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Resistance, Resistivity and I–V Characteristics

Key idea: Define resistance, use V = IR and R = ρL/A, interpret resistor and filament-lamp I–V characteristics, and explain the temperature effect in metals.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Apply resistance and resistivity, interpret I–V characteristics and explain temperature effects.

1. Definitions (Must Know)

A. Resistance, R

The resistance of a component is the ratio of the potential difference across it to the current in it:

R = V/I

Unit: ohm (Ω), where 1 Ω = 1 V A⁻¹.

B. Electrical conductor

An electrical conductor is a material with many mobile charge carriers, so it conducts electricity easily.

In metals, the charge carriers are free electrons.

C. Electrical insulator

An electrical insulator has (almost) no mobile charge carriers, so it does not conduct electricity under normal conditions.

D. Resistivity, ρ

Resistivity, ρ, is a material property that links resistance to geometry:

R = ρL/A

where L is length and A is cross-sectional area.

Unit: Ω m.

2. Key Ideas (What Earns Marks)

  • Use V = IR for an ohmic component at constant physical conditions.
  • Metals conduct because they have many mobile electrons; insulators have very few mobile carriers.
  • Resistivity is a material property (depends on material and temperature).
  • For a uniform wire at fixed temperature:
    • R ∝ L
    • R ∝ 1/A
  • Typical metals: resistivity increases with temperature (more collisions).
  • The gradient of an I–V graph is I/V = 1/R, not R.
Avoid a common phrase

Don’t say “electrons slow down so current decreases” unless the circuit conditions are stated. In many questions, the supply voltage is fixed, so increased resistivity increases R and reduces current via I = V/R.

3. Detailed Explanations

A. Ohmic and non-ohmic I–V characteristics

An ohmic resistor at constant temperature has I ∝ V, so its I–V graph is a straight line through the origin. A filament lamp heats as current increases; its resistance rises, so the graph’s gradient decreases at larger |V|.

I–V characteristics: ohmic resistor and filament lamp

Current against potential difference for an ohmic resistor and a filament lamp, showing the lamp curve becoming less steep as its temperature rises.

Scroll across the graph to read all labels.

Current against potential difference for an ohmic resistor and a filament lamp, showing the lamp curve becoming less steep as its temperature rises.Current against potential difference for an ohmic resistor and a filament lamp, showing the lamp curve becoming less steep as its temperature rises.
Both characteristics pass through the origin and are symmetric for reversed polarity. The lamp becomes hotter and more resistive as current magnitude increases.
Open full-size graph
View figure data
Values and uncertainty for I–V characteristics: ohmic resistor and filament lamp
SeriesPotential difference, V (V)Potential difference, V uncertaintyCurrent, I (A)Current, I uncertainty
Ohmic resistor-6-1
Ohmic resistor-3-0.5
Ohmic resistor00
Ohmic resistor30.5
Ohmic resistor61
Filament lamp-6-0.72
Filament lamp-3-0.52
Filament lamp-1-0.25
Filament lamp00
Filament lamp10.25
Filament lamp30.52
Filament lamp60.72

At an operating point, resistance is always R = V/I. It is not generally found from the reciprocal tangent gradient for a non-linear component.

B. Conduction in metals (drift + collisions)

In a metal, free electrons move randomly. When a potential difference is applied, an electric field is set up and electrons gain a small net drift velocity.

Electrons frequently collide with lattice ions. These collisions are why:

  • energy is transferred to the lattice (heating),
  • the drift speed is small even when the current is large (because number density is huge).

Link to the drift current equation:

I = nAv|q|

(see Drift Velocity & Current).

C. Why insulators do not conduct (in this model)

In an insulator, electrons are tightly bound to atoms, so there are very few mobile charge carriers.

With no carriers to drift, current is (approximately) zero.

D. Resistivity and geometry

For a uniform wire:

R = ρL/A

Interpretation:

  • longer wire ⇒ more collisions ⇒ larger R,
  • thicker wire (larger A) ⇒ more parallel paths ⇒ smaller R.

E. Temperature dependence for metals

For typical metals, increasing temperature increases lattice vibration.

This increases collision frequency, reducing electron mobility (for a given field), so resistivity ρ increases.

In filament lamps, the temperature rise can be large, so R increases significantly as current increases (non-ohmic behaviour).

4. Common Mistakes

  • Mixing up resistivity ρ (Ω m) with resistance R (Ω).
  • Calling the gradient of an I–V graph the resistance. For axes I against V, the straight-line gradient is 1/R.
  • Using A in mm² without converting to m².
  • Claiming “insulators have no electrons” (they do; the issue is that electrons are not free to move).
  • Forgetting that ρ usually depends on temperature.

5. Exam Tips

  • If a question changes wire geometry: R₂/R₁ = (ρ₂/ρ₁)(L₂/L₁)A₁/A₂ For the same material at the same temperature, ρ₂ = ρ₁.
  • If you are given diameter d: A = π(d/2)²
  • For “temperature effect” explanations in metals: use “more lattice vibration → more collisions → lower mobility → higher ρ”.

