Physics · Electricity

Resistivity and wire resistance

Homogeneous conductor, uniform cross-section and specified temperature in a one-dimensional ohmic model. Contact resistance, current crowding and strong temperature profiles require separate treatment.

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Formula

Calculate

R = ρ \cdot \frac{l}{A}

R in Ω · ρ in Ω·mm²/m · l in m · A in mm²

Variables & units

Variables & units
Symbol	Meaning	Unit
$R$	Resistance of the conductor	Ω
$ρ$	Resistivity (copper: 0.0178)	Ω·mm²/m
$l$	Length of the conductor	m
$A$	Cross-sectional area of the conductor	mm²

LaTeX

R = \rho \cdot \frac{l}{A}

QUANTA STUDYquanta-study.de

Closed loop with 18 m outgoing and return conductors, load and source. The graph below shows total loop resistance against prescribed temperature. — In the supplied linear model, resistance of both wires together rises from 0.864 ohm at 20 °C to 1.03248 ohm at 70 °C. Loop length is 36 m.

Resistivity and wire resistance calculator

Find the resistance, resistivity, length or cross-section of a wire, also from its diameter and at a different temperature.

Calculation model

Find

Resistivity ρ

Material value from the table, e.g. copper 0.0178 Ω·mm²/m at 20 °C; watch the unit.

Conductor length ℓ

Length the current actually flows through, e.g. 50 m; with an outgoing and return wire count both.

Cross-sectional area A

Conducting cross-section without insulation, e.g. 1.5 mm² for a household cable.

Filled with example values. Type your own and the result updates straight away.

Result: Resistance R

R = 0.5933 Ω

With ρ = 0.0178 Ω·mm²/m, ℓ = 50 m and A = 1.5 mm², R = 0.5933 Ω.

Working · rearrange for R, substitute, evaluate

1. Rearrange for R
$R=\frac{\rho\ell}{A}$
2. Substitute values (in SI units)
- ρ = 0.0178 Ω·mm²/m = 1.78 · 10⁻⁸ Ω·m
- A = 1.5 mm² = 1.5 · 10⁻⁶ m²
$\frac{1.78\times10^{-8}\cdot 50}{0.0000015}=0.593333$
3. Result
R = 0.5933 Ω

Quick tips

Twice the length, twice the resistance; twice the cross-section, half the resistance. Half the diameter means a quarter of the area, so four times the resistance.
Unit check: with ρ in Ω·mm²/m, ℓ in m and A in mm², the mm² and m cancel and Ω is left. No conversion to m² needed.

Assumptions and limits

Homogeneous conductor with constant conductive area at a specified uniform temperature; one-dimensional ohmic DC approximation. Positive material and geometry values. Use the actually traversed conductor length; contacts and current redistribution are separate. This is not complete cable sizing.

Derivation & background

Resistivity depends on material state and temperature. A homogeneous conductor gives $R=\rho\frac\ell A$ . Tabulated values depend on their conditions and are not universal exact material constants. The retained wire case explicitly supplies $\rho=0.0178\,\mathrm{\Omega\,mm^2/m}$ as a calculation input. A supplied linear temperature model can be used over a limited interval; geometric changes may require separate treatment. Doubling length doubles R; doubling area halves R for the same material state. At 20 °C, OpenStax Table 9.1 gives silver 0.0159, copper 0.0168, aluminium 0.0265 and iron 0.0971 in $\frac{\Omega{}\cdot{}mm^{2}}{m}$ . Its copper coefficient is 0.0039 per kelvin, while constantan is only 0.00003 per kelvin.

Exam blueprint

Validity range

Derivation steps

The local field-current-density relation becomes component resistance through conductor geometry.

1Choose a homogeneous conductor with specified length, constant cross-section and specified temperature.
2The local ohmic relation is $E=\rho J$ , with $\rho$ describing the material state.
3The uniform one-dimensional model gives $U=E\ell$ and $I=JA$ .
4Substitution gives $U=\rho\ell\frac IA$ ; comparison with $U=RI$ yields $R=\rho\frac\ell A$ .
5For different sections, add their resistances. Slowly varying geometry gives approximately $R=\int\frac{\rho(x)}{A(x)}\,\mathrm dx$ .
6Check area units, outgoing and return conductors, and temperature; contact resistance and current crowding require additional models.

Rearrangements

Cross-section

A = \frac{\rho \cdot l}{R}

Sizing cables for a maximum permitted resistance.

Length

l = \frac{R \cdot A}{\rho}

For example the wire length of a coil from a resistance measurement.

Resistivity

\rho = \frac{R \cdot A}{l}

Identifying a material from resistance and geometry.

Task variant

A single uniform 20 m conductor has supplied resistivity $\rho{}=\frac{0.0178 \Omega{}\cdot{}mm^{2}}{m}$ at a specified temperature. What area gives at most 0.2 Ω? Return conductor and contacts are not included here.

Approach: $A\ge\frac{\rho\ell}{R_{\max}}$ .
The values give at least $A=1.78\,\mathrm{mm^2}$ .
An example area of $2.5 mm^{2}$ gives about 0.1424 Ω in the stated model and meets this resistance limit.
It does not establish complete real cable sizing; material state, return path, contacts and further operating requirements need separate assessment.

An iron wire (ρ = 0.10 Ω·mm²/m, A = 0.5 mm²) has $R = 4 \Omega{}$ . How long is it?

