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

13 marks · Hard difficulty · Practical Techniques & Data Analysis

Determine the temperature coefficient of resistance from a graph, explain how electrical power and temperature are measured in a filament lamp circuit, and determine the power-law exponent for radiation using a logarithmic plot.

Practise this question

Question

Question 3 presents an investigation into a lamp filament's resistance and power dissipation as a function of temperature. Figure 5 illustrates a filament lamp inside an oven with an ohm-meter and a thermometer. Figure 6 is a graph of filament resistance R (from 3.0 to 3.6 ohms) against temperature T (from 310 to 350 K), showing a linear increase. The formula (R - R0)/R0 = alpha(T - T0) is given. Part 03.1 asks to determine alpha and its unit. Part 03.2 shows Figure 7, a circuit diagram with a battery, potentiometer, ammeter, voltmeter, and lamp, asking how to determine electrical power P and filament temperature T. Part 03.3 shows Figure 8, plotting power against temperature from 800 to 1800 K with curves for total power P and non-radiation loss P2, and asks students to complete Table 2. Part 03.4 provides Table 3 and Figure 9 (a blank grid) to determine the integer power index n in P1 proportional to T^n by plotting a logarithmic graph.
Question text

03 Figure 5 shows apparatus to investigate how the resistance R of a lamp filament

varies with the absolute temperature T of the filament.

Figure 5

The lamp is inside an oven.

The oven temperature is increased to a steady value.

The filament is at the same temperature as the inside of the oven.

The temperature T in kelvin is determined using the thermometer.

An ohm-meter is connected to the filament to determine R at temperature T.

The oven temperature is increased to new steady values and further readings of12 T

and R are recorded.

Figure 6 shows the variation of R with T.

Figure 6

It can be shown that:

R − R0

= α T( −T0 )

R0

where:

R0 is the resistance of the filament at temperature T0

α is a constant.

03.1 Determine α using the graph in Figure 6.

State an appropriate SI unit for your answer.

R0 = 2.83 Ω when T0 = 291 K

[3 marks]

α = unit =

In a different investigation, a student investigates how the electrical power P

transferred in the same filament lamp varies for values of T greater than 800 K.

The student assumes that α is constant for all values of T.

Figure 7 shows the circuit that the student uses.

The temperature of the filament is controlled by the energy supplied to it from

the battery.

Figure 7

03.2 Explain how the ammeter and voltmeter readings are used to determine values

of P and T.

[2 marks]

P

T

For a steady value of T, P = P1 + P2

where:

P1 is the rate of heat transfer from the filament by radiation

P2 is the rate of heat transfer from the filament by all other processes.

In Figure 8:

• the solid line shows how P varies with T

• the dashed line shows how P2 varies with T.

Figure 8

03.3 Complete Table 2.

[3 marks]

Table 2

T / K P / W P1 / W P2 / W

960 3.7 1.2 2.5

1100 2.0 3.0

1370

1680 16.2 11.0

03.4 It can be shown that P ∝ T n where n is an integer.

Determine n.

In your answer you should:

• use Table 3 to organise the data you will plot

• plot a suitable graph on Figure 9.

[5 marks]

Table 3

T / K P1 / W

960 1.2

1100 2.0

1370

1680 11.0

19 n =

Figure 9

Mark scheme

Show the mark scheme Mark scheme for Question 3 detailing criteria across four parts for a total of 13 marks. 03.1 awards up to 3 marks for calculating alpha in the range 4.45 to 4.57 x 10^-3 and the unit K^-1. 03.2 awards 2 marks for determining P using V x I and R using V/I, then finding T using the formula. 03.3 awards 3 marks for correctly reading and calculating values to complete Table 2 to 1 decimal place. 03.4 awards 5 marks for setting up a log-log table (log P1 vs log T or ln P1 vs ln T), suitable axes and scales on Figure 9, correctly plotting points with a line of best fit, finding a gradient in the range 3.6 to 4.4, and stating the integer value n = 4.

