AQA AS Level Physics Paper 2, June 2025: Question 26

1 mark · Easy difficulty · Multiple Choice

Determine the current in wire W using Kirchhoff's first law at the junctions.

Practise this question

Question

A circuit diagram showing two connected junctions with various currents. At the left junction, 3 A enters from the bottom, 4 A leaves from the top, and 1 A leaves to the left. A horizontal wire connects the left junction to the right junction. At the right junction, 3 A enters from the top and 4 A enters from the bottom, with a horizontal wire labelled W extending to the right. Multiple-choice options are A: 1 A, B: 2 A, C: 5 A, D: 7 A.

Mark scheme

Show the mark scheme Mark scheme table indicating question number 26 has correct answer C (5 A), with assessment objective AO1.

How to answer it

Kirchhoff's First Law in Wire Networks

What this question tests

This question assesses your ability to apply Kirchhoff's First Law (the conservation of electric charge) at junctions within an electrical circuit network. You must account for current directions using arrowheads to deduce unknown branch currents.

Question 26 (Multiple Choice)

Topic: Electricity • Charge Conservation • Junction Rule

✅ Correct Answer

Correct Option: C (5 A)

Mark Scheme Breakdown:
• AO1 (Knowledge & Application): 1 Mark for selecting C (5 A).

💡 Key Knowledge

  • Kirchhoff's First Law: The algebraic sum of currents entering a junction equals the sum of currents leaving the junction (ΣIin = ΣIout).
  • Physical Basis: Direct consequence of the conservation of electric charge; charge cannot be created, destroyed, or accumulate at a node.

📐 Step-by-Step Solution

Method 1: Junction-by-Junction Analysis

  1. 1 Analyse the Left Junction:
    • Current entering from below = 3 A
    • Current leaving upwards = 4 A
    • Current leaving to the left = 1 A
    Total known current leaving = 4 A + 1 A = 5 A .
    Applying ΣIin = ΣIout:
    3 A + Imiddle = 5 A ⇒ Imiddle = 2 A flowing from right to left into this junction.
  2. 2 Analyse the Right Junction:
    • Current entering from above = 3 A
    • Current entering from below = 4 A
    • Current leaving to the left (towards left junction) = 2 A
    Total current entering = 3 A + 4 A = 7 A .
    Applying ΣIin = ΣIout:
    7 A = 2 A + IW ⇒ IW = 7 A − 2 A = 5 A (flowing rightwards).

Method 2: Whole System ("Super-node") Approach

Treat both junctions and the connecting wire as a single closed system:

  • Total external current flowing IN: 3 A (bottom-left) + 3 A (top-right) + 4 A (bottom-right) = 10 A
  • Total external current flowing OUT: 1 A (left) + 4 A (top-left) + IW = 5 A + IW
  • Conservation of charge: 10 A = 5 A + IW ⇒ IW = 5 A

🧠 Exam Technique

  • Label intermediate wires: Draw an arrow and write the value on the middle horizontal wire before attempting to solve for W.
  • Double-check with Method 2: In MCQs, treating multiple connected nodes as one boundary is faster and prevents intermediate sign errors.

❌ Common Errors & Pitfalls

  • Option D (7 A): Students simply add the two currents at the right junction (3 A + 4 A) while ignoring the connecting wire to the left junction.
  • Option A (1 A): Incorrectly assuming current travels left-to-right in the central wire, subtracting 2 A from 3 A.
  • Direction Errors: Confusing the arrows pointing into a node with arrows pointing away from it.

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.