AQA AS Level Physics Paper 1, June 2025: Question 6

9 marks · Medium difficulty · Extended Answer

Describe an experiment to produce the current-voltage characteristic of a thermistor, explain its behaviour with temperature, and determine the resistance of a fixed resistor in series.

Practise this question

Question

Figure 9 shows a graph of potential difference V in volts against current I in milliamperes for a thermistor T. The voltage increases steeply and nearly linearly from 0 to about 2.8 V at 1.0 mA, peaks at around 3.5 V at 1.8 mA, and then decreases gradually to around 2.3 V at 9.0 mA. Question 06.1 is a 6-mark prompt asking to discuss the characteristic from 0 to 9 mA, including experimental method, resistance behaviour, and temperature explanation. Question 06.2 shows Figure 10, a series circuit with a 10.0 V battery, a fixed resistor R, and thermistor T, asking to determine the resistance of R when the current is 1.0 mA.
Question text

06 Figure 9 shows the variation of potential difference V with current I for a thermistor T.

Figure 9

06.1 Discuss, for the range 0 to 9 mA, the characteristic of T shown by Figure 9.

In your answer you should:

• describe how to produce a characteristic experimentally

• describe, in terms of resistance, the behaviour of T

• explain, with reference to temperature, the behaviour of T.

[6 marks]

06.2 Figure 10 shows thermistor T in series with a fixed resistor R and a battery.

*20*The variation of V with I shown in Figure 9 applies to T in this circuit.

Figure 10

The battery has an emf of 10.0 V and negligible internal resistance.

The current in T is 1.0 mA.

Determine the resistance of R.

[3 marks]

resistance = Ω

Mark scheme

Show the mark scheme The mark scheme for 06.1 provides a level-of-response grid (levels 1–3, up to 6 marks) covering three key areas: Area 1 (producing the characteristic experimentally with ammeter in series, voltmeter in parallel, and varying pd/current), Area 2 (behaviour in terms of resistance: initially constant, then decreasing), and Area 3 (explanation with temperature: self-heating causes temperature rise which decreases resistance). For 06.2 (3 marks), three alternative options are given using reading V = 2.8 V (acceptable 2.7 to 2.9 V) at 1.0 mA: subtracting to find pd across R (7.2 V) and dividing by 1.0 mA, or calculating equivalent resistances, yielding an answer of 7200 ohms.

Question Answers Additional Comments/Guidance Mark AO

06.1 The mark scheme gives some guidance as to what Area 1 Produce characteristic experimentally 6 2 × AO1

statements are expected to be seen in a 1- or 2-mark (L1),

3- or 4-mark (L2) and 5- or 6-mark (L3) answer. Guidance • Use of voltmeter in parallel with test component 2 × AO2

provided in section 3.10 of the ‘Mark Scheme Instructions’ • Use of (milli)ammeter in series with test

component 2 × AO3

document should be used to assist marking this question.

• Method of varying current or pd

Mark Criteria • Description of taking measurements of V and I

for different values (of I)

All three areas (as outlined alongside) covered with • Accept an appropriate circuit diagram for first

at least two aspects covered in some detail. three bullets.

6 marks can be awarded even if there is an error Area 1 is limited to partial coverage where

and/or parts of one aspect missing. temperature is varied by a method other than

A fair attempt to analyse all three areas. If there are changing current.

5 several errors or missing parts, then 5 marks should Area 2 Behaviour of thermistor

be awarded.

• Initially constant resistance (up to about 1 mA)

Two areas successfully discussed, or one discussed • Resistance then decreases (as current

4 and two others covered partially. Whilst there will be increases)

gaps, there should only be an occasional error. • Rapidly decreasing resistance at greater

One area discussed and one discussed partially, or current (greater than about 2 mA)

3 all three covered partially. There are likely to be Idea that the resistance is represented by

several errors and omissions in the discussion. gradient of line limits area 2 to partial.

Only one area discussed or makes a partial attempt Area 3 Explanation

at two areas.

