Resistivity

Understanding material properties that determine resistance

SPEC 3.5.1.3 • ELECTRICITY

What is Resistivity?

The resistance of a wire depends on three factors: its length, cross-sectional area, and the material it's made from.

Definition

Resistivity (ρ) is a property of a material that quantifies how strongly it opposes the flow of electric current. It is independent of the dimensions of the conductor.

Key Concept

Resistivity allows us to compare how well different materials conduct electricity, regardless of their shape or size. A material with low resistivity (like copper) is a good conductor.

The Resistivity Equation

Formula

ρ = \frac{R A}{L}

SymbolQuantitySI Unit

ρ

ResistivityΩ m (ohm-metres)

R

ResistanceΩ (ohms)

A

Cross-sectional aream²

L

Lengthm

Rearrangement

To find resistance:

R = \frac{ρ L}{A}

Resistance is proportional to length and inversely proportional to area.

Understanding the Equation

Longer Wire

More material for electrons to pass through = more collisions = higher R

Larger Area

More pathways for electrons = less crowding = lower R

Typical Resistivity Values

MaterialTypeResistivity (Ω m) at 20°C

CopperConductor1.7 × 10⁻⁸

AluminiumConductor2.8 × 10⁻⁸

ConstantanAlloy4.9 × 10⁻⁷

SiliconSemiconductor2.3 × 10³

GlassInsulator~10¹⁰ to 10¹⁴

Note

Good conductors have very low resistivity (10⁻⁸ Ω m), while insulators have extremely high resistivity (10¹⁰ Ω m or more). Semiconductors fall in between.

Worked Example01

A copper wire has length 2.0 m and diameter 0.50 mm. The resistivity of copper is 1.7 × 10⁻⁸ Ω m. Calculate the resistance of the wire.

1

Convert diameter to radius in metres: r = 0.50/2 = 0.25 mm = 2.5 × 10⁻⁴ m

2

Calculate cross-sectional area: A = πr² = π × (2.5 × 10⁻⁴)² = 1.96 × 10⁻⁷ m²

3

Apply the formula: R = ρL/A = (1.7 × 10⁻⁸ × 2.0) / (1.96 × 10⁻⁷)

4

R = 3.4 × 10⁻⁸ / 1.96 × 10⁻⁷ = 0.17 Ω

Answer: R = 0.17 Ω

Worked Example02

A wire of length 1.5 m has resistance 4.0 Ω and cross-sectional area 8.0 × 10⁻⁷ m². Calculate the resistivity of the wire material.

1

Write the resistivity formula: ρ = RA/L

2

Substitute values: ρ = (4.0 × 8.0 × 10⁻⁷) / 1.5

3

Calculate: ρ = 3.2 × 10⁻⁶ / 1.5 = 2.1 × 10⁻⁶ Ω m

Answer: ρ = 2.1 × 10⁻⁶ Ω m

Worked Example03

Wire A has length L and diameter d. Wire B is made of the same material but has length 2L and diameter 2d. Find the ratio of their resistances R_A : R_B.

1

For wire A: R_A = ρL/A_A where A_A = π(d/2)² = πd²/4

2

For wire B: A_B = π(2d/2)² = π(d)² = πd² (4 times larger)

3

R_B = ρ(2L)/(πd²) = 2ρL/(πd²)

4

R_A/R_B = [ρL/(πd²/4)] / [2ρL/(πd²)] = (4ρL/πd²) × (πd²/2ρL) = 4/2 = 2

Answer: R_A : R_B = 2 : 1

Worked Example04

A wire has resistance 5.0 Ω. It is stretched uniformly to twice its original length. Calculate its new resistance.

1

Key insight: Volume remains constant when wire is stretched. V = LA

2

If length doubles (L' = 2L), then area must halve (A' = A/2) to keep V constant.

3

New resistance: R' = ρL'/A' = ρ(2L)/(A/2) = 4ρL/A = 4R

4

R' = 4 × 5.0 = 20 Ω

Answer: R' = 20 Ω

Effect of Temperature: Metals

When a metal is heated, its resistance increases. This change is governed by the temperature coefficient of resistance (α). Metals have a positive temperature coefficient.

Why?

At higher temperatures, metal ions have more thermal energy and vibrate with greater amplitude. This causes more frequent collisions with conduction electrons, impeding their motion and reducing the current for a given p.d.

Key Features

Graph shows roughly linear increase

Does not pass through origin (has resistance R₀ at 0°C)

Number of charge carriers stays constant

Effect of Temperature: Thermistors

An NTC (Negative Temperature Coefficient) thermistor has resistance that decreases as temperature increases.

Circuit Symbol:

NTC Thermistor

Why?

Thermistors are made from semiconductors. At higher temperatures, bound electrons gain enough energy to move into the conduction band, releasing more charge carriers. This dramatically increases conductivity and reduces resistance.

Metals vs Thermistors

Metal Conductors

Positive temperature coefficient (α > 0)

R increases with T

Free electrons already present

Lattice vibrations impede electron flow

Roughly linear relationship

NTC Thermistors

Negative temperature coefficient (α < 0)

R decreases with T

Semiconductor material

More carriers released at higher T

Non-linear (exponential) relationship

Filament Lamp Application

A filament lamp is most likely to break when switched on. When cold, the filament has low resistance, so a large current surge occurs. This thermal shock causes expansion stress. Once hot, R is higher, I is lower, and the filament is stable.

