Wire Resistance & Resistivity

Pick a material, a length, a thickness and a temperature. The page works out what that wire's resistance is, what it costs you in voltage and heat at the current you set, and what the same job would weigh in six other metals.

Worked example — Wire Resistance & Resistivity

Cross-sectional area	1.5001 mm²
Resistivity at this temperature	1.68 × 10⁻⁸ Ω·m
Resistance of the whole length	2.2399 Ω
Resistance per metre	1.12 × 10⁻² Ω/m
Voltage lost along it	19.478 V
Power turned into heat in it	169.4 W
Resistance ÷ its own value at 20 °C	1.0000
What the conductor weighs	2.688 kg

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Wire Resistance & Resistivity

A wire's resistance is R = ρL/A: the resistivity ρ of the material it is made of, times how long it is, divided by its cross-sectional area. Resistivity is measured in ohm-metres and belongs to the substance alone — copper is 1.68 × 10⁻⁸ Ω·m whether it is a hair or a bus bar, and the shape of the wire is entirely in L and A. Area reaches the diameter through a square, A = πd²/4, so thickness counts twice. Resistivity itself moves with temperature, and this page uses the first-order form ρ(T) = ρ₂₀(1 + α(T − 20)).

What each symbol means

ρ: resistivity, in ohm-metres. A property of the material and nothing else; the wire's shape is not in it
L: length of the conductor, in metres. In a cable that means there AND back, which is the step the worked problem is about
A: cross-sectional area, in m². πd²/4 for a round wire, so doubling the diameter divides the resistance by four
α: the temperature coefficient, per kelvin. Around 0.004 for a pure metal and ten times smaller for a resistance alloy
R: resistance of that particular piece of wire, in ohms — the only quantity here that depends on every one of the others

Assumes: Direct current, a conductor of one material with the same round cross-section along its whole length, and a resistivity that follows a straight line through its value at 20 °C. The figures are for pure annealed metals; commercial cable is alloyed and work-hardened and runs a few per cent higher, and nichrome is a family of compositions rather than one, so treat its 1.10 × 10⁻⁶ Ω·m as the middle of a range. The mass column is the conductor only, with no insulation, no steel core and no sag.
Breaks when: Far from 20 °C, which is the honest limit of the straight line above. Extrapolate this model to a lamp filament at 2,800 °C and it returns 13.5 times the room-temperature resistance; the true figure for tungsten is larger, and that is exactly why Resistors in Series and Parallel says a filament conducts better cold than hot and quotes no number for it. It breaks in four more places. Alternating current pushes the charge towards the surface, so a thick conductor stops using its middle and the effective area is smaller than πd²/4. Semiconductors and carbon run the coefficient the other way and conduct better as they warm. A superconductor takes ρ to nothing at all, at which point R = ρL/A has no work left to do. And at the ends of any real wire there is contact resistance, which belongs to the joint rather than to the metal and is what actually burns out a loose connection.

Only one metal here weighs less than copper 🖖

Aluminium is the worse conductor — 1.577 times copper's resistivity — so matching a copper cable takes 57.7% more cross-section and a wire 25.6% thicker. Then weigh it. Aluminium is 3.32 times less dense, and 1.577 ÷ 3.32 is 0.475: the same resistance, the same length, at 47.5% of the weight. Read the last column of the table below and look for a number under 1. There is one. Silver conducts better than copper and still comes out heavier; so do gold, tungsten, iron and nichrome. That one ratio is why every overhead line you have ever driven under is aluminium, and why the copper is indoors where nothing has to hold it up.

Ten times the voltage, a hundredth of the heat 🖖

Press An overhead line: 100 km of 400 mm² aluminium, 6.6248 Ω. Send 100 MW down it. The current is the power divided by the voltage and the loss is I²R, so the loss falls as the square of the voltage. At 400 kV the line carries 250 A and burns 0.414 MW, which is 0.414% of what is being sent. At 40 kV it carries 2,500 A and burns 41.4 MW. At 10 kV the arithmetic asks for 662.5 MW from a line delivering 100, which is its way of saying the line cannot do the job at all. Nothing about the wire changed in any of that — only the number on the substation.

Nichrome barely notices the heat, and copper does 🖖

Press A 100 m extension lead, then The same lead, hot. Only the temperature moves, and the resistance goes from 2.2399 Ω to 2.9639 Ω, so the 169.4 W it was wasting becomes 224.1 W. Between 0 °C and 100 °C copper gains 44.0% of its resistance. Nichrome over the same hundred degrees gains 4.0% — eleven times flatter, and the whole reason the alloy exists, because a heating element has to glow and still draw what it was built to draw. The second panel puts all seven side by side, each against its own value at 20 °C, so the only thing left to compare is the slope.

