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I/V Graph Of Thermistor

Key idea: Learn how a thermistor’s resistance changes with temperature (NTC/PTC), what its I–V curve shows, and how to find R = V/I at a point (O Level).

  • SEC G3 Physics 2027
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Learning objectives

  • State current as rate of charge flow measured in amperes
  • Distinguish conventional current from electron flow
  • Apply charge equals current multiplied by time
  • Define source e.m.f. as work done per unit charge around a circuit
  • Calculate total e.m.f. for sources in series
  • Define component potential difference as work done per unit charge
  • State resistance as potential difference divided by current
  • Apply resistance equals potential difference divided by current
  • Apply wire-resistance proportionalities for length and cross-sectional area
  • Describe the effect of temperature on metallic resistance
  • Sketch and interpret required current–voltage characteristics

1. Definition

Optional (Not required in 6091)

Thermistors are examinable in D.C. Circuits as input transducers in potential dividers, but the I–V graph of a thermistor is not required in the 6091 O Level syllabus.

If you’re revising for exams, focus on: LDR & Thermistor in Potential Dividers.

A thermistor is a resistor whose resistance changes with temperature.

  • NTC thermistor: resistance decreases when temperature increases.
  • PTC thermistor: resistance increases when temperature increases.

2. Key Ideas

  • When current increases, the thermistor warms up.
  • For an NTC thermistor, warming up makes resistance decrease, so current increases faster at higher voltages.
  • On an I vs V graph, this often looks like a curve that becomes steeper as V increases (for an NTC thermistor).
  • Resistance at a point: R = V/I (use that point’s values). Resistance is not the gradient.

3. Detailed Explanations

A. Why an NTC thermistor gives a curve

  1. Increase the p.d. across the thermistor.
  2. Current increases, so the thermistor warms up.
  3. For an NTC thermistor, higher temperature means lower resistance.
  4. Lower resistance makes the current increase even more, so I does not increase proportionally with V.

B. Comparing common I–V shapes (I vs V)

ComponentTypical I–V shapeKey reason
Metallic conductor (constant temperature)straight line through originresistance constant
Filament lampcurve that becomes shallowerheating increases resistance
NTC thermistorcurve that becomes steeperheating decreases resistance

I–V graph for an NTC thermistor (typical shape)

A thermistor I–V curve that becomes steeper at higher voltages because heating reduces resistance for an NTC thermistor.

Scroll across the graph to read all labels.

A thermistor I–V curve that becomes steeper at higher voltages because heating reduces resistance for an NTC thermistor.A thermistor I–V curve that becomes steeper at higher voltages because heating reduces resistance for an NTC thermistor.
For an NTC thermistor, increasing current can heat it up, decreasing its resistance and making the I–V curve steeper at higher voltages.
Open full-size graph
View figure data
Values for I–V graph for an NTC thermistor (typical shape)
Potential difference (V)NTC thermistor
00
10.02
20.1
30.22
40.36
50.5
60.6

4. Common Mistakes

  • Saying “the gradient is the resistance” (use R = V/I at a point).
  • Mixing up NTC and PTC (NTC: temperature increases → resistance decreases).
  • Confusing an NTC thermistor curve with a filament lamp curve (lamp resistance increases when it heats).

5. Exam Tips

  • In 6091, you usually do not need the thermistor I–V graph. If it appears, describe the curve and link it to temperature change.
  • For an NTC thermistor, use: “current increases → thermistor heats → resistance decreases → graph becomes steeper”.
  • If asked for resistance at a point, calculate R = V/I at that point.

6. Worked Examples

Modelled example 1

Resistance at a point

Core

Problem

At one point on an I–V curve, V = 3.0 V and I = 0.20 A. Find the resistance at that operating point.

Study the worked solution

  1. Use the coordinates of the operating point

    Method

    Calculate the ratio V/I using the two coordinates.

    Reason

    Resistance at an operating point is defined by the potential difference across the component divided by its current.

    Working

    R = V/I = 3.0/0.20 = 15 Ω

Guided practice 2

Comparing resistance at two points

About 4 min

Problem

At A, (V,I) = (2.0 V,0.10 A). At B, (V,I) = (6.0 V,0.60 A). Calculate the resistance at each point and use the values to describe the change.

Complete the guided steps

Unit: Ω
Unit: Ω
Resistance

Hints

Hint 1: treat each point separately
Apply R = V/I once at A and once at B.
Hint 2: compare the results
A smaller V/I ratio means a smaller resistance.
View solution step by step

  1. Calculate each point resistance

    Reason

    A non-linear component does not have one constant resistance for its whole curve.

