AQA A-Level Physics Paper 1, June 2025: Question 28

1 mark · Medium difficulty · Multiple Choice

Determine the voltmeter and ammeter readings in a circuit when a switch in parallel with one resistor is closed.

Practise this question

Question

A circuit diagram containing a battery with negligible internal resistance connected in series with an ammeter, resistor R1, and resistor R2. A voltmeter is connected in parallel across R1. A switch is connected in parallel across R2. The text states that R1 is double R2, and when the switch is open, the voltmeter reads 12 V and the ammeter reads 12 mA. A multiple-choice table gives four options (A, B, C, D) for the voltmeter and ammeter readings when the switch is closed.
Question text

28 In a circuit, the resistance of resistor R1 is double the resistance of resistor R2.

The internal resistance of the battery is negligible.

When the switch is open, the voltmeter reads 12 V and the ammeter reads 12 mA.

What are the readings on the voltmeter and ammeter when the switch is closed?

[1 mark]

Voltmeter reading / V Ammeter reading / mA

A 4 12

B 18 12

C 18 18

D 24 24

Mark scheme

Show the mark scheme Mark scheme for question 28 showing the correct answer as option C, with voltmeter reading 18 V and ammeter reading 18 mA, assessed under AO2.

28 C 18 18 AO2

How to answer it

DC Circuits: Potential Dividers & Short Circuits

What this question tests

This question assesses your ability to analyse direct current (DC) circuits in multiple states:

  • Potential divider relationships: How voltage splits across resistors in series in direct proportion to resistance (V ∝ R).
  • Ohm's Law: Applying V = I R across individual components and the circuit as a whole.
  • Battery EMF: Recognising that for a cell with negligible internal resistance, the terminal p.d. remains constant and equal to its EMF (ε).
  • Effect of switching / short circuits: Understanding that closing a switch in parallel with a resistor eliminates potential difference across it ( V = 0 ) and removes its resistance from the circuit.

Question 28 (1 Mark)

Multiple Choice Analysis

✅ Correct Answer: C (18 V, 18 mA)

When the switch is closed:

  • Voltmeter reading: 18 V
  • Ammeter reading: 18 mA
Award 1 mark for selecting option C (AO2).

💡 Key Knowledge

  • Series Voltage Rule: In a series loop, ε = V₁ + V₂ .
  • Voltage Ratio: Since I is constant in series, V₁ / V₂ = R₁ / R₂ .
  • Short Circuit: A closed ideal switch in parallel with a component has R = 0 Ω , so all current bypasses the component, dropping its p.d. to 0 V .
  • Zero Internal Resistance: Total circuit EMF is unaffected by changes in current.

Step-by-Step Calculation

📐 Circuit Derivation & Values

1 Analyse Initial State (Switch Open):

  • Resistors R₁ and R₂ are connected in series.
  • Ammeter reading ( I_open ) = 12 mA = 12 × 10⁻³ A.
  • Voltmeter across R₁ ( V₁ ) = 12 V.
  • Calculate R₁: R₁ = V₁ / I = 12 V / (12 × 10⁻³ A) = 1000 Ω (1 kΩ) .

2 Find Battery EMF (ε):

  • The question states: R₁ = 2 R₂ &implies; R₂ = R₁ / 2 = 500 Ω .
  • Voltage across R₂: V₂ = I × R₂ = (12 × 10⁻³ A) × 500 Ω = 6 V .
  • Alternative shortcut: Since R₁ = 2 R₂ , V₁ = 2 V₂ &implies; V₂ = 12 / 2 = 6 V .
  • Battery EMF: ε = V₁ + V₂ = 12 V + 6 V = 18 V .

3 Analyse Final State (Switch Closed):

  • Closing the switch creates a zero-resistance bypass across R₂ (shorting it out).
  • Potential difference across R₂ becomes 0 V .
  • The voltmeter is across R₁, which is now the only resistive component across the battery.
  • New Voltmeter Reading: V' = ε = 18 V .
  • New Ammeter Reading: I' = ε / R₁ = 18 V / 1000 Ω = 0.018 A = 18 mA .

Exam Technique & Common Pitfalls

🧠 Exam Technique: Ratio Method

You don't even need to calculate the exact resistances to solve this in under 30 seconds:

  • Open switch: R₁ has 2 parts of resistance, R₂ has 1 part (total 3 parts).
  • 2 parts = 12 V &implies; each part is 6 V &implies; Total EMF = 18 V.
  • Closed switch: Total resistance drops from 3 parts to 2 parts (a factor of 2/3).
  • Since I ∝ 1/R , current increases by a factor of 3/2:
    12 mA × 1.5 = 18 mA .

❌ Common Errors to Avoid

  • Confusing the ratio (Selecting A): Thinking R₂ = 2 R₁ instead of R₁ = 2 R₂ , leading to V₂ = 24 V and incorrect EMF.
  • Assuming constant current (Selecting B): Forgetting that shorting R₂ decreases total circuit resistance from 1500 Ω to 1000 Ω, which must cause the current to increase.
  • Assuming 12 V is the cell EMF (Selecting D): Mistaking the voltmeter reading for the supply EMF or doubling it blindly.

Topics

Physics · 3.5 Electricity

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.