Problem solved in full

A 2 kW heater at the end of 100 metres of flex 6 steps
A 2 kW heater is plugged into a 100 m extension lead of 1.5 mm² copper flex, on a 230 V supply. Work out how much of that 2 kW actually reaches the heater, before reading on. Press A 100 m extension lead to see the first half of it on the tool.

Open this problem in the tool
1. The label says 1.5 mm², which is the area of one conductor. As a diameter that is 1.382 mm, and the area the tool prints is 1.5001 mm².
  
  $A = \frac{π d ^{2}}{4} = \frac{π \cdot 1 . 38 2 ^{2}}{4} = 1.5001 m m^{2} = 1.5001 \times 1 0^{- 6} m^{2}$
2. Here is the step almost everyone misses. A 100 m lead carries the current 100 m out and 100 m back, so the conductor is 200 m long.
  
  $L = 2 \times 100 = 200 m$
3. Now ρL/A, with copper at 1.68 × 10⁻⁸ Ω·m: 2.2399 Ω.
  
  $R = \frac{ρ L}{A} = \frac{1 . 68 \times 1 0 ^{- 8} \cdot 200}{1 . 5001 \times 1 0 ^{- 6}} = 2.2399 Ω$
4. Take the current the heater's rating implies, 8.696 A. That would lose 19.478 V along the lead and leave 169.4 W in it — the two figures the tool prints.
  
  $I = \frac{2000}{230} = 8.696 A \Rightarrow V = 19.478 V, P = I^{2} R = 169.4 W$
5. But that current is wrong, and the reason is the previous lesson: the lead and the heater are in series, so the supply sees both. A 2 kW heater on 230 V is 26.45 Ω, the pair comes to 28.6899 Ω, and the real current is 8.0168 A.
  
  $R_{load} = \frac{23 0 ^{2}}{2000} R_{load} + R_{lead} I = \frac{230}{28 . 6899} 26.45 Ω 28.6899 Ω 8.0168 A$
6. Redo the two figures at the current that actually flows. The heater is left with 212.04 V across it and delivers 1,699.9 W; the lead takes 144.0 W.
  
  $V_{load} = I R_{load} P_{load} = I^{2} R_{load} P_{lead} = I^{2} R_{lead} 212.04 V 1, 699.9 W 144.0 W$
Answer

The heater delivers 1,699.9 W of its 2,000, so 300.1 W of heating simply never happens. Only 144.0 W of that shortfall is warming the cable — the other 156.1 W is not drawn from the supply at all, because the heater is a resistance and a resistance at 212 V takes less current than the same resistance at 230. That is the part the naive calculation gets wrong in the safe direction: it overstates the current, and so overstates the loss in the lead by 18%. The danger it understates is elsewhere, in the 144 W sitting inside a coiled reel with nowhere to go.

$\frac{1 , 699 . 9}{2 , 000} = 0.84992, 000 - 1, 699.9 = 300.1 W$

Learning path

What Ohm’s law leaves out

References (3)

Where the resistivity of an element and its temperature dependence are tabulated. This is the source for the SHAPE of what this page computes; the figures it prints are the standard handbook set, and the note below says so: P. D. Desai, T. K. Chu, H. M. James and C. Y. Ho, "Electrical Resistivity of Selected Elements." Journal of Physical and Chemical Reference Data 13(4), 1984, pp. 1069-1096. DOI 10.1063/1.555723
The combination rules and the three forms of resistive power the last two steps of the worked problem use. Cited on Ohm's Law and on Resistors in Series and Parallel as well, deliberately: one circuits text across the chapter. J. W. Nilsson and S. A. Riedel, Electric Circuits, 10th ed. Pearson/Prentice Hall, 2015. ISBN 978-0-13-376003-3.
Insight block 1 as a live engineering decision rather than a physics exercise. Modern overhead lines trade the same two properties this page divides against each other, and the paper costs out the result on ultra-high-voltage spans: Y. Zhang, H. Sun, X. Zhang, Z. Chen, L. Zhou and X. Huang, "Economic analysis of aluminium conductor composite core and conventional aluminium conductor steel reinforced conductors in overhead ultra-high voltage transmission lines." IET Generation, Transmission & Distribution 17(5), 2023, pp. 1098-1110. DOI 10.1049/gtd2.12719

By the 42Math team · Reviewed against published sources. Corrections welcome.