AQA AS Level Physics Paper 2, June 2022: Question 2
10 marks · Medium difficulty · Practical Techniques & Data Analysis
Determine the resistance per unit length and resistivity of a copper wire using a potentiometer bridge circuit, and calculate wire dimensions and circuit modifications.
Practise this questionQuestion
Question text
02 Figure 3 shows a circuit used to find the resistance per unit length of a copper wire.
Figure 3
The copper wire is fixed with tape to a metre ruler that has 2 mm graduations.
Contact P is placed on the wire close to one end of the ruler and held firmly in place
using a bulldog clip.
When contact Q is placed on the wire as shown in9 Figure 3 the voltmeter shows
a non-zero reading.
Q is moved along the wire until the voltmeter reading is zero.
Figure 4 shows enlarged views of the position of P and the new position of Q.
Figure 4
02.1 Determine, in m, the length x of copper wire between P and Q.
[1 mark]
10 x = m
02.2 When the voltmeter reading is zero:
R1 R3
=
R2 R4
where R4 is the resistance of the copper wire between P and Q.
Determine, in Ω m−1, the resistance per unit length of the copper wire.
R1 = 2.2 MΩ
R2 = 3.9 kΩ
R3 = 75 Ω
[2 marks]
resistance per unit length = Ω m−1
02.3 The diameter d of the copper wire is approximately 0.4 mm.
Suggest:
• a suitable measuring instrument to accurately determine d
• how to reduce the effect of random error on the result for d.
[3 marks]
02.4 Determine the resistivity ρ of copper.
*10* diameter d of the copper wire = 0.38 mm
[2 marks]
ρ = Ω m
The copper wire is replaced with a constantan wire of diameter 0.38 mm.
resistivity of constantan
= 30
resistivity of copper
02.5 Suggest one change to the circuit to make the voltmeter read zero for the same value
of x as in Question 02.1.
[1 mark]
02.6 Calculate, in mm, the diameter of a constantan wire that has the same resistance
per unit length as the copper wire.
[1 mark]
diameter = mm
END OF SECTION A
Section B
Answer all questions in this section.
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidance Mark AO
02.1 0.879 (m) 1 AO3
02.2 Correct answer gives 0.15(1) (Ω m−1). 2 2 ×
AO3
correctly determines R4 OR divides their incorrect R4 by their 1 Correct R4 = 0.13(3)
02.1 1
12 Allow a correction to m if their 02.1 is in mm
correct R4
evaluates 2 2 Condone 3 sf answer
their 02.1
– SICS – – JUNE 2022
02.3 Treat references to zero error as neutral unless 3 3 ×
micrometer screw gauge explicitly linked to reducing random error. AO1
OR For allow ‘micrometer’ or ‘screw gauge’ or
digital (vernier) callipers 1 travelling microscope.
Reject ‘(vernier) callipers’.
repeat measurements at different points (along the wire)
Accept “readings” for “measurements”.
OR
Repeat “experiment” is insufficient. 11
repeat measurements in different directions / orientations
OR
repeat measurements AND reject / discard anomalies 2
For 3 some mention of repeat (measurements)
calculate average / mean (from repeated measurements) 3
owtte must be seen in body of answer
– SICS – – JUNE 2022
d 2
02.4 use of A = × 0.382 2 2 ×
1 For allow POT in d: either A = OR
4 AO3
A = π × 0.192 OR A = 1.1(3) (× 10−7) seen
For expected answer is 1.7 × 10−8 (Ω m)
12 ρ = their 02.2 × 1.1(3) × 10−7 (Ω m) If no other mark given, allow 1 mark for 6.8 × 10−8
(Ω m)
02.5 decrease R1 / 2.2 MΩ by a factor of 30 Unless quantitative change identified, must give 1 AO3
new resistance, e.g.
