AQA AS Level Physics Paper 2, June 2025: Question 7

1 mark · Medium difficulty · Multiple Choice

Identify which circuit change requires the sliding contact of a balanced bridge circuit to move towards X to restore zero current.

Practise this question

Question

A circuit diagram showing a cell connected across two parallel branches. The top branch contains two resistors in series labelled M and N. The bottom branch consists of a long uniform metal wire labelled XY, with end X on the left and end Y on the right. An ammeter is connected between the junction of resistors M and N and a movable contact at point P on wire XY. The ammeter initially reads zero. The question asks which change to the circuit causes the balance point P to shift towards X.

Mark scheme

Show the mark scheme Mark scheme table row for question 7 showing the correct answer as option A ('N was replaced with a resistor of greater resistance.') with assessment objective AO2.

How to answer it

⚡ AQA AS Level Physics • Section A / Multiple Choice

Balanced Potential Divider Circuit (Metre Bridge)

What this question tests

This question assesses your understanding of potential dividers in parallel (a Wheatstone bridge arrangement), the relationship between wire length and resistance ( R ∝ L ), and how circuit modifications affect the balance point where no current flows ( I = 0 ).

Question Analysis & Full Solution

Question 07 • 1 Mark

✅ Correct Answer

A: N was replaced with a resistor of greater resistance.

Award 1 mark for selecting option A (Assessment Objective AO2).

💡 Key Knowledge

  • Null Deflection Condition: The ammeter reads zero when the potential at the junction between M and N equals the electric potential at point P.
  • Ratio Principle: For two parallel branches connected to the same source:
    RM / RN = RXP / RPY
  • Uniform Wire: For a uniform wire of constant resistivity and cross-sectional area, resistance is directly proportional to length ( R ∝ L ). Thus:
    RXP / RPY = LXP / LPY

📐 Step-by-Step Deduction

  1. Identify the initial balance condition:
    The potential difference across resistor M matches the potential difference across wire segment XP:
    VM / Vtotal = RM / (RM + RN) = LXP / LXY
  2. Analyze the effect of shifting P towards X:
    Moving P closer to X reduces the length LXP , which means the fraction LXP / LXY decreases.
  3. Determine what must happen to the resistor branch:
    For balance to be restored, the fraction RM / (RM + RN) must also decrease.
  4. Evaluate how to decrease this fraction:
    To make RM / (RM + RN) smaller:
    • RM must decrease, OR
    • RN must increase (increasing the denominator).
    Therefore, replacing N with a larger resistor requires P to move towards X.

🧠 Exam Technique: Elimination Strategy

  • Rule out B instantly: Replacing wire XY with a wire of greater total resistance changes its resistance per metre uniformly. The ratio of lengths LXP / LPY required for balance remains completely unchanged.
  • Rule out C and D instantly: Changing the cell's emf (C) or internal resistance (D) changes the terminal potential difference across both parallel branches equally. Because the balance point depends purely on the ratio of resistances, the terminal p.d. cancels out completely.

❌ Common Misconceptions & Traps

  • Confusing the direction of movement: Students often think increasing RN makes the balance point move towards Y. Remember: higher RN means N takes a greater share of the total p.d., leaving M with a smaller share. Segment XP must therefore also have a smaller share of length.
  • Thinking cell properties alter balance: A common error is assuming that a cell with internal resistance "steals" voltage and shifts the balance point. Null methods are independent of supply voltage!

Topics

Physics · 3.5 Electricity

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.