$\ell=\frac{RA}{\rho}=\frac{4\,\Omega\cdot0.5\,\mathrm{mm^2}}{0.10\,\mathrm{\Omega\,mm^2/m}}=20\,\mathrm m$ .

Common mistakes

Mixing $mm^{2}$ and $m^{2}$ .

Stay consistent: with ρ in $\frac{\Omega{}\cdot{}mm^{2}}{m}$ use A in $mm^{2}$ ; with ρ in $\Omega{}\cdot{}m$ use A in $m^{2}$ .

Forgetting the outgoing and return conductor in cables.

For voltage drop, add the resistances of the actual outgoing and return conductors. With equal lengths, areas, materials and temperatures, loop resistance is twice one conductor’s resistance.

Reading ρ as mass density.

Here ρ is the electrical resistivity, a different quantity with the same symbol.

Ignoring the temperature dependence.

Tabulated values need a reference temperature. Heating increases resistance for many metals with a positive temperature coefficient; this does not hold for every conductor.

Exam context

Tasks combine the formula with Ohm’s law and power loss: choosing cross-sections, voltage drop of long lines and sizing heating wires.

Formula cluster

DC circuits

Extends Ohm’s law by geometry and material dependence.

U = R·I

Ohm's Law

Rges = R₁+R₂ ; 1/Rges = 1/R₁+1/R₂

Resistors in Series and Parallel

Worked example

Given one 50 m wire, $A=1.5\,\mathrm{mm^2}$ and the supplied $\rho=0.0178\,\mathrm{\Omega\,mm^2/m}$ , $R=\rho\frac\ell A\approx0.593\,\Omega$ . An identical return conductor doubles loop resistance.

Applications

Sizing cable cross-sections (house wiring), heating wires, strain gauges, resistance thermometers (Pt100), overhead lines

How do you calculate with Resistivity (Conductor Resistance)?

Here is how to work through a typical Resistivity (Conductor Resistance) (R = ρ·l/A) task step by step:

1
Task
A single uniform 20 m conductor has supplied resistivity $\rho{}=\frac{0.0178 \Omega{}\cdot{}mm^{2}}{m}$ at a specified temperature. What area gives at most 0.2 Ω? Return conductor and contacts are not included here.
Solution path
1. Approach: $A\ge\frac{\rho\ell}{R_{\max}}$ .
2. The values give at least $A=1.78\,\mathrm{mm^2}$ .
3. An example area of $2.5 mm^{2}$ gives about 0.1424 Ω in the stated model and meets this resistance limit.
4. It does not establish complete real cable sizing; material state, return path, contacts and further operating requirements need separate assessment.
2
Task
An iron wire (ρ = 0.10 Ω·mm²/m, A = 0.5 mm²) has $R = 4 \Omega{}$ . How long is it?
Solution path
$\ell=\frac{RA}{\rho}=\frac{4\,\Omega\cdot0.5\,\mathrm{mm^2}}{0.10\,\mathrm{\Omega\,mm^2/m}}=20\,\mathrm m$ .

Understand and apply the equation

Two wires of the same material can have very different resistance. Resistivity describes the material state; length and cross-section turn it into resistance of a particular conductor. Temperature, outgoing and return paths, and changes in geometry therefore matter. A stretched model conductor shows why fixed quantities must be stated explicitly.

What is resistivity and how do you calculate wire resistance?

Compare two equally long wires at the same temperature: a larger cross-section allows more current at equal voltage. A longer wire with unchanged area needs more voltage for the same current. This is a geometric statement within the ohmic model, not a substitute for the material’s temperature specification or a check of contacts and current distribution.[1]

\begin{aligned}R&=\rho\frac{\ell}{A}\\{}[\rho]&=\Omega\,\mathrm m\end{aligned}

Homogeneous straight conductor with constant cross-section, uniform temperature and approximately uniform current density.[1]

ρ is neither mass density nor the total resistance of the object. Composition, state and temperature determine it; calling it a material constant does not mean temperature independent. A is conductive cross-section perpendicular to current, excluding insulation. Contact and interface resistances are separate contributions. The local relation E=ρJ applies in the stated linear regime.[1][2]

Derive from field and current density

The local material relation connects field and current density. Length converts field into voltage; cross-section converts current density into current. Keep these steps separate to understand why area appears in the denominator and why its unit must be converted correctly. Ohm’s law then gives the resistance between the conductor ends.[1]

\begin{aligned}U&=E\ell\\{}I&=JA\\{}\frac{U}{I}&=\frac{\rho J\ell}{JA}=\rho\frac{\ell}{A}\end{aligned}

Uniform E and J in the ohmic model; division is defined for a nonzero probing current.[1]

Length dependence corresponds to conductor segments traversed in succession; area dependence corresponds to additional equivalent current paths alongside each other. This explains doubling resistance at double length and halving it at double area. Double diameter instead gives four times the circular area and one quarter of the resistance, with other conditions unchanged.[1][3]

Explore further: Account for nonuniform conductors segment by segment

R=\int_0^\ell\frac{\rho(x)}{A(x)}\,\mathrm dx

One-dimensional approximation for the same continuous current path, excluding separate contact contributions.[1][3]

