AQA A-Level Physics Paper 1, June 2024: Question 6
11 marks · Medium difficulty · Short Answer
Calculate the resistance of a variable resistor in a circuit containing a thermistor, determine a constant B in the thermistor temperature equation, explain the need to control current, and describe how to demonstrate conservation of charge and energy.
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Question text
06 The circuit in Figure 5 is used as part of a temperature sensor.
The battery has an emf of 6.5 V and negligible internal resistance.
Figure 5
The initial temperature of the thermistor is 22 °C.
At this temperature the resistance of the thermistor is 350 Ω and the circuit current
is 12 mA.
06.1 Calculate the resistance of the variable resistor.
[2 marks]
14 resistance = Ω
06.2 The resistance R of the thermistor at temperature θ in K is given by:
B −
R = R e 0
where R0 is the resistance at the initial temperature θ0 in K, and B is a constant.
The temperature of the thermistor is increased to 318 K.
The variable resistor is adjusted so that the circuit current is again 12 mA.
The potential difference across the thermistor is now 3.2 V.
Determine B.
State an appropriate unit for your answer.
[5 marks]
B = 15 unit =
06.3 Explain why the current in the thermistor needs to be controlled.
[2 marks]
06.4 Explain how ammeters and voltmeters can be used in the circuit in Figure 5 to
demonstrate the conservation of charge and the conservation of energy.
Refer to points X, Y and Z in your answer.
[2 marks]
Mark scheme
Show the mark scheme
Question Answers Additional comments/Guidelines Mark AO
06.1 Appropriate use of V = IR e.g. for MP1: 2 AO2
• determines total circuit resistance
• determines pd across thermistor
• use of = 𝐼𝐼(𝑅𝑅XY+𝑅𝑅YZ)
Condone POT error in MP1
Variable resistor resistance = 190 (Ω) Expect to see total circuit resistance of 542
(Ω)
Expect to see pd across resistor of 2.3 V.
Condone POT error in MP1
Calculator value: 191.67 Ω
06.2 Evidence of 22 oC converted to K Allow ecf from MP1. Expect 295 K 5 AO2
3.2 3.2 Expect to see 267 Ω
Determines R OR OR −3
12 12×10
𝑅𝑅
Evidence of use of ln( ) with their values Condone use of R0 = 190 Ω in MP3
𝑅𝑅0
B = 1110 K
Accept 1100
Accept answers that round to 1110 or 1120
Allow ecf only from temperature conversion.
Do not accept k for K
Current causes thermistor temperature to change
06.3 Allow a clear description of thermal runaway 2 AO3
Thermistor resistance decreases as temperature for both marks.
increases
06.4 Uses ammeter(s) (in series) to show current at X = current 2 AO1
at Y = current at Z Do not allow ‘currents across’.
OR
Do not accept ‘battery pd’ unless it is clearly
Uses voltmeter(s) to show that emf/terminal pd / 6.5 V = pd
being measured.
across XY + pd across YZ
If points XYZ are not referred it must be clear
where the meters are attached.
Links current readings to (conservation of) charge
AND
Links pd readings to (conservation of) energy
Total 11
How to answer it
A-Level Physics Study Guide: Temperature Sensor Circuits
What this question tests
This question assesses your ability to apply Ohm's law in a potential divider circuit, manipulate exponential temperature-resistance equations for thermistors, understand self-heating effects in sensors, and link electrical measurements to fundamental conservation laws (charge and energy).
Calculate the resistance of the variable resistor
✅ Correct Answer
190 Ω
📐 Calculation Steps
- Find total circuit resistance using Ohm's Law: R_total = V / I = 6.5 / (12 × 10⁻³) = 542 Ω
- Subtract thermistor resistance from total resistance: R_variable = R_total - R_thermistor = 542 - 350 = 192 Ω (using exact total) or 190 Ω depending on intermediate rounding.
❌ Common Errors
- Forgetting to convert milliamperes to amperes ( 12 mA = 12 × 10⁻³ A ), leading to power-of-ten (POT) errors.
Determine constant B and state appropriate unit
✅ Correct Answer
B = 1110 K (or 1100 to 1120 K)
Unit: K (Kelvin)
💡 Key Knowledge
- Always convert Celsius to Kelvin by adding 273.15 (or 273 ). Initial temp: 22 + 273 = 295 K .
- Logarithms must be used to bring the constant B out of the exponent: ln(R / R_0) = B((1 / θ) - (1 / θ_0)) .
📐 Step-by-Step Calculation
- Convert temperatures: θ_0 = 295 K , θ = 318 K .
- Determine new thermistor resistance R at 318 K using V = 3.2 V and I = 12 mA :
R = 3.2 / (12 × 10⁻³) = 266.7 Ω - Rearrange the exponential equation using natural logs:
ln(266.7 / 350) = B × ((1 / 318) - (1 / 295)) - Calculate values:
-0.2724 = B × (0.003145 - 0.003390)
-0.2724 = B × (-0.000245) - Solve for B:
B = -0.2724 / -0.000245 = 1110 K
🧠 Exam Technique
Make sure you do not write lowercase k for the unit of B . Temperature units must be capital K .
Explain why current needs to be controlled
💡 Key Knowledge & Correct Answer
- Passing too large a current through the thermistor causes internal heating (Joule heating / self-heating).
- As the thermistor heats up due to the current, its resistance changes independently of the external ambient temperature being measured, leading to inaccurate temperature sensor readings.
❌ Common Errors
Students often vague out and say "it protects the circuit" without linking the current directly to temperature change via thermistor resistance characteristics.
Explain how meters demonstrate conservation laws
✅ Correct Answer
- Conservation of Charge: Place an ammeter in series at point X, Y, and Z. The readings will be equal, showing current (charge per second) is not used up in a single loop.
- Conservation of Energy: Place voltmeters across components XY and YZ, and across the battery. The sum of the potential differences across XY and YZ equals the emf of the battery.
🧠 Exam Technique
You must explicitly reference points X, Y, and Z in your answer and specify where the meters are attached. Avoid imprecise phrasing like "currents across" or "battery pd" unless clearly measured.
Topics
Physics · 3.5 Electricity
Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.