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Resistance Resisitivity

Reference data and engineering information about resistance resisitivity for electrical applications.

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Overview

Engineering reference data for Resistance Resisitivity in electrical engineering.

Key Formulas

Ohm's Law

V=IRV = IR

Voltage = Current × Resistance.

Power

P=VI=I2R=V2/RP = VI = I^2R = V^2/R

Electrical power.

Energy

E=PtE = Pt

Energy = Power × Time.

Variables

SymbolDescriptionUnit
VVVoltageV
IICurrentA
RRResistanceΩ
PPPowerW

Common Resistivity Values

Conductors at 20°C

Good conductors have low resistivity. Values below are for temperatures around 20°C.

5 rows
Resistivity of Common Electrical Conductors
Material
Resistivity (Ω m)
Note
Aluminum (pure)2.65e-8Relative to Cu: 1.6
Carbon1e-7
Copper (annealed)1.724e-8Reference: 1.00
Iron (pure)1e-7Relative to Cu: 5.68
Silver1.6e-8Relative to Cu: 0.94

Source: engineeringtoolbox.com

Insulators

Good insulators have high resistivity.

7 rows
Resistivity of Common Insulators
Material
Resistivity (Ω m)
Bakelite1×10¹²
Glass1×10¹⁰ – 1×10¹¹
Mica9×10¹²
Paraffin wax (pure)1×10¹⁶
Polystyrene1×10¹⁴
Porcelain1×10¹²
Water, distilled1×10¹⁰

Source: engineeringtoolbox.com

Resistivity Comparison (Relative to Copper)

This table shows how the resistivity of various metals compares to that of annealed copper (which is set as 1.00). A lower number indicates better conductivity.

8 rows
Resistivity Compared to Annealed Copper
Material
Relative Resistivity
Silver0.94
Gold1.4
Aluminum (pure)1.6
Chromium1.8
Zinc3.4
Brass3.7 – 4.9
Nickel5.1
Steel7.6 – 12.7

Source: engineeringtoolbox.com

Practical Example: Wire Resistance Calculation

The fundamental relationship R=ρL/AR = \rho L / A allows you to calculate the resistance of a wire if you know its material properties and dimensions.

Example 1: 10 m of 17-gauge Copper Wire Given:

  • Length, L=10mL = 10\,\text{m}
  • Cross-sectional area, A=1.04mm2=1.04×106m2A = 1.04\,\text{mm}^2 = 1.04 \times 10^{-6}\,\text{m}^2
  • Copper resistivity, ρ=1.7×108Ωm\rho = 1.7 \times 10^{-8}\,\Omega\cdot\text{m}

Calculation: R=ρLA=(1.7×108Ωm)(10m)1.04×106m20.16ΩR = \frac{\rho L}{A} = \frac{(1.7 \times 10^{-8}\,\Omega\cdot\text{m})(10\,\text{m})}{1.04 \times 10^{-6}\,\text{m}^2} \approx 0.16\,\Omega

Example 2: Effect of Reducing Wire Gauge (to 24-gauge) The same 10 m length of copper wire is now 24-gauge with A=0.205mm2A = 0.205\,\text{mm}^2.

R=(1.7×108Ωm)(10m)0.205×106m20.83ΩR = \frac{(1.7 \times 10^{-8}\,\Omega\cdot\text{m})(10\,\text{m})}{0.205 \times 10^{-6}\,\text{m}^2} \approx 0.83\,\Omega

This demonstrates how significantly resistance increases as the cross-sectional area decreases.

Unit Conversion

Conversions for the SI unit of resistivity (ohm metre, Ω·m): 1Ωm=102Ωcm=39.37Ωinch=3.28Ωfoot1\,\Omega\cdot\text{m} = 10^2\,\Omega\cdot\text{cm} = 39.37\,\Omega\cdot\text{inch} = 3.28\,\Omega\cdot\text{foot}

References