Resistance, resistivity, I–V characteristics and temperature

Key idea: H2 Physics lessons on circuit representation, material behaviour, networks and RC transients.

  • GCE A-Level H2 Physics 2027

Learn the idea

Big question: Why do I–V curves change with material and temperature?

Resistance is V/I at a stated operating point; resistivity gives R = ρl/A for a uniform conductor. An ohmic conductor has a straight I–V line at constant temperature. A filament lamp heats and its resistance rises, while a semiconductor diode conducts strongly mainly in forward bias. Temperature changes carrier scattering or carrier density depending on the material.

Separate resistance from resistivity

Resistance at an operating point is R = V/I. For a uniform conductor, R = ρl/A, where resistivity ρ is a material property at a stated temperature while l and A describe geometry.

In a metal, rising temperature increases lattice vibration and carrier scattering, so resistivity rises. An NTC thermistor instead gains many mobile carriers as it warms, so its resistance falls.

Check your understanding: A wire's length doubles and diameter doubles at constant temperature. What is the resistance factor?

Area becomes four times larger, so R changes by 2/4 = 1/2.

Read each I–V curve at its operating point

An ohmic conductor has V proportional to I at constant temperature. On an I-vertical graph, a filament lamp becomes less steep as self-heating raises resistance; a diode carries little reverse current and rises sharply in forward bias.

A curved component still has operating resistance V/I. Gradient gives a different quantity depending on axis choice, so label axes before discussing steepness.

Check your understanding: Does a straight tangent on one part of a curve prove ohmic behaviour?

No. Ohmic behaviour requires the whole relevant V–I relation to be proportional at constant temperature.

A-Level current–voltage characteristicsFour qualitative current against potential difference graphs. An ohmic conductor is a straight line through the origin. A filament lamp becomes less steep as voltage magnitude rises. A diode carries almost no reverse current and rises steeply in forward bias. An NTC thermistor becomes steeper as self-heating reduces resistance.Ohmic conductorVIFilament lampVISemiconductor diodeVINTC thermistorVIforward risewarmer, lower R
Scroll diagram horizontally to read all labels.
These are I-against-V graphs, so the local gradient represents conductance. The lamp heats and becomes less conducting; the NTC thermistor heats and becomes more conducting.

Key ideas to keep

  • Gradient depends on which quantity is plotted vertically.
  • A curved graph still has a resistance V/I at each point.
  • Resistivity is a material property at a stated temperature; resistance also depends on geometry.

Worked example

Compare four I–V characteristics from their physics

Question: Compare the I–V shapes of an ohmic resistor, filament lamp, semiconductor diode and NTC thermistor.

  1. Step 1: Fix the axes

    Why: The meaning of steepness changes if I and V are swapped.

    Working: Take I vertically and V horizontally throughout.

  2. Step 2: Separate metallic behaviours

    Why: An ohmic resistor stays at constant temperature while a lamp self-heats.

    Working: The resistor is straight through the origin; the lamp becomes less steep as lattice scattering raises resistance.

  3. Step 3: Explain semiconductor behaviours

    Why: Carrier availability and junction direction control their curves.

    Working: A diode is strongly asymmetric; an NTC thermistor becomes steeper as heating raises carrier density and lowers resistance.

Answer: At constant temperature an ohmic resistor gives a straight line through the origin. A lamp's gradient decreases in an I-against-V graph as heating raises metal resistivity. A diode has negligible reverse current and a steep forward rise after its turn-on region. An NTC thermistor heats as current rises; greater carrier number density lowers resistance, making the curve progressively steeper.

Check: Every curve has a physical explanation, rather than being four shapes to memorise.

Practise with support

Try this

A wire's length doubles and diameter halves at constant ρ. State the resistance factor.

Hint: Apply the diameter change to area before using R = ρl/A.

Check your answer

Area is proportional to diameter squared, so it becomes one quarter. R = ρl/A therefore changes by 2/(1/4) = 8.

Practise independently

Your turn

Explain temperature effects in a metal and NTC semiconductor using the specific microscopic quantities required by the syllabus.

Check your answer

In a metal, higher temperature increases lattice scattering, so carrier drift velocity for a given field falls and resistivity rises. In an NTC semiconductor, heating releases many more charge carriers, increasing number density enough to lower resistivity despite increased scattering.

Common mistakes

Common mistake

Resistance is always the gradient of an I–V graph.

What is wrong with this reasoning?

Show better thinking

Resistance is V/I at an operating point. Graph-gradient interpretation depends on which quantity is plotted vertically and on linearity.

Common mistake

Metals and NTC thermistors change resistance for the same microscopic reason.

What is wrong with this reasoning?

Show better thinking

Metal heating reduces drift velocity through greater scattering; NTC heating greatly increases carrier number density.

Exam guidance

Label axes and explain the microscopic temperature effect instead of saying only that resistance changes.

Exam-style practice [8 marks]

A 2.0 m wire of radius 0.15 mm has ρ = 4.0 × 10⁻⁷ Ω m. Find R. Then contrast its heated I–V curve with an NTC thermistor's.

Plan before you answer

  • Convert the radius before finding area.
  • Use R = ρl/A.
  • Contrast metal scattering with NTC carrier density.
Mark your answer and compare the model

Marking points

Tick each point only if your answer states it clearly.

Model answer

A = π(1.5 × 10⁻⁴)² = 7.07 × 10⁻⁸ m², so R = ρl/A = 11.3 Ω. Metal heating reduces drift velocity and raises resistance; NTC heating increases carrier number density and lowers resistance.

Check what stayed with you

Recall question

At constant ρ, length triples and area doubles. State the resistance factor.

Check the answer

R = ρl/A, so the factor is 3/2.

Try this next

Continue to the next lesson in this topic.

Internal resistance, terminal p.d. and output power

Syllabus and review details

This lesson covers the listed H2 Physics 9478 outcomes. Topic 16 states no explicit exclusions. I–V graph gradient language always names the plotted axes; resistance is V/I at an operating point rather than automatically a graph gradient. Metal temperature dependence is explained through drift velocity and NTC behaviour through carrier number density. Terminal p.d. is E − Ir while a source supplies current. Capacitor rules are not copied from resistor rules. The exponential forms assume constant R and C, an ideal source or isolated discharge loop, and τ = RC.

  • GCE A-Level H2 PhysicsTopic 16(c) / Topic 16(d) / Topic 16(e) / Topic 16(f) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027