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Resistivity of Metals Chart & Calculator

Resistivity values for common metals, plus resistance and temperature correction, step by step
Find the resistance of 50 m of copper with a 10 mm^2 cross-section
Compare copper and aluminium conductors of the same size over 50 m
Correct a 0.084 ohm copper conductor from 20 C to 75 C
Find the nichrome wire length for a 48 ohm heating element of 0.5 mm diameter

What Electrical Resistance Actually Is

Resistance is how strongly a particular object opposes the flow of charge, defined by

R=VI[Ω]R = \frac{V}{I} \qquad [\Omega]

with VV in volts and II in amperes. One ohm is one volt per ampere. It is a property of the object — this length of this wire.

Resistivity ρ\rho is the corresponding property of the material, independent of shape, and the two are joined by the geometry:

R=ρLAρ=RALR = \frac{\rho L}{A} \qquad \rho = \frac{RA}{L}

  • ρ\rho — resistivity, ohm-metres (Ω·m)
  • LL — length, metres (m)
  • AA — cross-sectional area, square metres (m²)

Physically, resistance is the scattering of conduction electrons by lattice vibrations and by impurities. Heating the metal increases the vibrations, which is why metals get more resistive as they warm.

The operating assumption: a uniform, homogeneous conductor at a stated temperature carrying DC.

Resistivity of Metals Chart

Values at 20 °C, with the temperature coefficient α\alpha per °C referenced to 20 °C:

Materialρ\rho (Ω·m)α\alpha (per °C)
Silver1.59×10−81.59 \times 10^{-8}0.00380.0038
Copper1.68×10−81.68 \times 10^{-8}0.003930.00393
Gold2.44×10−82.44 \times 10^{-8}0.00340.0034
Aluminium2.65×10−82.65 \times 10^{-8}0.004290.00429
Tungsten5.60×10−85.60 \times 10^{-8}0.00450.0045
Iron9.71×10−89.71 \times 10^{-8}0.00500.0050
Nichrome1.10×10−61.10 \times 10^{-6}0.00040.0004

Correct to another temperature with the linear relation

ρT=ρ20[1+α(T−20)]\rho_T = \rho_{20}\left[1 + \alpha (T - 20)\right]

Nichrome is roughly 6565 times more resistive than copper and barely changes with temperature, which is exactly what a heating element needs.

Treat these as nominal textbook values. Real conductors vary with alloy, temper and purity, and a design that depends on the number should use the supplier's certified data, with any current-carrying decision made under the applicable wiring code.

Common Mistakes to Avoid

  • Using a resistivity without its temperature — the chart is at 20 °C, and a conductor running hot is measurably more resistive.
  • Leaving the area in mm² — 1010 mm² is 1×10−51 \times 10^{-5} m². Skipping this is a factor of a million.
  • Swapping resistance and resistivity — resistivity is Ω·m and belongs to the material; resistance is Ω and belongs to the object.
  • Assuming aluminium can simply replace copper — for the same resistance an aluminium conductor needs roughly 58%58\% more cross-sectional area.
  • Applying the linear α\alpha far outside its range — it is a first-order fit valid near the reference temperature, not down to cryogenic or up to melting temperatures.
  • Forgetting that heating is self-reinforcing — current heats the conductor, which raises resistance, which raises the drop.
  • Reading nichrome's low α\alpha as low resistivity — its resistivity is very high; it is the change with temperature that is small.

