AQA AS Level Physics Paper 2, November 2021: Question 20

1 mark · Medium difficulty · Multiple Choice

Calculate the new ammeter reading when the variable resistor in a circuit with a cell of emf 12 V and internal resistance is changed from 10 ohms to 5 ohms.

Practise this question

Question

A multiple-choice question showing a circuit diagram with a cell of emf 12 V and non-negligible internal resistance connected in series with an ammeter and a variable resistor. A voltmeter is connected across the cell. Below the diagram, text states that when the variable resistor is 10 ohms, the voltmeter reads 10 V and the ammeter reads 1.0 A. It asks for the new ammeter reading when the variable resistor is changed to 5 ohms, with four options A (1.4 A), B (1.7 A), C (2.0 A), and D (2.4 A).
Question text

20 In the circuit shown, the cell has an emf of 12 V and an internal resistance which is not

negligible.

When the resistance of the variable resistor is 10 Ω the voltmeter reads 10 V and the

ammeter reads 1.0 A.

The resistance of the variable resistor is changed to 5 Ω.

What is the new reading on the ammeter?

[1 mark]

A 1.4 A

B 1.7 A

C 2.0 A

D 2.4 A

Mark scheme

Show the mark scheme The mark scheme table shows question number 20 with the correct answer option B, corresponding to a reading of 1.7 A.

20 B 1.7 A

How to answer it

Question 20: Internal Resistance and Terminal PD

AQA AS-Level Physics • Electricity

What this question tests

This question assesses your understanding of sources of electromotive force (emf), internal resistance, terminal potential difference, and how changing load resistance affects current in a single-loop DC circuit.

Question Part 20

Determining the new ammeter reading

✅ Correct Answer

Option B (1.7 A)

Worth 1 mark

💡 Key Knowledge

  • Equation linking emf and internal resistance: E = I(R + r) or E = V + Ir
  • Emf ( E ) is a constant property of the cell ( 12 V ).
  • Internal resistance ( r ) remains constant for a given cell.

🧠 Exam Technique

Two-stage problems involving unknown constants (like r ) require you to use the initial conditions to find the constant before calculating the new scenario.

❌ Common Errors

Assuming the terminal potential difference stays at 10 V when the variable resistor is changed. The terminal pd drops because more current flows, increasing the "lost volts" across r .

📐 Step-by-Step Calculation

  1. Find the internal resistance (r) using initial conditions:
    Using E = V + Ir
    12 = 10 + (1.0 × r)
    r = 2.0 Ω
  2. Calculate the new total resistance of the circuit:
    New variable resistor resistance, R = 5.0 Ω
    Total resistance = R + r = 5.0 + 2.0 = 7.0 Ω
  3. Calculate the new current (ammeter reading):
    Using I = E / (R + r)
    I = 12 / 7.0 = 1.714 A → 1.7 A (to 2 s.f.)

Topics

Physics · 3.5 Electricity

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.