AQA A-Level Physics Paper 1, June 2023: Question 29

1 mark · Medium difficulty · Multiple Choice

Calculate the percentage decrease in resistance when a resistor dissipating 100 W is replaced after the supply voltage is reduced from 25 V to 20 V to maintain the same power dissipation.

Practise this question

Question

Multiple choice question 29 featuring text about a resistor dissipating 100 W connected across a 25 V supply, then connected across a 20 V supply maintaining the same power dissipation. Four options are provided: A 20, B 36, C 64, and D 80, each with an answer box.
Question text

29 A resistor dissipates 100 W when connected across a 25 V supply with negligible internal

resistance.

The supply output is reduced to 20 V and the resistor is replaced so that the power

dissipated is still 100 W.

What is the percentage decrease in resistance?

[1 mark]

A 20

B 36

C 64

D 80

Mark scheme

Show the mark scheme Mark scheme indicating question number 29 has the correct answer B with a value of 36.

29 B 36

How to answer it

Power Dissipation and Percentage Decrease in Resistance

Question 29 • Multiple Choice • 1 Mark

What this question tests

This question assesses your ability to apply electrical power equations relating potential difference, resistance, and power ( P = V² / R ), manipulate algebraic expressions for resistance, and calculate percentage decreases under changing circuit conditions.

Exam Question Breakdown

✅ Correct Answer: Option B (36%)

The correct option is B because the initial resistance is 6.25 Ω and the final resistance is 4.00 Ω , resulting in a 36% decrease.

💡 Key Knowledge

  • Power formula involving voltage and resistance: P = V² / R
  • Rearranging for resistance: R = V² / P
  • Percentage decrease formula: (Change / Original) × 100%

🧠 Exam Technique

Since power ( P ) remains constant at 100 W , you can bypass calculating actual numerical resistance values if you spot the proportional relationship: R ∝ V² . This allows you to work entirely with voltage ratios to save precious time in multiple-choice questions.

❌ Common Errors

  • Inverted percentage: Dividing the change by the final resistance instead of the initial resistance.
  • Wrong power equation: Using P = I²R or P = IV without correctly accounting for the changing current.

📐 Step-by-Step Calculation

  1. Find initial resistance (R₁):
    Using R₁ = V₁² / P  →  R₁ = 25² / 100 = 625 / 100 = 6.25 Ω
  2. Find final resistance (R₂):
    Using R₂ = V₂² / P  →  R₂ = 20² / 100 = 400 / 100 = 4.00 Ω
  3. Calculate percentage decrease:
    Percentage Decrease = ((R₁ - R₂) / R₁) × 100%
    Percentage Decrease = ((6.25 - 4.00) / 6.25) × 100% = (2.25 / 6.25) × 100% = 36%
Alternative Ratio Method: Since R ∝ V² , R₂ / R₁ = (V₂ / V₁)² = (20 / 25)² = (0.8)² = 0.64 . This means R₂ is 64% of R₁ , so the resistance has decreased by 100% - 64% = 36% .

Topics

Physics · 3.5 Electricity

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.