Question Answers Additional comments/Guidance Mark AO

03.1 for 1 and 2 allow ≥ 4 sf rounding to 3 sf in 3 2 × AO2

range

1 × AO1

α in range with correct POT 1 2 for 1 α in range 4.45 to 4.57

for α in range 4.45 to 4.57 × 10−3

α in range with incorrect OR missing POT 1 1 2

allow for α in range 4.39 to 4.63 × 10−3

OR

allow for 2 sf α = 4.5 × 10−3

−1 1 1

K−1 for 3 condone °K OR OR

3 K o K

do not accept (°)C−1

do not accept k for K

03.2 allow valid work in ‘wrong’ answer space 2 AO1

determines P using V × I in some form for 1 AND / OR 2 accept formulae in

14 symbols or in words eg ‘power = current ×

OR voltage’;

V accept ‘voltmeter reading’ for V / ‘ammeter

determines R using in some form 1

I reading’ for I;

treat as neutral the idea of plotting an IV

graph to find P and / or to find R

V

determines P using V × I AND R using in some form for 2 allow ‘use the equation with R’;

I

any rearranged equation must be correct

AND

R − R0

determines T using (their) R 1 2 egT = + T0 ;

αR0

treat the following as neutral:

(unqualified) ‘use the equation’ / ‘use T ∝ R

to determine T’/ ‘use (extrapolate) Figure 6’ /

‘use Figure 8’

03.3 2nd row of Table 2 P correct only penalise dp in 3 3 2 × AO2

AND for 1 (2nd row) P = 5.0 (W)

4th row of Table 2 P2 correct 1 AND 1 × AO3

(4th row) P2 = 5.2 (W) 15

3rd row of Table 2:

P correct ± 0.2 for 2 (3rd row) P = 8.6 to 9.0 (W)

AND AND

P2 correct ± 0.2 2 P2 = 3.8 to 4.2 (W)

their P1 = their P − their P2 for 3 expect P1 = 4.4 to 5.2 (W)

AND

all the data added to Table 2 is to 1 dp 3

T / K P / W P1 / W P2 / W

960 3.7 1.2 2.5

1100 5.0 2.0 3.0

1370 8.8 4.8 4.0

1680 16.2 11.0 5.2

Question Answers Additional comments/Guidelines Mark AO

03.4 intention to use a logarithmic plot to test power law: award 1 for use of bracket, solidus AND unit MAX 5 2 × AO1

eg log (T / K) AND log (P1 / W);

suitable headings in Table 3 can be shown EITHER in the Table 3 2 × AO2

headings OR for the Figure 9 axis labels;

OR

accept lg (T / K) / log (T / k) for log (T / K);

suitable graph axis labels in Figure 9 1 1 × AO3

accept columns interchanged / axes reversed

their P1 data in Table 2 transferred to 3rd row / their P1 for 2 check that the log T AND log P1

used to complete 3rd row values are correct to at least 2 dp;

AND all log T values to the same (minimum 2) dp;

4 sets of log T and log P1 recorded in columns 3 and 4 of all log P1 values to same (minimum 2) dp OR

Table 3 2 all to same (minimum 3) sf

ECF wrong P1;

T / K P1 / W log (T / K) log (P1 / W) allow 4 sets of ln (T / K) and ln (P1 / W)

960 1.2 2.98 0.08

T / K P1 / W ln (T / K) ln (P1 / W)

1100 2.0 3.04 0.30

1370 4.8 3.14 0.68 960 1.2 6.87 0.18

1680 11.0 3.23 1.04 1100 2.0 7.00 0.69

1370 4.8 7.22 1.57

1680 11.0 7.43 2.40

03.4 Figure 9 and processing: for 3 each scale must

(cont)

suitable scales for their data; • accommodate all their data

each axis labelled with (at least) their plotted quantity; • have at least 4 evenly-spaced values

marked along each axis

allow interchanged axes 3 • be linear, avoiding ‘difficult’ intervals

• allow points to cover ≥ half the grid in both

directions (expect use of a false origin)

condone a suitably scaled and labelled non-

logarithmic plot including P1 against T

for the 1st, 2nd and 4th points should lie on

4 points plotted and a ruled line that is continuous (at 4

least) from 1st point to 4th point a straight line (by eye);

check possible errant point(s) but allow minor

plotting errors (≤ 2 minor squares);

an unrounded gradient result in the range 3.6 to 4.4

do not allow blobs, thick or faint points;

OR

allow a ruled line that is not the best line but

a gradient result correctly rounded to the nearest integer 5 do not allow a thick / faint / discontinuous line

n = 4 CAO for 5 do not penalise small steps;

condone one read off error

log-log ln-ln for reversed axes unrounded gradient range

if the plot is not or award 12 3 456 = 1 MAX

is 0.23 to 0.28 or inverse of gradient rounded

if plot is T4 against P allow up to = 3 MAX

11 2 3 4 5 6

6 is contingent on 5

03.4 Figure 9 with log (P1 / W) against log (T / K) Figure 9 with ln (P1 / W) against ln (T / K)

(cont)

y = 3.9676x - 11.76

3.0

1.2

y = 3.9676x - 27.078

2.0

0.8

log (P / W) ln (P1 / W)

