AQA A-Level Physics Paper 3 (3A), June 2025: Question 2

12 marks · Medium difficulty · Practical Techniques & Data Analysis

Analyze repeat timing data to find percentage uncertainty, determine capacitance and resistance from a graph of period against resistance, and design a resistor combination for a specified timing range.

Practise this question

Question

Question text

02 Figure 3 shows a circuit that causes a loudspeaker to emit ‘clicks’ at a constant rate.

The time T between successive clicks depends on the resistances R1 and R2 and the

capacitance C.

You do not need to know any details about the electronic system.

Figure 3

A student uses a stopwatch to determine T.

Table 1 shows the student’s repeated measurements of 20T.

Table 1

20T / s 20T / s 20T / s 20T / s 20T / s

19.22 18.91 19.367 19.04 19.45

02.1 Suggest why the student measures 20T.

[1 mark]

02.2 Suggest why the student repeats their measurements of 20T.

[1 mark]

02.3 Determine the percentage uncertainty in the student’s result for T.

*06*Assume that all the data in Table 1 are valid.

[2 marks]

percentage uncertainty in T = %

The student determines T for different values of R2.

The values of R1 and C are not changed.

The student produces the graph in Figure 4.

Figure 4

02.4 It can be shown that

T = (R1 + 2R2)C ln2

Determine C and R1.

[5 marks]

C = F

10 R1 = Ω

02.5 The student has access to a range of fixed and variable resistors.

The student wants to modify the circuit in Figure 3 so that T can be varied

continuously from a minimum of 0.70 s to a maximum of 1.25 s.

Show how the student can achieve this by combining two resistors to make R2.

You should use information from Figure 4.

*09* [3 marks]

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidance Mark AO

02.1 (measuring 20T) allow (idea of) ‘reducing the effect of 1 AO1

uncertainty due to ‘stop-start error’ / ‘human

(idea of) reduces the effect of random error (on T) reactions’ / ‘human error’;

OR accept ‘to improve precision’ / ‘to reduce

reduces the percentage uncertainty relative error’ / reduces the percentage error;

condone ‘reduces the uncertainty in T ‘

‘to calculate an average’ is neutral;

to reduce ‘uncertainty’ / ‘measurement error’ /

‘human error’ / ‘human reactions’ / ‘stop-start

error’ are neutral

‘easier to measure’ / ‘improve accuracy’ /

comments about systematic error are neutral

02.2 (repeating the (20T) measurements) allow ‘to avoid miscounting cycles’; 1 AO1

(idea of) a check for / enables the detection of / elimination accept ‘to improve accuracy’ / ‘to reduce

of anomalies (in 20T) statistical error’ / ‘account for systematic

error’

‘to reduce the effect of anomalies’ / ‘to

calculate an average’ are neutral;

‘to reduce (effect of) uncertainty’ /

‘measurement error’ / ‘human error’ are

neutral

‘improve precision’ / comments about random

error are neutral

02.3 average (20)T for 1 2 AO2

OR average 20T = 19.2 (s)

uncertainty in (20)T 1 (use evidence of calculator value 19.196 to

confirm that all 5 sets were used)

19.45 −18.91

uncertainty in 20T = 0.27 OR ;

allow 19.45 − 19.2(0) OR 19.2(0) − 18.91

average T = 0.96 (s) (calculator value 0.9598)

uncertainty in T = 0.0135 / 0.014 (s)

0.973 − 0.946

OR ;

allow 0.973 − 0.96 OR 0.96 − 0.946

percentage uncertainty based on a correct method for 2 accept any > 2 sf that rounds to 1.4%;

= 1.4 % 2 condone max 4 sf;

do not allow 0.014

1 is for the process, not the result

02.4 gradient evaluated from ∆T divided by ∆R2 with ∆R2 ≥ 6 kΩ; 5 AO2

10 allow one read off error;

allow POT error in substituted data

evidence of valid simultaneous equations with full

expect gradient ≈ 5.5 × 10−5 (s Ω−1);

substitution using read-offs for T and R2 with ∆R2 ≥ 6 kΩ 1

award 2 3 for C in range with correct POT;

attempts to determine C by a valid method 2 2 is for the process, not the result: see page

for 3 allow ECF for use of their R1 ;

C in range 3.8 × 10−5 to 4.1 × 10−5 (F) no ECF for POT OR a calculation error in

gradient

award 4 5 for R1 in range with correct POT;

R 4 is for the process, not the result etc;

attempts to determine 1 by a valid method 4

do not allow working off grid to determine R1

for 5 allow ECF for use of their C or their

gradient

R in range 2.2 to 2.5 × 104 (Ω)

min 2 sf for 3 and 5 ;

allow ≥ 3 sf rounding to 2 sf in range;

don’t penalise twice for results based on

simultaneous equations with R2 values in kΩ

Question Additional comments/Guidelines for 2 Additional comments/Guidelines for 4

