IGCSE · 4 · extended response
A thermistor alarm with a reversed response
Electricity and magnetism · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Thermistor calibration
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Resistance / kΩ
Horizontal axis (x): Temperature / °C
Read the exact data as a table
| Temperature / °C | Resistance / kΩ |
|---|---|
| 10 | 8 |
| 20 | 4 |
| 30 | 2 |
| 40 | 1 |
| 50 | 0.5 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A 6.0 V supply is connected to a 2.0 kΩ fixed resistor in series with an NTC thermistor. Output voltage is measured across the thermistor by a high-resistance input. Its resistance–temperature data are supplied. The alarm activates when output falls below 2.0 V. Supplied data (Temperature / °C; Resistance / kΩ): (10, 8), (20, 4), (30, 2), (40, 1), (50, 0.5).
- (a)
Find output voltage at 20°C and 40°C, and identify the temperature at which the threshold is reached.
3 marks - (b)
Predict how the response changes if output is taken across the fixed resistor, and explain why the thermistor is not an ohmic resistor with one universal resistance.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Find output voltage at 20°C and 40°C, and identify the temperature at which the threshold is reached.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
For a series divider, Vout = 6.0 Rtherm/(2.0 kΩ+Rtherm). At 20°C, this is 6(4)/(2+4) = 4.0 V.
- 2
At 40°C, Vout = 6(1)/(2+1) = 2.0 V. The NTC resistance decreases as temperature rises, so the thermistor voltage also decreases.
- 3
The strict 'below 2.0 V' trigger occurs above 40°C, assuming monotonic response and negligible self-heating. A real alarm needs tolerance and hysteresis specifications.
(b)
Predict how the response changes if output is taken across the fixed resistor, and explain why the thermistor is not an ohmic resistor with one universal resistance.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
The two divider voltages sum to 6.0 V, so the fixed-resistor voltage is 6.0−Vtherm.
- 2
As the thermistor resistance falls, total resistance falls and current rises; the fixed-resistor voltage IRfixed therefore rises.
- 3
Resistance depends strongly on temperature. Even at one bath temperature a large measuring current can cause self-heating, so a calibration requires controlled conditions and suitably low power.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- For a series divider, Vout = 6.0 Rtherm/(2.0 kΩ+Rtherm). At 20°C, this is 6(4)/(2+4) = 4.0 V.
- At 40°C, Vout = 6(1)/(2+1) = 2.0 V. The NTC resistance decreases as temperature rises, so the thermistor voltage also decreases.
- The strict 'below 2.0 V' trigger occurs above 40°C, assuming monotonic response and negligible self-heating. A real alarm needs tolerance and hysteresis specifications.
- The two divider voltages sum to 6.0 V, so the fixed-resistor voltage is 6.0−Vtherm.
- As the thermistor resistance falls, total resistance falls and current rises; the fixed-resistor voltage IRfixed therefore rises.
- Resistance depends strongly on temperature. Even at one bath temperature a large measuring current can cause self-heating, so a calibration requires controlled conditions and suitably low power.
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