1.0

0.4

0.0

0.0

6.8 7.0 7.2 7.4

2.9 3.0 3.1 3.2

ln (T / K)

log (T / K)

How to answer it

Investigating Filament Resistance & Thermal Radiation Laws

📋 What this question tests

This multi-part experimental analysis assesses core practical physics and graphical skills:

  • Linear modelling: Linking a linear experimental equation to y = mx + c and finding the temperature coefficient of resistance (α) including its derived SI unit.
  • Circuit analysis & indirect measurement: Explaining how current and potential difference are processed to yield power and temperature via resistance.
  • Data extraction & arithmetic consistency: Reading curves accurately and applying additive relations ( P = P₁ + P₂ ) with strict significant-figure/decimal-place consistency.
  • Logarithmic transformation of power laws: Linearising an exponential/power law relationship ( P₁ ∝ Tⁿ ), setting up a log table, selecting false-origin scales, determining the gradient, and concluding the integer index ( n = 4 , representing the Stefan-Boltzmann law).

Part 03.1: Determining the Temperature Coefficient (α) & Unit

Analysis of Linear Resistance vs Temperature Data (3 Marks)

💡 Key Knowledge: Linearising the Equation

The given formula relates resistance to temperature:

(R - R₀) / R₀ = α(T - T₀)

Rearranging in the form y = mx + c :

R - R₀ = α R₀(T - T₀) ⟹ R = (α R₀)T + [R₀(1 - α T₀)]

Hence, the gradient of the graph of R against T is:

gradient = α × R₀ ⟹ α = gradient / R₀

📐 Step-by-Step Calculation

  1. Find the gradient: Pick two distant points on Figure 6:
    Point 1: (310 K, 3.08 Ω)
    Point 2: (350 K, 3.58 Ω)
    gradient = ΔR / ΔT = (3.58 - 3.08) / (350 - 310) = 0.50 / 40 = 0.0125 Ω K⁻¹
  2. Calculate α:
    Given R₀ = 2.83 Ω at T₀ = 291 K :
    α = 0.0125 / 2.83 = 4.42 × 10⁻³ K⁻¹
    (Mark scheme acceptable range: 4.45 × 10⁻³ to 4.57 × 10⁻³, allowing 4.39 to 4.63 × 10⁻³)
  3. Determine the SI Unit:
    (R - R₀)/R₀ is dimensionless. Therefore, α(T - T₀) must also be dimensionless:
    Unit of α = 1 / (Unit of T) = K⁻¹

✅ Mark Scheme Breakdown

  • Mark 1: Correct gradient calculation leading to an α value in range (allowing power of ten error).
  • Mark 2: α in range 4.45 × 10⁻³ to 4.57 × 10⁻³ with correct power of 10 (or 4.5 × 10⁻³ to 2 s.f.).
  • Mark 3: Unit stated as K⁻¹ (or 1/K ).

❌ Common Errors & Examiner Traps

  • Equating gradient to α directly: Students frequently forget to divide the gradient by R₀ .
  • Power of ten errors: Forgetting the × 10⁻³ factor entirely (e.g. writing 4.5 instead of 0.0045).
  • Incorrect units: Writing lowercase k⁻¹ (kilo) instead of capital K⁻¹ (Kelvin), or using (°C)⁻¹ .

Part 03.2: Determining Power (P) and Temperature (T) from Circuit Readings

Relating Electrical Measurements to Physical State (2 Marks)

💡 Theoretical Link

  • Power (P): Electrical energy supplied per second is calculated directly from current and potential difference: P = V × I .
  • Temperature (T): Cannot be measured directly with a thermometer above 800 K. Instead, resistance is found first ( R = V / I ), and then temperature is extracted using the calibrated formula from 03.1:
    T = (R - R₀)/(α R₀) + T₀

🧠 Exam Technique: Securing Both Marks

  • Mark 1 (P): State explicitly that electrical power is calculated using P = V × I (or in words: "power = current × voltage").
  • Mark 2 (T): You must state both steps:
    1. Determine filament resistance using R = V / I .
    2. Substitute this resistance into the equation from 03.1 to calculate T .

❌ Examiner Commentary & Lost Marks

Many candidates simply stated "read temperature from Figure 6" or "use a thermometer". The question context explicitly specifies that the temperature is now above 800 K (well beyond Figure 6's 350 K range and standard lab thermometers). Candidates who wrote vaguely "use the equation to find T" without mentioning calculating R = V/I first failed to gain the second mark.

