AQA A-Level Physics Paper 1, June 2025: Question 5

6 marks · Medium difficulty · Short Answer

Calculate the terminal potential difference and internal resistance of cell combinations, and deduce the number of cells in a battery module.

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Question text

05 This question is about the batteries used to store energy in electric vehicles.

Each battery is a series of modules. Each module is an array of identical cells.

Table 2 shows the properties of one of these cells.

Table 2

emf 3.66 V

internal resistance 30.0 mΩ

maximum current 6.77 A

05.1 Calculate the terminal potential difference of the cell at its maximum current.

[1 mark]

terminal potential difference = V

05.2 Figure 6 shows an array of 20 of these cells connected to form a module.

Figure 6

Calculate the internal resistance of the module shown in Figure 6.

[2 marks]

internal resistance = Ω

05.3 Figure 7 shows the battery for an electric vehicle.

This battery has an emf of 352 V and gives a maximum current of 500 A.

*10*The battery consists of a series of identical modules.

These modules are different from the module shown in Figure 6.

Figure 7

The individual cells used in each module have the properties shown in Table 2.

Deduce the number of cells in each module.

[3 marks]

number of cells =

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidance Mark AO

05.1 VT = 3.46 (V) E = V + Ir 1 AO1

3.66 − (6.77 × 0.0300) = 3.46

Do not accept rounding errors.

Calculator value = 3.4569

Accept 3.5

05.2 Resistance of cells in series = 5 × 0.0300 Ω = 0.150 Ω Condone method that uses 5 sets of 4 2 1 × AO1

parallel cells in series. 1 × AO2

Condone POT error in MP1

In parallel: their series resistance ÷ 4 (= 0.0375 Ω) 1

Accept 0.038

r = 0.0375 (Ω) 2

05.3 emf per module = 352 ÷ 16 = 22.0 V 1 Expect to see: 3 AO3

Number of cells in one row = 6

Number of cells in one row = their emf per module ÷ 3.66 MAX 2 if answer not a whole number

16 Number of rows in parallel = 500 6.77 (= 73.9) = 74

÷ 2 If no other mark given: award max 1 for 96

cells (divided by 16) to give 6 per module

Number of cells = 444

Total 6

How to answer it

Electric Vehicle Battery Packs: EMF, Internal Resistance & Cell Networks

AQA A-Level Physics • Electricity • Combination of Cells & Internal Resistance

What this question tests

Terminal potential difference: Applying V = ε - Ir (or ε = V + Ir ) taking unit prefixes into account (mΩ to Ω).
Network resistance: Combining internal resistances in series branches and parallel arrangements.
Modular circuit analysis: Deducing the internal matrix (series rows vs. parallel branches) of large energy storage batteries from total EMF, maximum output current, and circuit topology.

Question 05.1 • 1 Mark

Terminal Potential Difference of a Single Cell

Calculate the terminal potential difference of the cell at its maximum current.

📐 Step-by-Step Calculation

Identify given data from Table 2:
EMF, ε = 3.66 V
Internal resistance, r = 30.0 mΩ = 30.0 × 10⁻³ Ω = 0.0300 Ω
Current, I = 6.77 A
Apply EMF equation:
V = ε - Ir
Substitute values:
V = 3.66 - (6.77 × 0.0300)
V = 3.66 - 0.2031 = 3.4569 V
Round appropriately:
V = 3.46 V (or 3.5 V )

✅ Correct Answer

Terminal potential difference = 3.46 V

(Accept 3.5 V)

Mark Scheme:
[1 Mark] for 3.46 (V) .
• Do not accept rounding errors (e.g., truncating incorrectly to 3.45 V).

❌ Common Errors

Prefix neglect: Forgetting to convert 30.0 mΩ into 0.0300 Ω , resulting in a negative or nonsensical potential difference ( 3.66 - 203.1 = -199.4 V ).
Truncation error: Writing 3.45 V instead of rounding 3.4569 V up to 3.46 V .

🧠 Exam Technique

Always inspect the order of magnitude. In chemical cells, lost volts ( Ir ) should normally be a small fraction of the total EMF. A terminal p.d. moderately lower than 3.66 V confirms a sensible calculation.

Question 05.2 • 2 Marks

Internal Resistance of a Series-Parallel Module

Calculate the internal resistance of the module shown in Figure 6.

