R = ρ·L/A

Wire resistance with voltage drop and power loss

R = ρ·L/A; given a known current, the same resistance also yields voltage drop and ohmic power loss for cable sizing.

MINTSI
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Inputs

Ohmic resistance of the conductor at the reference temperature of ρ.

Material property; copper about 0.0175 Ω·mm²/m at 20 °C, aluminum about 0.0282 Ω·mm²/m at 20 °C.

Electrical length; add both go and return conductors where relevant.

Metal cross-section normal to current flow.

Current used to compute voltage drop and power loss.

Ohmic voltage drop along the conductor at current I.

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Result

Select a target and calculate.

Calculation

R = ρ · L / A; Vdrop = I · R

R = ρ·L/A; given a known current, the same resistance also yields voltage drop and ohmic power loss for cable sizing.

Understand the inputs
  • Conductor resistance ROhmic resistance of the conductor at the reference temperature of ρ.
  • Resistivity ρMaterial property; copper about 0.0175 Ω·mm²/m at 20 °C, aluminum about 0.0282 Ω·mm²/m at 20 °C.
  • Conductor length LElectrical length; add both go and return conductors where relevant.
  • Conductor area AMetal cross-section normal to current flow.
  • Operating current ICurrent used to compute voltage drop and power loss.
  • Voltage drop VdropOhmic voltage drop along the conductor at current I.
Example

20 m of copper cable (ρ ≈ 0.0175 Ω·mm²/m) with a 1.5 mm² cross-section gives R ≈ 0.233 Ω; at 10 A this is a voltage drop of about 2.33 V (P = I²·R ≈ 23.3 W).

Assumptions and limits

DC, homogeneous cross-section and constant temperature; contact resistance and high-frequency skin effect are excluded.

Technical article

Understand Wire resistance with voltage drop and power loss

This calculator finds a conductor's DC resistance from resistivity, length and cross-section, and derives voltage drop and ohmic power loss at a known current — the key figures for cable sizing.

What does this quantity describe?

The resistance of a homogeneous conductor is proportional to its length L and the material's resistivity ρ, and inversely proportional to its cross-section A: R = ρ·L/A. At a known operating current I this gives the voltage drop Vdrop = I·R and the power dissipated as heat P = I²·R.

Picture the conductor as a narrow pipe carrying a viscous fluid: a long, thin pipe resists flow more than a short, thick one; the conductor material corresponds to how rough the pipe wall is.

Formula and variables

R = ρ · L / A; Vdrop = I · R

  • R = ρ·L/A
  • Vdrop = I·R
  • P = I²·R
Symbol / inputMeaning
Conductor resistance ROhmic resistance of the conductor at the reference temperature of ρ.
Resistivity ρMaterial property; copper about 0.0175 Ω·mm²/m at 20 °C, aluminum about 0.0282 Ω·mm²/m at 20 °C.
Conductor length LElectrical length; add both go and return conductors where relevant.
Conductor area AMetal cross-section normal to current flow.
Operating current ICurrent used to compute voltage drop and power loss.
Voltage drop VdropOhmic voltage drop along the conductor at current I.

Choose the inputs correctly

Resistivity ρ, length L and area A give R; adding operating current I gives the voltage drop. ρ is a material property at a reference temperature, usually 20 °C; for other operating temperatures, the linked temperature-dependence calculator gives the correction.

How to use the calculator

Select the target quantity. For resistance, enter ρ, L and A. For cable sizing, choose area A as the target and work back from the allowable voltage drop via R = Vdrop/I, or use voltage drop Vdrop directly as the target to check the effect of an existing cross-section.

Worked example

20 m of copper cable (ρ ≈ 0.0175 Ω·mm²/m at 20 °C) with a 1.5 mm² cross-section gives R = 0.0175 · 20 / 1.5 ≈ 0.233 Ω. At 10 A this is a voltage drop of 2.33 V and a power loss of P = I²·R ≈ 23.3 W.

Understand the result and units

A voltage drop of 2.33 V on a 230 V load is about 1% of the rated voltage, within typical tolerance limits for final circuits; the same 2.33 V on a 12 V low-voltage system would instead be nearly 20% and generally unacceptable.

Resistivity is often given in Ω·mm²/m, conveniently matching cable cross-sections in mm², less commonly in Ω·m; 1 Ω·mm²/m equals 10⁻⁶ Ω·m. Length is given in m, area in mm² and resistance in Ω.

How this differs from the plain conductor resistance calculator

This calculator extends the bare R = ρ·L/A formula with operating current as an input, to directly derive voltage drop and power loss — the practical figures that matter for cable sizing. If only the bare resistance without a current input is needed, the simpler conductor resistance calculator is sufficient.

Typical applications

The relationship is used to select cable cross-sections for long runs, to estimate transmission losses in wiring, and to size windings in coils and transformers, where winding resistance sets ohmic losses.

Assumptions, limits and common mistakes

The model applies to DC at a constant temperature uniform across the conductor cross-section and excludes contact and transition resistance at terminals and connectors. At high-frequency AC, the skin effect pushes current toward the conductor's outer surface, so effective AC resistance exceeds the DC value computed here; at 50/60 Hz this effect is negligible for common conductor sizes.

Common mistake: A common mistake is entering only the one-way run length for a circuit with a return conductor instead of double that length, underestimating voltage drop and power loss by a factor of two. Resistivity is also sometimes used without regard to actual operating temperature, even though ρ for metals rises noticeably with temperature.

Frequently asked questions

What does resistivity mean intuitively?

It is the resistance of a 1 m length with a 1 mm² cross-section of the given material; copper, at about 0.0175 Ω·mm²/m, is markedly lower than aluminum (about 0.0282 Ω·mm²/m) or steel.

Why do long, thin conductors have high resistance?

Because resistance is proportional to length and inversely proportional to cross-section — a long path with little available cross-section area impedes current flow more.

How does this differ from the plain conductor resistance calculator?

This calculator uses the same core R = ρ·L/A but adds operating current, voltage drop and power loss — relevant when checking whether a cable cross-section suffices for an application.

Does the formula hold at high-frequency AC?

Only partly: at high frequencies the skin effect concentrates current near the conductor's outer surface, so effective resistance exceeds the DC value computed here.

How do I size a cross-section for a maximum allowable voltage drop?

Compute the allowable resistance from R = Vdrop,max/I, set that as the target value for R and solve for A; or choose A directly as the target and adjust it until the shown voltage drop falls within the allowable range.