuit, the internal resistance must be twice the external resistance ( 2 × 2.0 Ω = 4.0 Ω ).

Topics

Physics · 3.5 Electricity

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.