AQA AS Level Physics Paper 2, June 2023: Question 31

1 mark · Medium difficulty · Multiple Choice

Select the correct circuit diagram used to determine the current-voltage characteristic of a filament lamp between 0 and 6.0 V using a potential divider setup.

Practise this question

Question

Multiple-choice question asking to identify the correct circuit for finding the current-voltage characteristic of a filament lamp from 0 to 6.0 V. Four circuits labeled A, B, C, and D are shown, each containing a battery, a variable resistor or potential divider, a lamp, an ammeter, and a voltmeter in various arrangements.
Question text

31 The current–voltage characteristic between 0 and 6.0 V is required for a filament lamp.

The lamp is connected in a circuit with a battery of emf 6.0 V and negligible internal

resistance.

Which circuit should be used?

[1 mark]

A

B

C

D

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer is option A, displaying the corresponding circuit diagram with a potential divider and properly placed ammeter and voltmeter.

31 A

How to answer it

Circuit Selection for Filament Lamp Characteristics

What this question tests

This question assesses your understanding of how to configure a potentiometer (potential divider circuit) to obtain a continuous current-voltage (I-V) characteristic curve starting from 0 V up to the maximum supply voltage (6.0 V). It also tests correct meter placement (ammeter in series, voltmeter in parallel) for component testing.

Question 3.1 [1 Mark

Selecting the Correct Circuit Diagram

✅ Correct Answer: Circuit A

Circuit A uses a potentiometer (potential divider) arrangement that allows the potential difference across the lamp to be smoothly varied all the way down to 0 V, combined with an ammeter in series and a voltmeter in parallel with the lamp.

💡 Key Knowledge: Potentiometers vs. Variable Resistors

  • Potentiometer (Potential Divider): Connects across the full supply voltage, allowing output potential difference to start at 0 V and increase up to the source emf (6.0 V).
  • Variable Resistor in Series: Can limit current, but cannot reduce the potential difference across a component down to 0 V because the component always shares the voltage drop with the variable resistor.

🧠 Exam Technique

Eliminate options systematically:

  1. Look at the variable component: Circuits C and D use a simple variable resistor in series, so eliminate them immediately since you need a range starting from 0 V.
  2. Compare A and B: Circuit B places the ammeter in the main supply branch rather than directly in series with the lamp branch, which would measure total current instead of just the lamp current if other branches existed, or fails proper placement conventions. Circuit A correctly positions the ammeter in series with the lamp and the voltmeter across the lamp.

❌ Common Errors

  • Choosing circuits C or D because students confuse a potentiometer symbol with a rheostat (variable resistor).
  • Failing to realize that a complete I-V characteristic from 0 V requires a potential divider setup.
Mark Scheme Guidance: 1 mark awarded for selecting option A. No alternative marks or follow-throughs apply for multiple-choice questions.

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.