Exam Note

AQA only tests NTC (Negative Temperature Coefficient) thermistors, not PTC thermistors.

Thermistor Applications

Thermistors are used as temperature sensors in many devices:

Applications

Digital thermometers - measuring body or room temperature

Fire alarms - detecting rapid temperature rise

Car engines - monitoring coolant temperature

Refrigerators - maintaining set temperature

Ovens and incubators - precise temperature control

How It Works

The thermistor is often used in a potential divider circuit. As temperature changes, the resistance changes, altering the output voltage. This voltage is measured and converted to a temperature reading.

Worked Example05

A thermistor has resistance 5000 Ω at 20°C and 2000 Ω at 50°C. It is connected to a 10 V supply. Calculate the current at each temperature.

1

At 20°C: I = V/R = 10/5000 = 0.002 A = 2.0 mA

2

At 50°C: I = V/R = 10/2000 = 0.005 A = 5.0 mA

3

Note: Current increases as temperature increases (because R decreases)

Answer: 2.0 mA at 20°C, 5.0 mA at 50°C

Superconductivity

Definition

A superconductor is a material which, below a certain temperature (the critical temperature, T_c), has zero resistivity.

Key Points

Below T_c, resistance drops abruptly to zero

Current can flow indefinitely without energy loss

T_c depends on the material

Most T_c values are close to 0 K (−273°C)

Why Zero Resistance?

At very low temperatures, electrons can pair up to form Cooper pairs. These pairs move through the lattice without losing energy to collisions, resulting in zero resistance.

Applications of Superconductors

Strong Magnetic Fields

MRI scanners - medical imaging

Particle accelerators (e.g., LHC at CERN)

Maglev trains - magnetic levitation

No heating allows very high currents

Power Transmission

Zero energy loss in cables

More efficient power grids

Currently limited by cooling costs

Research into "high-T_c" materials

Limitation

Most superconductors require cooling to very low temperatures (often using liquid helium at 4 K), making widespread use expensive. "High-temperature" superconductors still need liquid nitrogen (~77 K).

Worked Example06

A power line has resistance 2.0 Ω and carries current 500 A. Calculate the power lost as heat. What would the power loss be if superconducting cables were used?

1

Power dissipated: P = I²R

2

With normal cable: P = (500)² × 2.0 = 250,000 × 2.0 = 500,000 W = 500 kW

3

With superconductor: R = 0, so P = I² × 0 = 0 W

Answer: 500 kW with normal cable, 0 W with superconductor

Worked Example07

A constantan wire (ρ = 4.9 × 10⁻⁷ Ω m) and copper wire (ρ = 1.7 × 10⁻⁸ Ω m) have the same length and diameter. Calculate the ratio of their resistances.

1

Since L and A are the same for both wires, R = ρL/A

2

R_constantan/R_copper = ρ_constantan/ρ_copper

3

Ratio = (4.9 × 10⁻⁷)/(1.7 × 10⁻⁸) = 4.9/0.17 = 28.8 ≈ 29

Answer: R_constantan : R_copper ≈ 29 : 1

Practice

Question 1: A nichrome wire has diameter 0.40 mm and resistivity 1.1 × 10⁻⁶ Ω m. Calculate the length of wire needed to make a 15 Ω heating element.

1

Calculate area: A = π(d/2)² = π(0.20 × 10⁻³)² = 1.26 × 10⁻⁷ m²

2

Rearrange: L = RA/ρ = (15 × 1.26 × 10⁻⁷)/(1.1 × 10⁻⁶)

3

L = 1.89 × 10⁻⁶ / 1.1 × 10⁻⁶ = 1.7 m

Answer: L = 1.7 m

Practice

Question 2: Two wires P and Q are made of materials with resistivities ρ and 2ρ respectively. Wire P has length L and area A. Wire Q has length 2L and area 4A. Find the ratio R_P : R_Q.

1

R_P = ρL/A

2

R_Q = (2ρ)(2L)/(4A) = 4ρL/4A = ρL/A

3

R_P : R_Q = (ρL/A) : (ρL/A) = 1 : 1

Answer: R_P : R_Q = 1 : 1

Practice

Question 3: Explain why constantan (an alloy) is often used for making standard resistors rather than pure copper, even though copper is a better conductor.

1

Constantan has a much higher resistivity than copper, so shorter lengths give useful resistance values.

2

More importantly, constantan has a very small temperature coefficient of resistance.

3

This means its resistance stays nearly constant over a wide temperature range.

4

Standard resistors need stable, predictable values regardless of operating conditions.

Answer: Constantan has higher ρ and very low temperature coefficient, making it ideal for stable resistors.

Summary

Key Equations

EquationDescription

ρ = \frac{R A}{L}

Resistivity definition (unit: Ω m)

R = \frac{ρ L}{A}

Resistance from resistivity

Key Concepts

Resistivity is a material property (independent of dimensions)

Metals: R increases with T (positive coefficient - lattice vibrations)

NTC Thermistors: R decreases with T (negative coefficient - more carriers)

Superconductors: R = 0 below critical temperature T_c

Stretched wire: Volume constant, so R increases by factor of (stretch ratio)²