For discrete uniform segments this becomes $\sum_j\frac{\rho_j\ell_j}{A_j}$ . An arithmetic mean area cannot simply be substituted into $\frac{\ell}{A}$ : narrow segments contribute more strongly. Abrupt geometry changes can produce highly nonuniform current density, limiting the one-dimensional model. At high frequencies, current redistribution changes effective area; DC resistance is not automatically sufficient.[1]

State temperature relative to a reference

\begin{aligned}&\rho(T)\approx\rho_0[1+\alpha(T-T_0)]\\{}&R(T)\approx R_0[1+\alpha(T-T_0)]\end{aligned}

The second approximation neglects geometry changes; α is the local temperature coefficient at the stated reference.[1]

Many metals have a positive coefficient near a reference temperature. This does not make the linear approximation universal across all temperatures and materials. Semiconductors, alloys and phase changes may behave very differently. Joule heating couples electrical calculation to heat rejection: computed dissipation does not determine temperature by itself. Temperature must be specified or found from a thermal balance.[1][4]

Examples: outgoing and return conductors, varying area and stretching

A two-wire loop at elevated temperature

Given: Outgoing and return conductors each $18\,\mathrm m$ , $A=0.75\,\mathrm{mm^2}$ . Model values: $\rho_{20}=0.018\,\mathrm{\Omega\,mm^2/m}$ , $\alpha=0.0039\,\mathrm{K^{-1}}$ ; both conductors at $70\,{}^\circ\mathrm C$ , $I=2.4\,\mathrm A$ .
Find: Find loop resistance, voltage loss and dissipation.

Approach: Include both conductor lengths, then apply the supplied temperature coefficient.
$\begin{aligned}R_0&=\rho_0\frac{\ell_{\mathrm{loop}}}{A}\\{}R_T&=R_0[1+\alpha(T-T_0)]\end{aligned}$
Include both outgoing and return paths.
$\begin{aligned}\ell_{\mathrm{loop}}&=36\,\mathrm m\\{}R_{20}&=0.018\frac{36}{0.75}=0.864\,\Omega\end{aligned}$
Apply the specified uniform temperature.
$R_{70}=0.864(1+0.0039\cdot50)=1.03248\,\Omega$
Use the same loop current for losses.
$\begin{aligned}\Delta U&=IR=2.477952\,\mathrm V\\{}P_{\mathrm{loss}}&=I^2R=5.9470848\,\mathrm W\end{aligned}$

Result: The model gives about 1.03 Ω, 2.48 V drop and 5.95 W line loss.

Check: Each identical conductor contributes half; ΔU×I gives the same 5.9470848 W. This is neither a temperature prediction nor an installation approval.

Two different cross-sections in one path

Given: $\rho=0.020\,\mathrm{\Omega\,mm^2/m}$ ; first segment $2\,\mathrm m$ at $0.40\,\mathrm{mm^2}$ , second $1\,\mathrm m$ at $0.20\,\mathrm{mm^2}$ ; constant temperature and ideal contacts.
Find: Find total resistance and the error from averaging areas.

Approach: Successive segments are electrically in series and must be calculated individually.
$R=\rho\frac{\ell_1}{A_1}+\rho\frac{\ell_2}{A_2}$
Calculate each segment with its own A.
$\begin{aligned}R_1&=\frac{0.020\cdot2}{0.40}=0.10\,\Omega\\{}R_2&=\frac{0.020\cdot1}{0.20}=0.10\,\Omega\end{aligned}$
Add along the continuous current path.
$R=0.20\,\Omega$
Compare the length-weighted arithmetic area as an incorrect substitute.
$\begin{aligned}\overline A&=\frac{2\cdot0.40+1\cdot0.20}{3}=\frac{1}{3}\,\mathrm{mm}^2\\{}R_{\mathrm{wrong}}&=\frac{0.020\cdot3}{\frac{1}{3}}=0.18\,\Omega\end{aligned}$

Result: Correct series resistance is 0.20 Ω; arithmetic averaging alone underestimates it by 10%.

Check: Here the shorter segment with half the area contributes the same resistance as the longer one. Inverse area explains this.

Previous comparisons specified a fixed area. In the following original geometric model, volume and resistivity are instead fixed, forcing area to change with length.[1]

Longer and thinner at fixed volume

Given: A uniform model conductor is 1.0 m long with area 0.50 mm² and supplied resistivity 0.020 Ω·mm²/m. Its length increases to 1.5 m at constant volume. Resistivity and temperature remain unchanged in this model; no actual elastic-strain behaviour is asserted.
Find: Find the new area and resistance ratio.

Approach: volume conservation determines new area before applying the resistance relation.
$\begin{aligned}A_1\ell_1&=A_0\ell_0\\A_1&=\frac{A_0\ell_0}{\ell_1}\\\frac{R_1}{R_0}&=\left(\frac{\ell_1}{\ell_0}\right)^2\end{aligned}$
Include both geometric changes.
$\begin{aligned}A_1&=\frac{0.50}{1.5}\,\mathrm{mm^2}=\frac13\,\mathrm{mm^2}\\R_0&=0.020\frac{1.0}{0.50}\,\Omega=0.040\,\Omega\\R_1&=0.020\frac{1.5}{\frac13}\,\Omega=0.090\,\Omega\end{aligned}$
Independently calculate the factor from the length ratio.
$\frac{R_1}{R_0}=1.5^2=2.25$

Result: Resistance increases by a factor of 2.25. A factor of 1.5 would require unchanged cross-section.

