AQA AS Level Physics Paper 2, June 2025: Question 30

1 mark · Medium difficulty · Multiple Choice

Determine the power dissipated in a resistor connected in parallel with another resistor across a 6.0 V cell.

Practise this question

Question

A circuit diagram showing a 6.0 V cell of negligible internal resistance connected in parallel to two branches: one branch containing a 20 ohm resistor and the other branch containing an unknown resistor R. The text states that the total current supplied by the cell is 1.1 A, and asks for the power dissipated in resistor R, with multiple-choice options A: 1.8 W, B: 3.3 W, C: 4.8 W, and D: 6.6 W.

Mark scheme

Show the mark scheme A mark scheme row for question 30 showing the correct answer as option C, with value 4.8 W and assessment objective AO2.

How to answer it

Power Dissipated in a Parallel Circuit

What this question tests

This multiple-choice question assesses your ability to apply Kirchhoff’s First Law (conservation of charge) and potential difference rules in parallel circuits, and to calculate electrical power using P = IV or P = V²/R .

Question 30

Multiple Choice — 1 Mark

Assessment Objective: AO2 (Application of knowledge and understanding in a practical context)

✅ Correct Answer

C : 4.8 W

The current branching through resistor R is 0.80 A at a potential difference of 6.0 V, giving a dissipated power of 4.8 W.

💡 Key Knowledge

  • Negligible internal resistance: The terminal potential difference equals the cell emf ( V = 6.0 V ) across every parallel branch.
  • Parallel branches: V_total = V₁ = V₂ = 6.0 V .
  • Kirchhoff’s Current Law: I_total = I₂₀ + I_R .
  • Power formulae: P = IV = I²R = V²/R .

📐 Step-by-Step Solutions

Method 1: Branch Currents (Most Direct)

  1. Calculate the current in the 20 Ω branch:
    Since the cell has negligible internal resistance, the p.d. across the 20 Ω resistor is 6.0 V.
    I₂₀ = V / R = 6.0 V / 20 Ω = 0.30 A
  2. Determine current through resistor R using Kirchhoff's First Law:
    The total current leaving the cell is 1.1 A.
    I_R = I_total - I₂₀ = 1.1 A - 0.30 A = 0.80 A
  3. Calculate power dissipated in R:
    P_R = V × I_R = 6.0 V × 0.80 A = 4.8 W

Method 2: Conservation of Energy (Alternative)

  1. Total power delivered by the cell:
    P_total = E × I_total = 6.0 V × 1.1 A = 6.6 W
  2. Power dissipated by the 20 Ω resistor:
    P₂₀ = V² / R = (6.0)² / 20 = 36 / 20 = 1.8 W
  3. Power dissipated in R:
    P_R = P_total - P₂₀ = 6.6 W - 1.8 W = 4.8 W

🧠 Exam Technique & Speed Tip

  • Inspect the distractors: Notice that each incorrect option represents a partial or miscalculated step:
    • A (1.8 W): Power in the 20 Ω resistor ( 36 / 20 ).
    • B (3.3 W): Result of incorrectly halving the total power ( 6.6 / 2 ).
    • D (6.6 W): Total power output of the cell ( 6.0 × 1.1 ).
  • In multiple-choice circuit questions, calculating the total power and subtracting the known branch power is often faster and leaves fewer chances for rounding mistakes.

❌ Common Errors

  • Assuming identical resistors: Splitting the current equally (0.55 A each) to get 3.3 W. Resistor R is not necessarily 20 Ω!
  • Selecting the wrong resistor's power: Calculating 1.8 W for the 20 Ω branch and accidentally ticking A without subtracting from total.
  • Selecting total circuit power: Calculating 6.0 × 1.1 = 6.6 W and stopping there (option D).

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.