6. Worked Examples

Modelled example 1

Resistance from resistivity

Core

Problem

A wire has length L = 2.0 m, area A = 1.0 × 10⁻⁶ m² and resistivity ρ = 1.7 × 10⁻⁸ Ω m. Find its resistance.
Study the worked solution

  1. Select the material-and-geometry relation

    Method

    Use R = ρ L/A.

    Reason

    Resistivity describes the material; length and area describe the sample geometry.

    Working

    R = (1.7 × 10⁻⁸)2.0/(1.0 × 10⁻⁶)

  2. Evaluate

    Method

    Obtain 3.4 × 10⁻² Ω.

    Reason

    All quantities are already in SI units.

    Working

    R = 3.4 × 10⁻² Ω

Guided practice 2

Comparing two wires (same material)

About 4 min

Problem

Wire 2 has twice the length and half the cross-sectional area of wire 1. Both have the same material and temperature. Find R₂/R₁.

Try this before viewing the solution

Hints

Hint 1: cancel the shared property
The same material at the same temperature has the same ρ.
Hint 2: form the ratio
Use R₂/R₁ = (L₂/L₁)(A₁/A₂).
View solution step by step

  1. Cancel resistivity

    Method

    Compare only length and area factors.

    Reason

    ρ₂/ρ₁ = 1 under the stated conditions.

    Working

    R₂/R₁ = (L₂/L₁)A₁/A₂

  2. Apply both changes

    Method

    Multiply the factor two by the reciprocal of one-half.

    Reason

    Resistance increases with length but decreases with area.

    Working

    R₂/R₁ = 2(1/(1/2)) = 4

Common misconception 3

Finding resistivity

Find and correct the mistake

Learner claim

A wire has R = 0.80 Ω, L = 5.0 m and A = 2.0 × 10⁻⁷ m². A learner rearranges R = ρ L/A as ρ = RL/A. Diagnose the error and find ρ.

Try this before viewing the solution

Correct rearrangement

View solution step by step

  1. Rearrange before substituting

    Method

    Use ρ = RA/L.

    Reason

    Multiplying R = ρ L/A by A/L isolates ρ.

    Working

    ρ = ((0.80)(2.0 × 10⁻⁷))/5.0

  2. Evaluate with the material unit

    Method

    Obtain 3.2 × 10⁻⁸ Ω m.

    Reason

    Resistivity has unit ohm metre, unlike resistance.

    Working

    ρ = 3.2 × 10⁻⁸ Ω m

Examiner practice 4

Effect of halving the diameter (same material)

3 marks

Examination question

Two wires have the same length, material and temperature. Wire 2 has half the diameter of wire 1. Find R₂/R₁. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Relate diameter to area

    1 mark

    Method

    Square the diameter factor.

    Reason

    A = π d²/4.

    Working

    A₂/A₁ = (1/2)² = 1/4

  2. Relate resistance to area

    1 mark

    Method

    Use R ∝ 1/A.

    Reason

    Resistivity and length are unchanged.

    Working

    R₂/R₁ = A₁/A₂

  3. State the factor

    1 mark

    Method

    Conclude that resistance quadruples.

    Reason

    Taking the reciprocal of one-quarter gives four.

    Working

    R₂/R₁ = 4

Challenge 5

Current change when resistivity changes (fixed voltage)

Minimal support

Independent transfer

A wire connected to a fixed voltage supply has unchanged geometry but its resistivity doubles. Assuming the wire dominates the circuit resistance, determine the current factor and justify both links.

Try this before viewing the solution

Hints

Hint 1: hold geometry fixed
First use R = ρ L/A, then apply Ohm’s law at fixed voltage.
View solution step by step

  1. Transfer the material change to resistance

    Method

    Double the resistance.

    Reason

    With fixed L and A, R ∝ ρ.

    Working

    R₂ = 2R₁

  2. Transfer resistance to current

    Method

    Halve the current.

    Reason

    At fixed V, I = V/R.

    Working

    I₂/I₁ = R₁/R₂ = 1/2

7. Mind Stretchers

Mind stretcher 1: Filament lamp resistance increaseExtension

Explain why a filament lamp’s resistance increases as current increases.

Show Answer

Larger current causes more heating (P = I²R). The filament temperature rises, lattice vibration increases, collision frequency increases, so resistivity (and hence resistance) increases.

Mind stretcher 2: Stretching a wire (constant volume)Extension

A wire of fixed resistivity is stretched so its length doubles. Assuming volume stays constant, what happens to its resistance?

Show Answer

If volume AL is constant, doubling L halves A.

Then:

R = ρL/A ⇒ R₂/R₁ = (L₂/L₁)A₁/A₂ = (2)(1/(1/2)) = 4

So resistance quadruples.

Mind stretcher 3: Optional (Enrichment)Extension

A. Other charge carriers (beyond this lesson)

Not all conductors use electrons:

  • electrolytes conduct via positive and negative ions,
  • gases conduct when ionised.

Continue with Semiconductors and NTC Thermistors.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027