Check: Volume remains $0.50\,\mathrm{mm^2}\cdot1.0\,\mathrm m=\frac13\,\mathrm{mm^2}\cdot1.5\,\mathrm m$ . Real deformation can change the material state too; that additional effect is explicitly excluded here.

View diagram: two simultaneous geometric changes

Fixed volume: resistance grows quadratically

QUANTA STUDYquanta-study.de

At fixed volume, area decreases inversely with length. The curve assumes constant temperature and resistivity; it is not a material strain measurement.[1]

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How do you convert Ω·m, Ω·mm²/m and wire diameter?

\begin{aligned}1\,\Omega\,\mathrm{mm}^2\,\mathrm m^{-1}&=10^{-6}\,\Omega\,\mathrm m\\{}A&=\frac{\pi d^2}{4}\\{}A&=\frac{\rho\ell}{R}\end{aligned}

A and ρ must use compatible area conventions; d is diameter, not radius.[1]

The introductory 50 m/1.5 mm² copper case with specified ρ=0.0178 gives 0.5933 Ω, about 0.59 Ω for that one conductor length. An equally long return conductor doubles loop resistance. The supplied ρ is a rounded practical model value, not an exact temperature-free constant. Typical-material tables must state temperature, purity and rounding; they do not replace data-sheet measurements.[1]

Five exercises with unit checks

Calculate · Foundation

A single model conductor with ρ=0.018 Ω·mm²/m is 4 m long with radius 0.30 mm. Find R.

Show hint

Use the radius directly in πr².

Learning goal: Calculate cross-section from radius.

Show solution and calculation

Approach: calculate circular area from radius before finding resistance.
A=π×0.30²=0.282743 mm²; R≈0.254648 Ω.
Diameter 0.60 mm gives the same area through $A=\frac{\pi d^2}{4}$ .

R=\frac{0.018\cdot4}{\pi(0.30)^2}\approx0.25465\,\Omega

Calculate · Application

In an ideal voltage-drop model: total conductor length 40 m, ρ=0.018 Ω·mm²/m, I=1.5 A, maximum drop 1.2 V. What area meets this condition alone?

Show hint

Rearrange $\Delta U=I\frac{\rho\ell}{A}$ for A.

Learning goal: Find area from a voltage-drop limit alone.

Show solution and calculation

Approach: total loop resistance must meet the specified voltage-drop limit.
A≥0.90 mm².
Check R=0.8 Ω and IR=1.2 V.
This is only the electrical model bound; ampacity, protection, installation and temperature remain undetermined.

A\ge\frac{1.5\cdot0.018\cdot40}{1.2}=0.90\,\mathrm{mm}^2

Calculate · Application

R₀=2.0 Ω at 20 °C, α=0.0040 K⁻¹. Find the model value at 45 °C.

Show hint

Use a 25 K difference, not absolute 318 K.

Learning goal: Apply linear temperature dependence.

Show solution and calculation

Approach: use temperature difference from the specified reference state.
R≈2.0(1+0.0040×25)=2.20 Ω.
Increase 10%; the reference expression does not predict behaviour at arbitrarily high temperatures.

R=2.0(1+0.0040\cdot25)=2.20\,\Omega

Interpret · Application

A single homogeneous model conductor has two 1 m sections: A1=1 mm² and A2=2 mm²; constant ρ=0.020 Ω·mm²/m. Compare its resistance with a 2 m conductor having arithmetic mean area 1.5 mm² everywhere.

Show hint

Sections are in series, not parallel.

Learning goal: Combine local resistances along a varying cross-section.

Show solution and calculation

Approach: Add local resistances.
The stepped conductor has 0.030 Ω; uniform arithmetic-mean area gives approximately 0.0266667 Ω.
Averaging area alone underweights the narrow section.
Equal material volume does not imply equal resistance; resistance depends on inverse area along the current path.

\begin{aligned}R_{\mathrm{stepped}}&=\rho\left(\frac{L_1}{A_1}+\frac{L_2}{A_2}\right)=0.030\,\Omega\\{}R_{\mathrm{uniform}}&=\rho\frac{2\,\mathrm m}{1.5\,\mathrm{mm^2}}\approx0.0266667\,\Omega\end{aligned}

[1]

Error analysis · Application

A wire has A=1 mm², ρ=2×10⁻⁸ Ω·m and length 3 m. Someone inserts A=10⁻³ m² and obtains 0.000060 Ω. Correct area and resistance and find the error factor.

Show hint

Square the millimetre conversion factor for area.

Learning goal: Convert squared area units correctly.

Show solution and calculation

Approach: One square millimetre is one millionth of a square metre.
Resistance is therefore 0.060 Ω.
The incorrect area is a thousand times too large, giving resistance a thousand times too small.
A length-prefix conversion cannot be reused unchanged for area.

\begin{aligned}1\,\mathrm{mm^2}&=(10^{-3}\,\mathrm m)^2=10^{-6}\,\mathrm{m^2}\\{}R&=\frac{(2\times10^{-8}\,\Omega\,\mathrm m)(3\,\mathrm m)}{10^{-6}\,\mathrm{m^2}}=0.060\,\Omega\end{aligned}

[1]

Line loss is one part of the calculation

Identifying a material alone does not specify ρ exactly. Contact drops, mechanical strain and temperature gradients may alter measured resistance. Real cable selection additionally depends on operating conditions and applicable protection requirements; this page provides a physical resistance model. Within its regime it distinguishes dissipation, relative geometry effects and temperature-dependent measurement signals. Electrical power describes ohmic dissipation but does not by itself determine an allowable conductor temperature.[1][4]

In brief

Material state and geometry jointly determine resistance. Check the supplied resistivity unit, the complete current loop and temperature. When length and cross-section both change, a simple one-factor comparison no longer applies.