• Initially temperature is constant so thermistor is

One of the three areas covered without significant ohmic

error. • Thermistor warms up due to increased current

0 No relevant analysis. • Increased temperature of thermistor causes

resistance of thermistor to decrease

06.2 Option 1 Any two from In each option: 3 1 × AO3

• (reads off V at 1.0 mA from their graph =) 2.8 (V) 1st bullet point range for voltage is 2.7 to 2.9 V 2 × AO2

• subtracts their pd from 10 (V) to determine pd across R Condone one power of ten error between MP1

and MP2

• divides their pd across R by 1.0 (mA)

(resistance =) 7200 (Ω)

Option 2 Any two from

• (reads off voltage across T (VT) at 1.0 mA from their graph =)

2.8 (V)

• finds resistance of T (RT) by use of V = IR with their read-off

for VT when I = 1 mA

• use of potential divider formula with their VT and their RT

OR

uses ratio of their pds to ratio of their resistances

(resistance =) 7200 (Ω)

Option 3 Any two from

• (reads off voltage across T (VT) at 1.0 mA from their graph =)

2.8 (V)

• (finds resistance of T (RT) by use of V = IR with their read-off

for VT when I = 1 mA) = 2800(Ω)

OR

(Uses V=IR to find the total resistance in the circuit) =

10000(Ω)

• Subtracts their thermistor resistance(RT) from their total

resistance

(resistance =) 7200 (Ω)

Total 9

How to answer it

Thermistor IV Characteristics & Circuit Calculations

WHAT THIS QUESTION TESTS

Core Skills & Specification Knowledge:

  • Practical Circuit Design: Setting up apparatus to measure V and I across a component using appropriate meters and variable control (variable power supply or potential divider).
  • NTC Thermistor Physics: Understanding self-heating effects, charge carrier release at higher temperatures, and why resistance R = V / I decreases even as current rises.
  • Graph Interpretation: Differentiating between local gradient ΔV / ΔI and true resistance V / I from a non-linear V-I characteristic curve.
  • DC Series Circuits: Applying Kirchhoff’s Second Law, Ohm’s Law, and unit conversions ( mA to A ) in a potential divider circuit.
PART 06.1 — 6 MARKS

Extended Response: Investigating & Explaining Thermistor Behaviour

Analysis of V-I characteristic curve (0 to 9 mA) with experimental setup

💡 Key Knowledge (3 Marked Areas)

Area 1: Experimental Method

  • Ammeter / milliammeter placed in series with thermistor T.
  • Voltmeter placed in parallel across thermistor T.
  • Means of varying V or I : variable DC power supply, rheostat, or potentiometer circuit.
  • Record pairs of V and I values across the full range.

Area 2: Resistance Behaviour

  • From 0 up to ~1.0 mA: Straight line graph, so resistance is constant (Ohmic).
  • Above ~1.0 mA: Resistance decreases as current increases.
  • Above ~2.0 mA: Resistance decreases very rapidly.

Area 3: Physical Explanation

  • Low currents: Minimal heating, temperature remains constant, so resistance remains constant.
  • Higher currents: Work done by current causes self-heating (increased temperature).
  • In an NTC semiconductor, higher temperature releases conduction electrons, drastically decreasing resistance.

🧠 Exam Technique & Structure

  • Use Headings: Structure your response using the 3 bullet points given in the question prompt (Experiment, Resistance, Temperature). This guarantees you cover all 3 mark areas.
  • Level 3 Requirement (5–6 marks): You must cover all three areas, with at least two covered in clear, sustained detail.
  • Circuit Diagrams: You can sketch the circuit diagram! A fully labelled diagram with a DC source, variable resistor/divider, series ammeter, and parallel voltmeter instantly secures Area 1.
  • Quote Graph Values: Reference key turning points, particularly the linear section from 0 to 1.0 mA and the peak voltage at ~1.8 mA .

✅ Model Answer Structure

1. Experimental Setup:
Connect thermistor T in series with a DC power supply, a milliammeter, and a variable resistor (or connect across a potential divider). Connect a high-resistance voltmeter directly in parallel across T. Vary the variable resistor to obtain various pairs of V and I from 0 to 9 mA.