02.4 their gradient 2 × their vertical intercept

C = R1 =

2×ln 2 their gradient

OR OR

their gradient their vertical intercept

C = R1 =

ln 4 ln 2 × their C

OR OR

their ∆T their T

C = R1 = − 2× their R2

2 × ln 2 × their ∆R2 ln 2× their C

(where their T and their R2 are corresponding values read-off

Figure 4)

correct manipulation of valid simultaneous equations

their T OR

C =

ln 2×(their R1 + 2(their R2))

TPR2Q − TQR2P

(where their R1 has been found by a method not involving C; R1 =

their T and their R2 are corresponding values read-off Figure TQ − TP

(based on read-offs from Figure 4 where their R2Q is their

value of R2 at time TQ etc)

02.5 marks can be awarded for suitable sketches allow A B for a switched 2 fixed resistor 3 1 × AO1

12 circuit that can produce both limiting values,

up to 12 for identifying that their R2 should 2 × AO3

• have minimum resistance 1 kΩ A

• have maximum resistance 11 kΩ B

• be a (fixed) resistor (or resistors) in series with a

variable resistor C 1 kΩ 10 kΩ

1 kΩ

one 1 kΩ resistor in series with one 10 kΩ variable

resistor

or valid labelled sketch 3

eg 11 kΩ

allow A B (R2) from 1 kΩ to 11 kΩ;

allow AB for deducing range = 10 kΩ;

1 kΩ 10 kΩ for C allow any arrangement that allows R2

to be varied continuously between limiting

values;

do not allow fixed resistor in parallel with a

variable resistor

for 3 CAO

Total 12

How to answer it

RC Timing Circuit, Uncertainties & Graphical Analysis

📌 What this question tests

This practical/skills-based question assesses core experimental physics competencies from AQA Paper 3 / Section A:

Experimental techniques in timing: Understanding why multiple periods are measured and why repeated readings are taken.
Uncertainty analysis: Calculating absolute uncertainty (half-range) and percentage uncertainty from repeated timing data.
Graphical linearisation: Linking a non-standard algebraic expression to the form y = mx + c , calculating gradients, interpreting axes with metric prefixes (kΩ), and extracting physical constants.
Circuit design & modification: Reading target values from a graph to specify a resistor network combining fixed and variable resistors.

Part 02.1 — Measuring Multiple Periods (20T)

Suggest why the student measures 20T [1 Mark]

✅ Correct Answers

Reduces the percentage uncertainty (or percentage error) in the measurement of T.
Reduces the effect of random errors (such as human reaction time or stopwatch start/stop delay).

❌ Common Errors & Examiner Traps

Writing simply "reduces uncertainty" or "reduces error" without stating percentage or effect of. Measuring 20T does not change absolute reaction time uncertainty!
Claiming it "improves accuracy" or "reduces systematic error" — reaction time is a random error, not a systematic one.
Writing "to calculate an average" (that explains repeats, not measuring 20 periods at once).

Mark Scheme Note: Award 1 mark for mentioning the reduction of percentage uncertainty OR reducing the effect of random error / human reaction time. "To improve precision" is condoned.

Part 02.2 — Repeating Measurements

Suggest why the student repeats their measurements of 20T [1 Mark]

✅ Correct Answers

To identify / detect / eliminate anomalies (outliers).
To check for miscounting of cycles/clicks.

🧠 Exam Technique: Know the Difference

Measuring 20T: Spreads human reaction time across 20 cycles → reduces percentage uncertainty.
Taking repeated sets: Enables detection and removal of anomalies before averaging. Never mix up the reasoning for these two distinct experimental steps!

Mark Scheme Note: Award 1 mark for detecting/eliminating anomalies or avoiding miscounting cycles. "To calculate an average" or "reduces human error" is neutral and scores 0.

Part 02.3 — Percentage Uncertainty Calculation

Determine the percentage uncertainty in the student's result for T [2 Marks]

📐 Step-by-Step Calculation

Data for 20T / s: 19.22, 18.91, 19.36, 19.04, 19.45

Calculate Mean Value of 20T:
Mean = (19.22 + 18.91 + 19.36 + 19.04 + 19.45) / 5 = 96.98 / 5 = 19.196 s ≈ 19.2 s
Calculate Absolute Uncertainty (half the range):
Uncertainty = (Max - Min) / 2 = (19.45 - 18.91) / 2 = 0.54 / 2 = 0.27 s
[1 mark awarded for either mean 20T or absolute uncertainty]
Calculate Percentage Uncertainty:
% uncertainty = (Uncertainty / Mean) × 100 = (0.27 / 19.196) × 100 = 1.4065% ≈ 1.4%
[1 mark awarded for final percentage based on valid method]

✅ Acceptable Values

Any answer rounding to 1.4% (e.g. 1.4% or 1.41%). Maximum 4 significant figures permitted.

❌ Common Traps

Dividing the uncertainty by 20 without also dividing the mean by 20. Note that:
% uncertainty in T = % uncertainty in 20T
Giving the answer as a fraction or decimal without multiplying by 100 (e.g. giving 0.014 scores 0 for the second mark).