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Refined question headings and added scientifically relevant paragraph links; retained existing derivations, examples, diagrams, exercises and sources.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible learning sequence, an additional model example, a quantitative diagram and a closing summary; retained existing examples, exercises and sources.

Cite this reference page

Quanta. Resistivity (Conductor Resistance). Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/spezifischer-widerstand

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Frequently asked questions about Resistivity (Conductor Resistance)

How do you calculate the resistance of a wire?+

For a homogeneous conductor with uniform cross-section and specified temperature conditions, resistance equals resistivity times length divided by conducting area. The area excludes insulation and is perpendicular to current. A circular section requires an unambiguous radius or diameter. The retained $\frac{50 m}{1.5} mm^{2}$ case with specified copper model value $\frac{0.0178 \Omega{}\cdot{}mm^{2}}{m}$ gives about 0.59 Ω for that single conductor. An equally long return conductor doubles loop resistance. Contact resistance and narrow sections are additional contributions. Temperature, material state and sometimes frequency limit the simple model; a material name alone does not give an exact constant resistance.

Why do thick cables have less resistance?+

A larger equivalent cross-section provides more parallel current paths. At unchanged length, temperature and composition, resistance therefore decreases inversely with conducting area. Doubling area halves resistance; doubling a round wire’s diameter instead quadruples area and reduces resistance to one quarter. Radius and diameter must not be confused. At fixed current, voltage drop and ohmic dissipation decrease with resistance. This geometry result assumes approximately uniform current density. High-frequency current redistribution and abrupt section changes can produce a different spatial current pattern, so the simple area dependence needs its stated regime.

What cable cross-section is needed for a long line?+

For a voltage-drop requirement alone, area follows from current, resistivity, total conductor length and permissible voltage drop. Total length includes outgoing and return paths. In the original model with 40 m, $\frac{0.018 \Omega{}\cdot{}mm^{2}}{m}$ , 1.5 A and at most 1.2 V drop, this condition requires at least $0.90 mm^{2}$ . This expressly is not a completed cable selection. Ampacity, heating, protection, installation, surroundings and applicable requirements are additional conditions. The assumed material value must also match operating state. The physical calculation provides one checkable bound; it cannot replace complete real protection and installation assessment.

What does resistivity tell you about a material?+

Resistivity characterises the local relationship between electric field and current density in the stated linear material model. It is not a wire-length property or mass density. Its SI unit is ohm metre; practical ohm square-millimetre per metre values use a scaling factor of one million. Purity, alloy composition, temperature and material state belong to a reliable value specification. A value such as 0.018 in these examples is expressly supplied and rounded. Material comparisons need compatible reference conditions. Resistance of the completed object then follows from this material value and conducting geometry.

Why does the resistance of metals rise with temperature?+

For many metals, stronger lattice vibrations impede charge transport at higher temperature. Near a stated reference, a positive temperature coefficient can approximate the increase linearly. The difference from reference temperature matters, not absolute temperature as a multiplier. With negligible geometry change, the same relative approximation applies to wire resistance. The original loop example uses 20 to 70 °C and the explicitly given coefficient 0.0039 per kelvin. Resistance rises from 0.864 Ω to 1.03248 Ω. Other materials or wider temperature regimes may behave differently; dissipation alone does not determine final temperature without a thermal balance.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 2, §9.3 Resistivity and Resistance

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Resistivity and wire resistance

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Symbol

Meaning

Unit

R

Resistance of the conductor

ρ

Resistivity (copper: 0.0178)

Ω·mm²/m

l

Length of the conductor

A

Cross-sectional area of the conductor

mm²

Derivation & background

Exam blueprint

Validity range

Derivation steps

The local field-current-density relation becomes component resistance through conductor geometry.

1Choose a homogeneous conductor with specified length, constant cross-section and specified temperature.
2The local ohmic relation is $E=\rho J$ , with $\rho$ describing the material state.
3The uniform one-dimensional model gives $U=E\ell$ and $I=JA$ .
4Substitution gives $U=\rho\ell\frac IA$ ; comparison with $U=RI$ yields $R=\rho\frac\ell A$ .
5For different sections, add their resistances. Slowly varying geometry gives approximately $R=\int\frac{\rho(x)}{A(x)}\,\mathrm dx$ .
6Check area units, outgoing and return conductors, and temperature; contact resistance and current crowding require additional models.

How do you calculate with Resistivity (Conductor Resistance)?

Here is how to work through a typical Resistivity (Conductor Resistance) (R = ρ·l/A) task step by step:

Task

Solution path

Approach: $A\ge\frac{\rho\ell}{R_{\max}}$ .
The values give at least $A=1.78\,\mathrm{mm^2}$ .
An example area of $2.5 mm^{2}$ gives about 0.1424 Ω in the stated model and meets this resistance limit.
It does not establish complete real cable sizing; material state, return path, contacts and further operating requirements need separate assessment.

