AQA AS Level Physics Paper 2, June 2023: Question 30

1 mark · Medium difficulty · Multiple Choice

Determine the range of potential difference observed on a voltmeter connected across a fixed resistor and a variable resistor in a series circuit powered by a 9.0 V battery.

Practise this question

Question

Multiple choice question 30 featuring a circuit diagram with a 9.0 V battery connected in series with a 3.0 k-ohm resistor, a variable resistor ranging from 0 to 3.0 k-ohm, and another 3.0 k-ohm resistor. A voltmeter is connected across the first fixed resistor and the variable resistor. Four options A, B, C, and D give different voltage ranges.
Question text

30 Three resistors are connected in series with a 9.0 V battery of negligible internal

resistance.

The resistance of the variable resistor is varied from 0 to 3.0 kΩ.

The range of potential difference observed on the voltmeter is

[1 mark]

A 0 to 6.0 V

B 3.0 V to 6.0 V

C 4.5 V to 6.0 V

D 4.5 V to 9.0 V

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer is C, corresponding to the range 4.5 V to 6.0 V.

30 C 4.5 V to 6.0 V

How to answer it

Determining the Range of Potential Difference in a Series Circuit

What this question tests

This question assesses your understanding of potential divider circuits, series resistor combinations, and how varying the resistance of a single component alters the output potential difference measured across a section of the circuit.

Question 30: Voltage Range Analysis

AS Level Physics - Electric Circuits

✅ Correct Answer

Option C: 4.5 V to 6.0 V

💡 Key Knowledge

  • In a series circuit, potential difference is shared in direct proportion to resistance ( V = IR ).
  • The voltmeter spans across the variable resistor AND the first fixed 3.0 kΩ resistor.
  • The total supply voltage is a fixed 9.0 V .

🧠 Exam Technique

Always evaluate the two extreme limits of a variable component (minimum resistance and maximum resistance) to find the absolute range of potential difference.

❌ Common Errors

Students often mistake the voltmeter as measuring only the variable resistor, or forget to include the fixed 3.0 kΩ resistor that shares the voltage within the voltmeter's loop.

📐 Step-by-Step Calculation

  1. Identify the components in the voltmeter loop: The voltmeter is connected across the first fixed resistor ( 3.0 kΩ ) and the variable resistor ( 0 to 3.0 kΩ ). Therefore, the measured voltage V is the sum of the potential differences across these two components.
  2. Calculate Lower Limit (Variable resistor = 0 kΩ):
    • Resistance across voltmeter loop = 3.0 kΩ + 0 kΩ = 3.0 kΩ
    • Total circuit resistance = 3.0 kΩ (fixed) + 0 kΩ (variable) + 3.0 kΩ (fixed) = 6.0 kΩ
    • Fraction of total voltage = 3.0 kΩ / 6.0 kΩ = 0.5
    • Lower voltage = 0.5 × 9.0 V = 4.5 V
  3. Calculate Upper Limit (Variable resistor = 3.0 kΩ):
    • Resistance across voltmeter loop = 3.0 kΩ + 3.0 kΩ = 6.0 kΩ
    • Total circuit resistance = 3.0 kΩ (fixed) + 3.0 kΩ (variable) + 3.0 kΩ (fixed) = 9.0 kΩ
    • Fraction of total voltage = 6.0 kΩ / 9.0 kΩ = 2/3
    • Upper voltage = (2/3) × 9.0 V = 6.0 V
  4. Conclusion: The observed range is 4.5 V to 6.0 V , matching Option C.
Mark Scheme Allocation: [1 mark] awarded for selecting C.

Topics

Physics · 3.5 Electricity

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.