2. Resistance Behaviour:
Resistance is given by R = V / I . Between 0 and ~1.0 mA, the graph is linear through the origin, indicating constant resistance (~2800 Ω). Beyond 1.0 mA, as current increases, resistance decreases continuously, dropping rapidly beyond the peak (~2 mA).

3. Explanation via Temperature:
At very low currents, heating is negligible; temperature is constant, obeying Ohm's law. As current rises, electrical power dissipated heats the thermistor. The elevated temperature liberates more charge carriers into the conduction band, causing resistance to fall.

❌ Common Errors & Misconceptions

  • The "Gradient = Resistance" Trap: Stating that resistance equals the gradient of the V-I curve. The mark scheme penalises this explicitly! The gradient is dV/dI , whereas resistance is R = V/I (the chord from the origin). When the gradient is negative (after 2 mA), resistance is not negative!
  • External Heating: Explaining an experiment where the thermistor is heated in a water bath with a Bunsen burner. The graph clearly shows current-driven self-heating in a circuit at ambient temperature.
  • Confusing with Filament Lamp: Claiming resistance increases due to lattice vibrations. A thermistor is a semiconductor—carrier generation vastly outweighs increased scattering.
Marking Scheme Guide:
• Level 3 (5–6 marks): All three areas covered; at least two in good detail. Logical presentation.
• Level 2 (3–4 marks): Two areas successfully covered, or one detailed and two partial.
• Level 1 (1–2 marks): Only one area addressed without major errors, or brief partial coverage of two areas.
PART 06.2 — 3 MARKS

Series Circuit Calculation: Resistor Value Determination

Calculating resistance of fixed resistor R when circuit current is 1.0 mA

📐 Step-by-Step Calculation (Method 1: pd Approach)

  1. Read voltage from Figure 9 at I = 1.0 mA:
    Locate 1.0 mA on the horizontal axis and read the vertical intercept:
    V_T = 2.8 V (allowed range: 2.7 V to 2.9 V).
  2. Find potential difference across fixed resistor R:
    Using Kirchhoff's Voltage Law ( V_supply = V_R + V_T ):
    V_R = 10.0 V - 2.8 V = 7.2 V
  3. Calculate resistance of R using Ohm's Law:
    Current is common in series ( I = 1.0 mA = 1.0 × 10⁻³ A ):
    R = V_R / I = 7.2 / (1.0 × 10⁻³) = 7200 Ω (or 7.2 kΩ )

💡 Alternative Calculation (Total Resistance Approach)

  1. Calculate total circuit resistance:
    R_total = V_supply / I = 10.0 / (1.0 × 10⁻³) = 10 000 Ω
  2. Calculate thermistor resistance at 1.0 mA:
    R_T = V_T / I = 2.8 / (1.0 × 10⁻³) = 2800 Ω
  3. Subtract to find R:
    R = R_total - R_T = 10 000 - 2800 = 7200 Ω

❌ Common Calculation Traps

  • Unit Conversion Error: Forgetting that current is in milliamperes ( mA ). Calculating 7.2 / 1.0 = 7.2 Ω loses the final mark.
  • Misreading the Graph Axis: On Figure 9, 5 small squares equal 1 V, meaning each small vertical grid square represents 0.2 V . At 1.0 mA, the line is 4 small squares above 2.0 V, giving exactly 2.8 V . Reading it as 2.4 V or 3.0 V is penalised outside tolerance.
  • Using 10 V directly across R: Ignoring the thermistor's share of the potential difference.
Mark Allocation (3 Marks):
• MP1: Correct read-off of V_T at 1.0 mA within 2.7 V to 2.9 V .
• MP2: Correct method to find V_R ( 10.0 - V_T ) and dividing by 1.0 mA (or subtracting resistances from R_total ).
• MP3: Correct final answer: 7200 Ω (or 7.2 kΩ ). Note: One power-of-ten error is condoned between MP1 and MP2, but penalized in final answer.

Topics

Physics · Practical skills · 3.5 Electricity · Experimental design · Data analysis

Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.