Task

An iron wire (ρ = 0.10 Ω·mm²/m, A = 0.5 mm²) has $R = 4 \Omega{}$ . How long is it?

Solution path

$\ell=\frac{RA}{\rho}=\frac{4\,\Omega\cdot0.5\,\mathrm{mm^2}}{0.10\,\mathrm{\Omega\,mm^2/m}}=20\,\mathrm m$ .

Understand and apply the equation

What is resistivity and how do you calculate wire resistance?

\begin{aligned}R&=\rho\frac{\ell}{A}\\{}[\rho]&=\Omega\,\mathrm m\end{aligned}

Homogeneous straight conductor with constant cross-section, uniform temperature and approximately uniform current density.[1]

Derive from field and current density

\begin{aligned}U&=E\ell\\{}I&=JA\\{}\frac{U}{I}&=\frac{\rho J\ell}{JA}=\rho\frac{\ell}{A}\end{aligned}

Uniform E and J in the ohmic model; division is defined for a nonzero probing current.[1]

Explore further: Account for nonuniform conductors segment by segment

R=\int_0^\ell\frac{\rho(x)}{A(x)}\,\mathrm dx

One-dimensional approximation for the same continuous current path, excluding separate contact contributions.[1][3]

State temperature relative to a reference

\begin{aligned}&\rho(T)\approx\rho_0[1+\alpha(T-T_0)]\\{}&R(T)\approx R_0[1+\alpha(T-T_0)]\end{aligned}

The second approximation neglects geometry changes; α is the local temperature coefficient at the stated reference.[1]

Examples: outgoing and return conductors, varying area and stretching

A two-wire loop at elevated temperature

Given: Outgoing and return conductors each $18\,\mathrm m$ , $A=0.75\,\mathrm{mm^2}$ . Model values: $\rho_{20}=0.018\,\mathrm{\Omega\,mm^2/m}$ , $\alpha=0.0039\,\mathrm{K^{-1}}$ ; both conductors at $70\,{}^\circ\mathrm C$ , $I=2.4\,\mathrm A$ .
Find: Find loop resistance, voltage loss and dissipation.

Approach: Include both conductor lengths, then apply the supplied temperature coefficient.
$\begin{aligned}R_0&=\rho_0\frac{\ell_{\mathrm{loop}}}{A}\\{}R_T&=R_0[1+\alpha(T-T_0)]\end{aligned}$
Include both outgoing and return paths.
$\begin{aligned}\ell_{\mathrm{loop}}&=36\,\mathrm m\\{}R_{20}&=0.018\frac{36}{0.75}=0.864\,\Omega\end{aligned}$
Apply the specified uniform temperature.
$R_{70}=0.864(1+0.0039\cdot50)=1.03248\,\Omega$
Use the same loop current for losses.
$\begin{aligned}\Delta U&=IR=2.477952\,\mathrm V\\{}P_{\mathrm{loss}}&=I^2R=5.9470848\,\mathrm W\end{aligned}$

Result: The model gives about 1.03 Ω, 2.48 V drop and 5.95 W line loss.

Check: Each identical conductor contributes half; ΔU×I gives the same 5.9470848 W. This is neither a temperature prediction nor an installation approval.

Two different cross-sections in one path

Given: $\rho=0.020\,\mathrm{\Omega\,mm^2/m}$ ; first segment $2\,\mathrm m$ at $0.40\,\mathrm{mm^2}$ , second $1\,\mathrm m$ at $0.20\,\mathrm{mm^2}$ ; constant temperature and ideal contacts.
Find: Find total resistance and the error from averaging areas.

Approach: Successive segments are electrically in series and must be calculated individually.
$R=\rho\frac{\ell_1}{A_1}+\rho\frac{\ell_2}{A_2}$
Calculate each segment with its own A.
$\begin{aligned}R_1&=\frac{0.020\cdot2}{0.40}=0.10\,\Omega\\{}R_2&=\frac{0.020\cdot1}{0.20}=0.10\,\Omega\end{aligned}$
Add along the continuous current path.
$R=0.20\,\Omega$
Compare the length-weighted arithmetic area as an incorrect substitute.
$\begin{aligned}\overline A&=\frac{2\cdot0.40+1\cdot0.20}{3}=\frac{1}{3}\,\mathrm{mm}^2\\{}R_{\mathrm{wrong}}&=\frac{0.020\cdot3}{\frac{1}{3}}=0.18\,\Omega\end{aligned}$

Result: Correct series resistance is 0.20 Ω; arithmetic averaging alone underestimates it by 10%.

Check: Here the shorter segment with half the area contributes the same resistance as the longer one. Inverse area explains this.

Previous comparisons specified a fixed area. In the following original geometric model, volume and resistivity are instead fixed, forcing area to change with length.[1]

Longer and thinner at fixed volume

Given: A uniform model conductor is 1.0 m long with area 0.50 mm² and supplied resistivity 0.020 Ω·mm²/m. Its length increases to 1.5 m at constant volume. Resistivity and temperature remain unchanged in this model; no actual elastic-strain behaviour is asserted.
Find: Find the new area and resistance ratio.