📐 Step-by-Step Calculation

Analyse the array topology:
Figure 6 shows 4 identical parallel branches.
Each branch has 5 cells in series (Total = 4 × 5 = 20 cells).
Calculate resistance of one series branch:
R_branch = 5 × r = 5 × 0.0300 Ω = 0.150 Ω
Calculate total equivalent resistance of the 4 parallel branches:
For 4 identical branches in parallel:
R_total = R_branch / 4 = 0.150 / 4 = 0.0375 Ω
Alternatively: 1 / R_total = 4 × (1 / 0.150) = 26.67 Ω⁻¹ → R_total = 0.0375 Ω

✅ Correct Answer & Marks

Internal resistance = 0.0375 Ω (or 0.038 Ω )

Mark Scheme Breakdown:
• Mark 1 (AO1): Resistance of cells in series = 5 × 0.0300 = 0.150 Ω OR dividing series resistance by 4.
• Mark 2 (AO2): Final value r = 0.0375 Ω (condone 0.038 Ω ).
Note: Examiner also condones treating it as 5 sets of 4 parallel cells in series: 5 × (0.0300 / 4) = 0.0375 Ω . Power-of-ten errors condoned in MP1 only.

💡 Key Knowledge

Identical resistors in series: R_series = n × r
Identical resistors in parallel: R_parallel = R / m (where m is the number of parallel branches).
Combining both: Total module resistance is (n / m) × r = (5 / 4) × 0.0300 = 0.0375 Ω .

❌ Common Errors

Miscounting the grid: Inverting rows and columns (e.g. calculating for 4 cells in series across 5 branches).
Treating all 20 cells in parallel: Dividing 0.0300 / 20 , completely ignoring the series connections.

Question 05.3 • 3 Marks

Deducing the Total Number of Cells in a Module

Deduce the number of cells in each module of the EV battery.

📐 Step-by-Step Calculation

Determine the number of modules in series:
Count the modules in Figure 7: 4 rows of 4 modules connected head-to-tail in series.
Total modules in series = 4 × 4 = 16 modules .
Find EMF of each module:
In a series chain, EMFs add up:
EMF per module = 352 V / 16 = 22.0 V
Calculate cells in series per branch (one row):
Since parallel branches share the same voltage, module EMF equals the EMF of a single series row:
Cells in one row = 22.0 V / 3.66 V = 6.01 → 6 cells
Calculate number of parallel rows required:
Each module carries the full battery current ( 500 A ) because modules are in series.
Each parallel cell branch can take a maximum of 6.77 A :
Number of parallel rows = 500 A / 6.77 A = 73.85 → 74 rows
(Must round up to 74 so current per branch does not exceed 6.77 A).
Calculate total cells per module:
Total cells = (cells per row) × (number of rows)
Total cells = 6 × 74 = 444 cells

✅ Correct Answer & Marks

Number of cells in each module = 444

Mark Scheme Breakdown:
• Mark 1 (AO3): EMF per module = 352 / 16 = 22.0 V
• Mark 2 (AO3): Number of cells in series row = 22.0 / 3.66 (= 6.01 → 6) OR Number of rows in parallel = 500 / 6.77 (= 73.9 → 74)
• Mark 3 (AO3): Final whole number 444 (from 6 × 74).
Penalty: Maximum 2 marks if final answer is not a whole number.
Special Case: If no other marks awarded, 1 mark max can be given for 96 cells total (using 352/3.66 = 96 cells in series divided by 16 = 6 cells).

🧠 Top-Level Exam Insights

Read the diagram carefully: The question states the battery consists of modules that are different from Figure 6. You cannot assume 4 branches or 5 cells per row!
Count modules correctly: Figure 7 shows a serpentine continuous series loop of 16 blocks (4 × 4).
Whole numbers only: Physical cells cannot exist in fractions. 6.01 must be rounded to 6 , and 73.85 branches must be rounded to 74 so current limits are respected.

❌ Common Errors

Assuming Figure 6 module architecture: Trying to use 20 cells per module or the resistance calculated in 05.2.
Rounding down branches: Choosing 73 rows means current per cell would be 500 / 73 = 6.85 A , which exceeds the rated maximum of 6.77 A.
Leaving answers as decimals: Leaving 6.01 × 73.9 = 444.1 loses the final mark.

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.