示例题目

Step 1: Area: A=10 mm2=1.0×10−5 m2A = 10\ \text{mm}^2 = 1.0 \times 10^{-5}\ \text{m}^2
Step 2: Copper: R=(1.68×10−8 Ω⋅m)(50 m)÷(1.0×10−5 m2)=(8.40×10−7)÷(1.0×10−5)=0.0840 ΩR = (1.68 \times 10^{-8}\ \Omega\cdot\text{m})(50\ \text{m}) \div (1.0 \times 10^{-5}\ \text{m}^2) = (8.40 \times 10^{-7}) \div (1.0 \times 10^{-5}) = 0.0840\ \Omega
Step 3: Aluminium: R=(2.65×10−8)(50)÷(1.0×10−5)=(1.325×10−6)÷(1.0×10−5)=0.1325 ΩR = (2.65 \times 10^{-8})(50) \div (1.0 \times 10^{-5}) = (1.325 \times 10^{-6}) \div (1.0 \times 10^{-5}) = 0.1325\ \Omega
Step 4: Ratio: 0.1325÷0.0840=1.580.1325 \div 0.0840 = 1.58, so aluminium is 58%58\% more resistive at the same size
Step 5: To match: AAl=10 mm2×1.58=15.8 mm2A_{\text{Al}} = 10\ \text{mm}^2 \times 1.58 = 15.8\ \text{mm}^2
Answer: Copper 0.08400.0840 Ω, aluminium 0.13250.1325 Ω; about 15.815.8 mm² of aluminium matches 1010 mm² of copper

Step 1: Temperature rise above the reference: 75 ∘C−20 ∘C=55 ∘C75\ ^\circ\text{C} - 20\ ^\circ\text{C} = 55\ ^\circ\text{C}
Step 2: αΔT=(0.00393 ∘C−1)(55 ∘C)=0.2162\alpha \Delta T = (0.00393\ ^\circ\text{C}^{-1})(55\ ^\circ\text{C}) = 0.2162
Step 3: R75=R20(1+0.2162)=(0.0840 Ω)(1.2162)R_{75} = R_{20}(1 + 0.2162) = (0.0840\ \Omega)(1.2162)
Step 4: R75=0.1022 ΩR_{75} = 0.1022\ \Omega
Step 5: That is a 21.6%21.6\% increase, which is why voltage-drop checks are done at the conductor's operating temperature
Answer: R75≈0.102R_{75} \approx 0.102 Ω, about 22%22\% above the 20 °C value

Step 1: A=πd2/4=π(0.50×10−3 m)2/4=π(2.50×10−7 m2)/4=1.963×10−7 m2A = \pi d^2/4 = \pi (0.50 \times 10^{-3}\ \text{m})^2/4 = \pi (2.50 \times 10^{-7}\ \text{m}^2)/4 = 1.963 \times 10^{-7}\ \text{m}^2
Step 2: Rearrange: L=RA/ρL = RA/\rho
Step 3: L=(48 Ω)(1.963×10−7 m2)÷(1.10×10−6 Ω⋅m)L = (48\ \Omega)(1.963 \times 10^{-7}\ \text{m}^2) \div (1.10 \times 10^{-6}\ \Omega\cdot\text{m})
Step 4: Numerator: 9.425×10−6 Ω⋅m29.425 \times 10^{-6}\ \Omega\cdot\text{m}^2, so L=8.57 mL = 8.57\ \text{m}
Step 5: On 240240 V: I=240 V÷48 Ω=5.0 AI = 240\ \text{V} \div 48\ \Omega = 5.0\ \text{A}
Step 6: P=VI=(240 V)(5.0 A)=1200 WP = VI = (240\ \text{V})(5.0\ \text{A}) = 1200\ \text{W}
Answer: L≈8.57L \approx 8.57 m of wire, drawing 5.05.0 A and dissipating 12001200 W on 240240 V

常见问题

Resistance is the ratio of the voltage across an object to the current through it, R = V/I, measured in ohms. Microscopically it is the scattering of conduction electrons by lattice vibrations and impurities, which is why metals become more resistive as they get hotter.

Resistivity is a material property in ohm-metres and does not depend on shape. Resistance is a property of a specific object in ohms and depends on length and cross-section as well as material, through R = rho L / A.

Silver, at about 1.59 x 10^-8 ohm-metres at 20 C, just below copper at 1.68 x 10^-8. Copper is used almost everywhere instead because it costs far less and does not tarnish into a poorly conducting surface layer.

For metals, use R_T = R_20 [1 + alpha (T - 20)] with alpha in reciprocal degrees Celsius. Copper's alpha of 0.00393 means a conductor at 75 C has about 22% more resistance than at 20 C. The relation is a first-order fit near the reference temperature.

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