Approach: volume conservation determines new area before applying the resistance relation.
$\begin{aligned}A_1\ell_1&=A_0\ell_0\\A_1&=\frac{A_0\ell_0}{\ell_1}\\\frac{R_1}{R_0}&=\left(\frac{\ell_1}{\ell_0}\right)^2\end{aligned}$
Include both geometric changes.
$\begin{aligned}A_1&=\frac{0.50}{1.5}\,\mathrm{mm^2}=\frac13\,\mathrm{mm^2}\\R_0&=0.020\frac{1.0}{0.50}\,\Omega=0.040\,\Omega\\R_1&=0.020\frac{1.5}{\frac13}\,\Omega=0.090\,\Omega\end{aligned}$
Independently calculate the factor from the length ratio.
$\frac{R_1}{R_0}=1.5^2=2.25$

Result: Resistance increases by a factor of 2.25. A factor of 1.5 would require unchanged cross-section.

View diagram: two simultaneous geometric changes

Fixed volume: resistance grows quadratically

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At fixed volume, area decreases inversely with length. The curve assumes constant temperature and resistivity; it is not a material strain measurement.[1]

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How do you convert Ω·m, Ω·mm²/m and wire diameter?

\begin{aligned}1\,\Omega\,\mathrm{mm}^2\,\mathrm m^{-1}&=10^{-6}\,\Omega\,\mathrm m\\{}A&=\frac{\pi d^2}{4}\\{}A&=\frac{\rho\ell}{R}\end{aligned}

A and ρ must use compatible area conventions; d is diameter, not radius.[1]

Five exercises with unit checks

Calculate · Foundation

A single model conductor with ρ=0.018 Ω·mm²/m is 4 m long with radius 0.30 mm. Find R.

Show hint

Use the radius directly in πr².

Learning goal: Calculate cross-section from radius.

Show solution and calculation

Approach: calculate circular area from radius before finding resistance.
A=π×0.30²=0.282743 mm²; R≈0.254648 Ω.
Diameter 0.60 mm gives the same area through $A=\frac{\pi d^2}{4}$ .

R=\frac{0.018\cdot4}{\pi(0.30)^2}\approx0.25465\,\Omega

Calculate · Application

In an ideal voltage-drop model: total conductor length 40 m, ρ=0.018 Ω·mm²/m, I=1.5 A, maximum drop 1.2 V. What area meets this condition alone?

Show hint

Rearrange $\Delta U=I\frac{\rho\ell}{A}$ for A.

Learning goal: Find area from a voltage-drop limit alone.

Show solution and calculation

Approach: total loop resistance must meet the specified voltage-drop limit.
A≥0.90 mm².
Check R=0.8 Ω and IR=1.2 V.
This is only the electrical model bound; ampacity, protection, installation and temperature remain undetermined.

A\ge\frac{1.5\cdot0.018\cdot40}{1.2}=0.90\,\mathrm{mm}^2

Calculate · Application

R₀=2.0 Ω at 20 °C, α=0.0040 K⁻¹. Find the model value at 45 °C.

Show hint

Use a 25 K difference, not absolute 318 K.

Learning goal: Apply linear temperature dependence.

Show solution and calculation

Approach: use temperature difference from the specified reference state.
R≈2.0(1+0.0040×25)=2.20 Ω.
Increase 10%; the reference expression does not predict behaviour at arbitrarily high temperatures.

R=2.0(1+0.0040\cdot25)=2.20\,\Omega

Interpret · Application

A single homogeneous model conductor has two 1 m sections: A1=1 mm² and A2=2 mm²; constant ρ=0.020 Ω·mm²/m. Compare its resistance with a 2 m conductor having arithmetic mean area 1.5 mm² everywhere.

Show hint

Sections are in series, not parallel.

Learning goal: Combine local resistances along a varying cross-section.

Show solution and calculation

Approach: Add local resistances.
The stepped conductor has 0.030 Ω; uniform arithmetic-mean area gives approximately 0.0266667 Ω.
Averaging area alone underweights the narrow section.
Equal material volume does not imply equal resistance; resistance depends on inverse area along the current path.

\begin{aligned}R_{\mathrm{stepped}}&=\rho\left(\frac{L_1}{A_1}+\frac{L_2}{A_2}\right)=0.030\,\Omega\\{}R_{\mathrm{uniform}}&=\rho\frac{2\,\mathrm m}{1.5\,\mathrm{mm^2}}\approx0.0266667\,\Omega\end{aligned}

[1]

Error analysis · Application

A wire has A=1 mm², ρ=2×10⁻⁸ Ω·m and length 3 m. Someone inserts A=10⁻³ m² and obtains 0.000060 Ω. Correct area and resistance and find the error factor.

Show hint

Square the millimetre conversion factor for area.

Learning goal: Convert squared area units correctly.

Show solution and calculation

Approach: One square millimetre is one millionth of a square metre.
Resistance is therefore 0.060 Ω.
The incorrect area is a thousand times too large, giving resistance a thousand times too small.
A length-prefix conversion cannot be reused unchanged for area.

\begin{aligned}1\,\mathrm{mm^2}&=(10^{-3}\,\mathrm m)^2=10^{-6}\,\mathrm{m^2}\\{}R&=\frac{(2\times10^{-8}\,\Omega\,\mathrm m)(3\,\mathrm m)}{10^{-6}\,\mathrm{m^2}}=0.060\,\Omega\end{aligned}

[1]

Line loss is one part of the calculation

In brief

Sources for this explanation

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October 2, 2026Refined question headings and added scientifically relevant paragraph links; retained existing derivations, examples, diagrams, exercises and sources.
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Frequently asked questions about Resistivity (Conductor Resistance)

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Resistivity and wire resistance

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Resistivity and wire resistance calculator

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Derivation & background

Exam blueprint

Validity range

Derivation steps

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Task variant

Common mistakes

Exam context

DC circuits

Worked example

Applications

How do you calculate with Resistivity (Conductor Resistance)?

