AQA A-Level Physics Paper 1, November 2020: Question 28

1 mark · Medium difficulty · Multiple Choice

Determine the current in the resistor connected in series with three identical cells of emf E and internal resistance r arranged in parallel.

Practise this question

Question

A multiple-choice question showing a circuit diagram with a resistor of resistance R connected in a loop with three identical cells, each with emf E and internal resistance r, connected in parallel. Four options A, B, C, D are given with algebraic expressions for the current in the resistor.
Question text

28 A resistor of resistance R and three identical cells of emf E and internal resistance r are

connected as shown.

What is the current in the resistor?

[1 mark]

3E

A ( )

3R + r

9E

B ( )

3R + r

E

C

R

3E

D

R

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 28 is option A.

28 A

How to answer it

Parallel Cells and Internal Resistance

Question 2.8 • [1 mark]

What this question tests

This question assesses your understanding of direct current (dc) circuits, specifically combining identical sources of electromotive force (emf) in parallel, treating internal resistance as part of a closed loop, and applying Kirchhoff's second law / terminal potential difference equations to find total current.

Question Part 2.8

Correct Option: A

✅ Correct Answer

Option A: 3E / (3R + r)

The total emf of identical cells connected in parallel is equal to the emf of a single cell ( E ), but their internal resistances combine in parallel to give r / 3 . Applying Ohm's law to the complete loop yields the correct expression.

💡 Key Knowledge

  • Emf in Parallel: Connecting identical cells in parallel does not increase the total emf ( E_total = E ). It only increases current capacity and reduces internal resistance.
  • Internal Resistance in Parallel: Three resistors of value r in parallel combine to give a total internal resistance of r / 3 .
  • Ohm's Law for Circuits: I = Total Emf / Total Resistance

🧠 Exam Technique

Don't fall into the trap of multiplying the emf by the number of cells just because they are stacked vertically. Always check whether components are in series or parallel configurations before summing their electrical properties.

❌ Common Errors

  • Multiplying Emf: Choosing Option B ( 9E / (...) ) by incorrectly multiplying the emf by 3, treating them like series cells.
  • Ignoring Internal Resistance: Choosing Option D ( 3E / R ) by forgetting internal resistance entirely.
  • Inverting Internal Resistance: Incorrectly treating the internal resistance as 3r instead of r / 3 .

📐 Step-by-Step Derivation

  1. Identify Total Emf: For n identical cells in parallel, Total Emf = E .
  2. Calculate Total Internal Resistance: Three internal resistances ( r ) in parallel give R_internal = r / 3 .
  3. Find Total Circuit Resistance: Add the load resistor R and total internal resistance: R_total = R + (r / 3) .
  4. Apply Circuit Equation:
    Current I = E / (R + r / 3)
    Multiply numerator and denominator by 3 to match the options:
    I = 3E / (3R + r)

Topics

Physics · 3.5 Electricity

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.