AQA AS Level Physics Paper 2, June 2023: Question 32

1 mark · Medium difficulty · Multiple Choice

Calculate the ratio of the power dissipated in resistor X to the power dissipated in resistor Z within a given circuit containing three resistors.

Practise this question

Question

A multiple-choice physics question showing a circuit diagram with a DC power supply connected in parallel to two branches: one branch contains two resistors in series labelled X and Y with resistance R each, and the other branch contains a single resistor labelled Z with resistance 2R. Below the diagram, text asks to find the ratio of the power in X to the power in Z, followed by four options: A (1/4), B (1/2), C (2), and D (4).
Question text

32 The diagram shows a circuit containing three resistors X, Y and Z.

X and Y each have resistance R.

Z has resistance 2R.

power in X

What is ?

power in Z

[1 mark]

A

B

C 2

D 4

Mark scheme

Show the mark scheme The mark scheme table shows question number 32 with the correct answer option B, indicating the ratio is 1/2.

32 B

How to answer it

Power Ratio in a Resistor Circuit

What this question tests

This question assesses your ability to apply Kirchhoff's laws, current division in parallel branches, and the electrical power equations (P = I²R or P = V²/R). You must evaluate how potential differences and currents distribute across series and parallel combinations containing algebraic resistances.

Question 3.2 — Multiple Choice (1 Mark)

Determining the Power Ratio

✅ Correct Answer

Option B (1/2)

The ratio of the power dissipated in resistor X to resistor Z is 1/2.

💡 Key Knowledge

  • Resistors X and Y are connected in series with each other, forming a branch with total resistance R + R = 2R .
  • Resistor Z is in a parallel branch with resistance 2R .
  • Power formula linked to current: P = I²R .

🧠 Exam Technique

For multiple-choice ratio questions, avoid substituting arbitrary numbers unless necessary. Use algebraic ratios to see how variables scale directly, which prevents arithmetic slips and saves time.

❌ Common Errors

Students often forget that current splits unequally or assume all resistors carry the same current. Another trap is mixing up P = I²R and P = V²/R when applied to parallel components.

📐 Step-by-Step Calculation

  1. Analyze Branch Resistances: The top branch contains X and Y in series, giving a combined resistance of R_top = R + R = 2R . The bottom branch contains resistor Z, which also has a resistance of R_bottom = 2R .
  2. Determine Current Distribution: Since both parallel branches have identical total resistance ( 2R ), the total current I from the supply splits equally between them. Therefore, the current through resistor X is I_X = I/2 , and the current through resistor Z is I_Z = I/2 (meaning I_X = I_Z ).
  3. Apply the Power Equation: Use P = I²R for both components:
    • Power in X: P_X = (I_X)² × R = (I/2)² × R = (I²/4)R
    • Power in Z: P_Z = (I_Z)² × (2R) = (I/2)² × 2R = (I²/4)(2R) = 2(I²/4)R
  4. Calculate the Ratio:
    P_X / P_Z = ((I²/4)R) / (2(I²/4)R) = 1 / 2
Mark Scheme Note: 1 mark awarded for selecting B.

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.