What is resistivity and how do you calculate wire resistance?

Derive from field and current density

Explore further: Account for nonuniform conductors segment by segment

State temperature relative to a reference

Examples: outgoing and return conductors, varying area and stretching

A two-wire loop at elevated temperature

Two different cross-sections in one path

Longer and thinner at fixed volume

Fixed volume: resistance grows quadratically

How do you convert Ω·m, Ω·mm2/m and wire diameter?

Five exercises with unit checks

A single model conductor with ρ=0.018 Ω·mm2/m is 4 m long with radius 0.30 mm. Find R.

In an ideal voltage-drop model: total conductor length 40 m, ρ=0.018 Ω·mm2/m, I=1.5 A, maximum drop 1.2 V. What area meets this condition alone?

R0=2.0 Ω at 20 °C, α=0.0040 K−1. Find the model value at 45 °C.

A single homogeneous model conductor has two 1 m sections: A1=1 mm2 and A2=2 mm2; constant ρ=0.020 Ω·mm2/m. Compare its resistance with a 2 m conductor having arithmetic mean area 1.5 mm2 everywhere.

A wire has A=1 mm2, ρ=2×10−8 Ω·m and length 3 m. Someone inserts A=10−3 m2 and obtains 0.000060 Ω. Correct area and resistance and find the error factor.

Line loss is one part of the calculation

In brief

Sources for this explanation

Frequently asked questions about Resistivity (Conductor Resistance)

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Resistivity (Conductor Resistance) for exams

Resistivity and wire resistance

Variables & units

Resistivity and wire resistance calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

DC circuits

Worked example

Applications

How do you calculate with Resistivity (Conductor Resistance)?

What is resistivity and how do you calculate wire resistance?

Derive from field and current density

Explore further: Account for nonuniform conductors segment by segment

State temperature relative to a reference

Examples: outgoing and return conductors, varying area and stretching

A two-wire loop at elevated temperature

Two different cross-sections in one path

Longer and thinner at fixed volume

Fixed volume: resistance grows quadratically

How do you convert Ω·m, Ω·mm2/m and wire diameter?

Five exercises with unit checks

A single model conductor with ρ=0.018 Ω·mm2/m is 4 m long with radius 0.30 mm. Find R.

In an ideal voltage-drop model: total conductor length 40 m, ρ=0.018 Ω·mm2/m, I=1.5 A, maximum drop 1.2 V. What area meets this condition alone?

R0=2.0 Ω at 20 °C, α=0.0040 K−1. Find the model value at 45 °C.

A single homogeneous model conductor has two 1 m sections: A1=1 mm2 and A2=2 mm2; constant ρ=0.020 Ω·mm2/m. Compare its resistance with a 2 m conductor having arithmetic mean area 1.5 mm2 everywhere.

A wire has A=1 mm2, ρ=2×10−8 Ω·m and length 3 m. Someone inserts A=10−3 m2 and obtains 0.000060 Ω. Correct area and resistance and find the error factor.

Line loss is one part of the calculation

In brief

Sources for this explanation

Frequently asked questions about Resistivity (Conductor Resistance)

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Resistivity (Conductor Resistance) for exams

How do you convert Ω·m, Ω·mm²/m and wire diameter?

A single model conductor with ρ=0.018 Ω·mm²/m is 4 m long with radius 0.30 mm. Find R.

In an ideal voltage-drop model: total conductor length 40 m, ρ=0.018 Ω·mm²/m, I=1.5 A, maximum drop 1.2 V. What area meets this condition alone?

R₀=2.0 Ω at 20 °C, α=0.0040 K⁻¹. Find the model value at 45 °C.

A single homogeneous model conductor has two 1 m sections: A1=1 mm² and A2=2 mm²; constant ρ=0.020 Ω·mm²/m. Compare its resistance with a 2 m conductor having arithmetic mean area 1.5 mm² everywhere.

A wire has A=1 mm², ρ=2×10⁻⁸ Ω·m and length 3 m. Someone inserts A=10⁻³ m² and obtains 0.000060 Ω. Correct area and resistance and find the error factor.

How do you convert Ω·m, Ω·mm²/m and wire diameter?

A single model conductor with ρ=0.018 Ω·mm²/m is 4 m long with radius 0.30 mm. Find R.

In an ideal voltage-drop model: total conductor length 40 m, ρ=0.018 Ω·mm²/m, I=1.5 A, maximum drop 1.2 V. What area meets this condition alone?

R₀=2.0 Ω at 20 °C, α=0.0040 K⁻¹. Find the model value at 45 °C.

A single homogeneous model conductor has two 1 m sections: A1=1 mm² and A2=2 mm²; constant ρ=0.020 Ω·mm²/m. Compare its resistance with a 2 m conductor having arithmetic mean area 1.5 mm² everywhere.

A wire has A=1 mm², ρ=2×10⁻⁸ Ω·m and length 3 m. Someone inserts A=10⁻³ m² and obtains 0.000060 Ω. Correct area and resistance and find the error factor.