Part 02.4 — Graphical Analysis to Determine C and R₁

Given T = (R₁ + 2R₂)C ln 2, determine C and R₁ [5 Marks]

💡 Mathematical Link to y = mx + c

Expand the given formula:

T = (2C ln 2) · R₂ + (R₁ C ln 2)

Comparing this to y = m x + c where the vertical axis is y = T and the horizontal axis is x = R₂ :

Gradient ( m ): m = 2 C ln 2
y-intercept ( c ): c = R₁ C ln 2

📐 Step-by-Step Solution

Step 1: Calculate the Gradient (m) from Figure 4
Take two well-separated points on the line of best fit (rule: ΔR₂ ≥ 6 kΩ):
Point 1 (intercept): (R₂ = 0 Ω, T = 0.64 s)
Point 2: (R₂ = 12.0 kΩ = 12 000 Ω, T = 1.30 s)
ΔT = 1.30 - 0.64 = 0.66 s
ΔR₂ = 12 000 - 0 = 12 000 Ω
Gradient = ΔT / ΔR₂ = 0.66 / 12 000 ≈ 5.50 × 10⁻⁵ s Ω⁻¹
[Mark 1: Valid gradient method using ΔR₂ ≥ 6 kΩ]
Step 2: Determine Capacitance C
Rearrange gradient = 2 C ln 2 :
C = gradient / (2 × ln 2) = (5.50 × 10⁻⁵) / (2 × 0.69315) = (5.50 × 10⁻⁵) / 1.3863 ≈ 3.97 × 10⁻⁵ F
[Mark 2: Valid formula linking C to gradient]
[Mark 3: C in acceptable range: 3.8 × 10⁻⁵ to 4.1 × 10⁻⁵ F]
Step 3: Determine Resistance R₁
Method A (using intercept): Read intercept c = 0.64 s .
Since c = R₁ C ln 2 , and gradient = 2 C ln 2 :
R₁ = 2 × (y-intercept) / gradient = 2 × 0.64 / (5.50 × 10⁻⁵) ≈ 23 270 Ω = 2.3 × 10⁴ Ω
Method B (using C and intercept directly):
R₁ = c / (C ln 2) = 0.64 / (3.97 × 10⁻⁵ × 0.69315) ≈ 23 260 Ω = 2.3 × 10⁴ Ω
[Mark 4: Valid method to determine R₁]
[Mark 5: R₁ in acceptable range: 2.2 × 10⁴ to 2.5 × 10⁴ Ω (or 22 kΩ to 25 kΩ)]

❌ Critical Unit Traps (Lost Marks)

Ignoring the kilo prefix on the x-axis: The axis reads R₂ / kΩ ! If you use 12 instead of 12 000, your gradient will be out by a factor of 10³, causing power-of-ten errors for both C and R₁.
Forgetting the factor of 2: Many students write gradient = C ln 2 instead of 2 C ln 2 , ending up with double the correct capacitance.

Part 02.5 — Resistor Network Design

Show how the student can achieve continuous variation of T from 0.70 s to 1.25 s [3 Marks]

📐 Step 1: Read Required Resistances from Figure 4

For minimum time T = 0.70 s : Reading horizontally across gives R₂ = 1.0 kΩ .
For maximum time T = 1.25 s : Reading horizontally across gives R₂ = 11.0 kΩ .
Required range of resistance: ΔR₂ = 11.0 kΩ - 1.0 kΩ = 10.0 kΩ .

✅ Circuit Specification & Diagram

To achieve a minimum of 1.0 kΩ and vary continuously up to 11.0 kΩ:

Connect a 1.0 kΩ fixed resistor in series with a 0–10 kΩ variable resistor (rheostat).

Circuit Representation:
──[ 1.0 kΩ fixed ]──[ 0–10 kΩ variable ]──

(A standard fixed resistor symbol in series with a resistor symbol having an arrow through it)

🧠 Mark Scheme Breakdown

Mark 1: Identify minimum resistance needed = 1.0 kΩ .
Mark 2: Identify maximum resistance needed = 11.0 kΩ (or total range = 10 kΩ).
Mark 3: Correct combination: a 1.0 kΩ fixed resistor in series with a 10 kΩ variable resistor (fully labelled sketch or clear written description).

❌ What Examiners Rejected

Connecting a fixed resistor in parallel with a variable resistor: parallel combinations change non-linearly and cannot produce a minimum of 1.0 kΩ and maximum of 11.0 kΩ with standard values.
Failing to mention that the resistors must be in series.
Using switched discrete fixed resistors without a variable resistor: the question specifically demands that T can be varied continuously.

Topics

Practical skills · Physics · 3.1 Measurements and their errors · 3.5 Electricity · 3.7 Fields and their consequences (A-level only) · Data analysis · Experimental design · Uncertainty and evaluation

Question and mark scheme from the AQA A-Level Physics examination, Paper 3 (3A), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.