OR
(new) R1 is 73 kΩ
increase R2 / 3.9 kΩ by a factor of 30
(new) R2 is 120 kΩ
OR
(new) R3 is 2.3 kΩ
increase R3 / 75 Ω by a factor of 30
02.6 2.1 (mm) allow > 2 sf answer rounding to 2.1 (mm) 1 AO2
Total 10
How to answer it
Determining Resistivity Using a Bridge Circuit
What this question tests
This question assesses your understanding of potential divider circuits (bridge arrangement), the factors affecting electrical resistance, correct use of high-precision measuring instruments, experimental techniques for reducing random errors, and calculations involving electrical resistivity (p = RA / l).
Length of Copper Wire ($x$)
✅ Correct Answer
0.879 m (Accept 0.878 m to 0.880 m)
💡 Key Knowledge
- Read the scale carefully, noting that the ruler has 2 mm graduations.
- Contact P is at approx 7.9 cm and Q is at approx 95.8 cm. Distance $x$ = Q - P.
Resistance per Unit Length
📐 Step-by-Step Calculation
- Find $R_4$ using the balance formula: When voltmeter reads zero, R₁ / R₂ = R₃ / R₄ . Rearranging gives R₄ = (R₂ × R₃) / R₁ .
- Substitute values: R₄ = (3900 × 75) / (2.2 × 10⁶) = 0.1329 Ω .
- Calculate resistance per unit length: Resistance per unit length = R₄ / x = 0.1329 / 0.879 = 0.151 Ω m⁻¹ .
❌ Common Errors
- Forgetting to convert megohms (MΩ) and kilohms (kΩ) into standard ohms before calculating.
- Inverting the potential divider ratio formula.
Measuring Diameter and Reducing Random Error
🧠 Exam Technique & Answers
- Instrument (1 mark): Use a micrometer screw gauge (or digital/vernier callipers - reject analogue vernier callipers).
- Random Error Reduction (1 mark): Repeat measurements at different points along the wire and/or in different orientations/directions.
- Processing (1 mark): Calculate the average (mean) of the repeated measurements after discarding any anomalies.
❌ Common Errors
- Writing "repeat the experiment" instead of repeating the individual diameter measurements.
- Failing to mention calculating a mean/average value.
Calculating Resistivity ( ρ )
📐 Step-by-Step Calculation
- Calculate cross-sectional area ($A$):
$A = \frac{\pi d^2}{4} = \frac{\pi \times (0.38 \times 10^{-3})^2}{4} = 1.13 \times 10^{-7}\text{ m}^2$. - Rearrange resistivity equation ( R = ρl / A ):
$\rho = \frac{R_4 \times A}{x} = (\text{Resistance per unit length}) \times A$. - Final evaluation:
$\rho = 0.151 \times 1.13 \times 10^{-7} = \mathbf{1.7 \times 10^{-8}\text{ Ω m}}$.
❌ Common Errors
- Squaring the radius directly without dividing diameter by 2, leading to a power of 4 error in area.
- Omitting unit conversions from millimetres to metres for the diameter.
Circuit Modifications
✅ Correct Answers (Any ONE)
- Decrease R₁ by a factor of 30 (new value: 73 kΩ).
- Increase R₂ by a factor of 30 (new value: 120 kΩ).
- Increase R₃ by a factor of 30 (new value: 2.3 kΩ).
💡 Key Knowledge
Since the constantan wire is 30 times more resistive than copper, its resistance for the same length $x$ will be 30 times greater. To keep the balance point ($x$) identical, the ratio of the fixed resistors in the bridge must change by a factor of 30.
Constantan Wire Diameter Calculation
✅ Correct Answer
2.1 mm
💡 Key Knowledge
- Resistance per unit length is proportional to 1 / A (or 1 / d² ).
- Since constantan's resistivity is 30 times greater, its cross-sectional area must also be 30 times greater to maintain the exact same resistance per unit length.
- $d_{\text{constantan}} = d_{\text{copper}} \times \sqrt{30} = 0.38 \text{ mm} \times \sqrt{30} \approx 2.08 \text{ mm} \rightarrow \mathbf{2.1 \text{ mm}}$.
Topics
Physics · Practical skills · Required Practicals · 3.5 Electricity · Experimental design · Data analysis · Uncertainty and evaluation · AS practicals (1–6)
Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.