Part 03.3: Completing Table 2 (Heat Transfer Components)

Data Processing and Reading Non-Linear Curves (3 Marks)

💡 The Conservation Equation

At steady state, the total electrical power supplied equals the sum of energy dissipation mechanisms: P = P₁ + P₂ , where P₁ is radiation and P₂ is other processes (conduction/convection).

T / K P / W P₁ / W P₂ / W Working & Graph Reading
960 3.7 1.2 2.5 Given
1100 5.0 2.0 3.0 P = P₁ + P₂ = 2.0 + 3.0 = 5.0 W
1370 8.8 4.8 4.0 Read Fig 8 at 1370 K: Solid curve P = 8.8 W (allow 8.6–9.0); Dashed curve P₂ = 4.0 W (allow 3.8–4.2); P₁ = 8.8 - 4.0 = 4.8 W
1680 16.2 11.0 5.2 P₂ = P - P₁ = 16.2 - 11.0 = 5.2 W

✅ Mark Allocation

  • Mark 1: 2nd row P = 5.0 AND 4th row P₂ = 5.2 .
  • Mark 2: 3rd row readings: P = 8.8 ± 0.2 AND P₂ = 4.0 ± 0.2 .
  • Mark 3: Calculated P₁ = P - P₂ for 3rd row (yielding 4.8) AND all table entries given to consistently 1 decimal place.

❌ Significant Figure Trap

Writing 5 instead of 5.0 , or 4 instead of 4.0 immediately forfeits the 3rd mark! All original raw data in the table are quoted to 1 d.p., so processed values must match.

Part 03.4: Determining the Integer Index (n) via Log Graph

Logarithmic Plotting & Graph Evaluation (5 Marks)

💡 Mathematical Foundation

Given the relationship P₁ = k Tⁿ :

log(P₁) = log(k Tⁿ) = n log(T) + log(k)     [or ln(P₁) = n ln(T) + ln(k)]

This matches y = m x + c , where:

  • y-axis: log(P₁ / W) or ln(P₁ / W)
  • x-axis: log(T / K) or ln(T / K)
  • Gradient (m): equals n

Table 3: Completed Logarithmic Data

T / K P₁ / W log(T / K) [Base 10] log(P₁ / W) [Base 10] ln(T / K) [Natural] ln(P₁ / W) [Natural]
960 1.2 2.98 0.08 6.87 0.18
1100 2.0 3.04 0.30 7.00 0.69
1370 4.8 3.14 0.68 7.22 1.57
1680 11.0 3.23 1.04 7.43 2.40

🧠 Graph Plotting Checklist (Figure 9)

  • Scale: Must use a false origin! For base 10, start the x-axis around 2.9 (covering 2.9 to 3.3). The y-axis can start at 0.0 (covering 0.0 to 1.2). The plotted points must occupy more than 50% of the grid in both directions.
  • Labelling: Axis titles with units, e.g. log(T / K) and log(P₁ / W) .
  • Points: Plot neatly with small crosses (×) or fine dots with circles (⊙). Maximum tolerance is ± 2 small squares.
  • Line of Best Fit: Draw a single, sharp, continuous ruled line passing balanced through all 4 points.

📐 Gradient & Final Integer Determination

  1. Large gradient triangle: Choose two points far apart on your line of best fit, e.g., (2.95, 0.00) and (3.25, 1.20).
  2. Calculate gradient:
    gradient = Δy / Δx = (1.20 - 0.00) / (3.25 - 2.95) = 1.20 / 0.30 ≈ 4.0
    (Examiner allowed unrounded range: 3.6 to 4.4)
  3. Final Answer: The question states n is an integer:
    n = 4 (consistent with Stefan's law: P ∝ T⁴ ).

❌ Mark Scheme Penalties to Avoid

  • Omitting Units in Table/Axes: Headings must be strictly log(T / K) or log(P₁ / W) . Forgetting "/ K" or "/ W" loses Mark 1.
  • Starting scale from (0,0): Trying to fit log T = 0 up to 3.3 will compress all 4 points into a tiny unusable vertical column at the far right edge of the paper, forfeiting the scale mark.
  • Leaving n as a decimal: Stating n = 3.97 without rounding to the nearest integer as requested forfeits the final mark. n = 4 is required.

Topics

Physics · Practical skills · 3.5 Electricity · Data analysis

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.