    Working

    R_A = 2.0/0.10 = 20 Ω, R_B = 6.0/0.60 = 10 Ω

  2. Interpret the change

    Working

    The resistance decreases from 20 Ω to 10 Ω, consistent with an NTC thermistor heating up.

Common misconception 3

Resistance is not the gradient

Find and correct the mistake

Learner response

At a point on a curved I–V graph, V = 5.0 V and I = 0.50 A. A learner draws a tangent and calls its gradient the resistance. Locate the first conceptual error, then find the resistance.

Diagnose before viewing the correction

Where is the first error?
Unit: Ω

View solution step by step

  1. Diagnose the graph-ratio error

    Method

    Reject the tangent gradient as the required resistance.

    Reason

    On an I-against-V graph the tangent gradient is Δ I/Δ V, while point resistance uses the coordinate ratio V/I.

    Working

    Read the voltage and current coordinates at the operating point and calculate R = V/I. On an I-against-V graph, even the origin-to-point slope is I/V, so its reciprocal gives resistance.

  2. Calculate the point resistance

    Reason

    The coordinate pair describes the component’s actual operating condition.

    Working

    R = 5.0/0.50 = 10 Ω

Examiner practice 4

Checking if it is ohmic

3 marks

Examination question

For a component, V increases from 2.0 V to 4.0 V while I increases from 0.10 A to 0.30 A. Determine whether the component is ohmic and explain using the data. [3 marks]

Write your answer before viewing the mark scheme

View solution step by step

  1. Compare the multiplication factors

    1 mark

    Method

    Compare the voltage and current multiplication factors.

    Reason

    An ohmic conductor at constant temperature has I ∝ V and therefore constant V/I.

    Working

    The potential difference doubles, while the measured current increases by a factor of three.

  2. State the ohmic prediction

    1 mark

    Reason

    If current were proportional to potential difference, doubling V would double I.

    Working

    The predicted current would be 0.20 A, not 0.30 A.

  3. Conclude from the interval

    1 mark

    Working

    These readings are not consistent with ohmic behaviour over this interval because I is not proportional to V, and V/I changes.

Challenge 5

Identifying the curve shape

Minimal support

New context

A temperature sensor has the readings below after each voltage has been applied long enough for its temperature to settle.

V/VI/A
1.00.020
3.00.220
5.00.500

Decide whether its I–V graph becomes steeper or shallower and identify whether its behaviour is consistent with an NTC thermistor or a filament lamp.

Try this without the worked method

Three possible current–voltage graph shapesOption A is a straight line through the origin. Option B rises but becomes shallower as voltage increases. Option C rises and becomes steeper as voltage increases. All graphs place current on the vertical axis and potential difference on the horizontal axis.AVI0BVI0CVI0
Scroll diagram horizontally to read all labels.
Choose the qualitative graph that best represents the supplied data. The axes use the same orientation; the sketches are not to scale.
Which graph matches the data?
Which component model is consistent with the data?

Hints

Hint 1: convert the table into resistance evidence
Compare V/I for the first and last readings.
Hint 2: connect resistance to curve shape
Falling resistance means each additional volt produces a larger current increase.
View solution step by step

  1. Extract the resistance trend from the table

    Reason

    The table is a different representation of the same I–V relationship.

    Working

    R_(1.0 V) = 1.0/0.020 = 50 Ω, R_(5.0 V) = 5.0/0.500 = 10 Ω

  2. Identify the curve and component model

    Working

    The I–V graph becomes steeper and is consistent with an NTC thermistor, not a filament lamp.

7. Mind Stretchers

Mind stretcher 1: Reducing heatingExtension

If you take I–V readings using small currents (so the thermistor hardly heats up), how would the graph change? Explain.

Show Answer

With little heating, the resistance changes less during the measurements. The I–V curve would be closer to a straight line (more “constant resistance” during the test).

Mind stretcher 2: Thermistor vs filament lampExtension

Both a filament lamp and an NTC thermistor heat up when current flows. Why do their I–V graphs curve in opposite ways?

Show Answer

For a filament lamp (metal), heating increases resistance, so the graph becomes shallower. For an NTC thermistor, heating decreases resistance, so the graph becomes steeper.

8. Practice and next step

Treat this page as supporting transfer, not a fourth required Topic 14 graph. Return to the three required I–V characteristics, then continue to thermistors in potential dividers, where the component is examinable.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027