The direct answer for calculating the DC resistance of a uniform wire is the resistance of a conductor formula: R = ρ(L/A). This equation dictates that resistance (R) increases linearly with the length (L) of the wire and the material's inherent resistivity (ρ), while decreasing inversely with the cross-sectional area (A). Whether you are sizing a 50-foot THHN feeder for a subpanel or winding a custom nichrome heating element, this formula is the baseline for predicting voltage drop and power loss.
The Core Equation and Symbol Definitions
At the bench or on the jobsite, you will see this formula written in a few different notations, but the fundamental physics remain identical. The standard SI formulation is:
R = ρ × (L / A)
Below is the definitive spec-sheet breakdown of every variable, including the strict SI units required to make the math work without throwing your decimal place off by a factor of a million.
| Symbol | Quantity | Standard SI Unit | Practical Jobsite Unit |
|---|---|---|---|
| R | Resistance | Ohms (Ω) | Milliohms (mΩ) |
| ρ (rho) | Electrical Resistivity | Ohm-meters (Ω·m) | Ω·cmil/ft (US) |
| L | Length of Conductor | Meters (m) | Feet (ft) |
| A | Cross-Sectional Area | Square meters (m²) | Circular mils (cmil) or mm² |
In North America, the NEC Chapter 9 Table 8 lists wire areas in circular mils (cmil). If you use the resistivity of annealed copper at 20°C in imperial units (approximately 10.37 Ω·cmil/ft), the formula becomes R = 10.37 × (L / A), where L is in feet and A is in cmil. This bypasses metric conversions entirely and is how voltage drop calculators are programmed in US electrical software.
Rearranged Forms for Any Unknown
You will rarely just solve for R. Often, you know the maximum allowable resistance (based on a 3% voltage drop limit) and need to find the required wire gauge (Area), or you are trying to identify an unknown alloy by measuring its resistivity. Here are the algebraic rearrangements solving for each variable:
- Solving for Resistivity (Material ID):
ρ = (R × A) / L - Solving for Length (Distance limit):
L = (R × A) / ρ - Solving for Area (Wire Sizing):
A = (ρ × L) / R
When sizing wire for a branch circuit, you typically rearrange for A. You plug in the resistivity of copper, the one-way length of the run, and the maximum resistance permitted by your voltage drop budget. The resulting Area tells you the minimum cross-section required, which you then round up to the next standard AWG size.
Assumptions, Limits, and Unit Traps
When the Formula Applies (and When It Doesn't)
The resistance of a conductor formula relies on three strict physical assumptions:
- Uniform Cross-Section: The wire must be a consistent cylinder. If the wire is tapered, kinked, or stranded with varying lay lengths, the simple A variable fails (requiring integral calculus for tapered shapes).
- Constant Temperature: Resistivity (ρ) is highly temperature-dependent. The standard ρ value for copper (1.68 × 10-8 Ω·m) is only valid at 20°C (68°F). If your wire is running hot inside a conduit in an attic, its actual resistance will be higher.
- DC or Low-Frequency AC: This formula calculates DC resistance. For AC power at 60Hz in wires larger than 1/0 AWG, skin effect forces current to the outer edge of the conductor, effectively reducing the usable Area (A) and increasing the AC resistance above the DC calculation.
The Unit Conversion Trap
The single most common reason students and hobbyists get wildly incorrect answers is failing to convert square millimeters (mm²) to square meters (m²). Standard SI unit rules dictate that if length is in meters, area must be in square meters.
Because 1 meter = 1,000 millimeters, 1 square meter = 1,000,000 square millimeters. Therefore, to convert mm² to m², you must multiply by 10-6. If you forget this and plug '2.5' directly into the area slot for a 2.5 mm² wire, your calculated resistance will be one million times too small.
Sanity Checking: Realistic Magnitudes
A thick copper wire has incredibly low resistance. If you calculate the resistance of a 10-foot run of 12 AWG copper and get 15 Ω, you made a math error. A realistic magnitude for standard building wire is in the milliohm (mΩ) range per foot. For example, 1,000 feet of solid 10 AWG copper at 75°C is roughly 1.2 Ω. If your final answer is in the hundreds or thousands of ohms for a standard copper feeder, check your decimal placement.
Worked Examples with Unit Tracking
Let's walk through two distinct scenarios. We will track units through every step to prove how they cancel out, leaving only Ohms (Ω).
Problem 1: Calculating Voltage Drop on a Metric Copper Feeder
Scenario: You are wiring a solar array using 50 meters of 4.0 mm² stranded copper wire. What is the total resistance of this single conductor at 20°C?
Knowns:
- L = 50 m
- A = 4.0 mm² = 4.0 × 10-6 m²
- ρ (copper at 20°C) = 1.68 × 10-8 Ω·m (Source: Georgia State University HyperPhysics)
Step-by-Step Solution:
- Write the formula:
R = ρ × (L / A) - Substitute values with units:
R = (1.68 × 10-8 Ω·m) × [50 m / (4.0 × 10-6 m²)] - Cancel the length units (m / m² leaves 1/m, which cancels the Ω·m to leave Ω):
R = (1.68 × 10-8 × 50) / (4.0 × 10-6) Ω - Calculate the numerator:
8.4 × 10-7 - Divide by the denominator:
(8.4 × 10-7) / (4.0 × 10-6) - Final Answer:
R = 0.21 Ω(or 210 mΩ)
Problem 2: Sizing a Nichrome Heating Element
Scenario: You are building a DIY reflow hotplate and need a heating element with exactly 12 Ω of resistance. You have a spool of 18 AWG Nichrome 80 wire. How many meters of wire must you cut?
Knowns:
- R = 12 Ω
- 18 AWG Area = 0.823 mm² = 0.823 × 10-6 m²
- ρ (Nichrome 80) ≈ 1.09 × 10-6 Ω·m (Note: Nichrome is roughly 65 times more resistive than copper)
Step-by-Step Solution:
- Rearrange formula to solve for Length:
L = (R × A) / ρ - Substitute values:
L = (12 Ω × 0.823 × 10-6 m²) / (1.09 × 10-6 Ω·m) - Notice how the 10-6 multipliers in the numerator and denominator cancel each other out perfectly, simplifying the math.
- Calculate:
L = (12 × 0.823) / 1.09 - Final Answer:
L = 9.06 meters
Frequently Asked Questions
How does temperature change the resistance of a conductor formula?
The base formula assumes a constant temperature (usually 20°C). To account for heat, you must apply the temperature coefficient of resistance (α). The adjusted formula becomes R = R20[1 + α(T - 20)]. For copper, α is approximately 0.00393 per °C. This means for every degree Celsius the wire heats up above 20°C, its resistance increases by roughly 0.4%. In a heavily loaded conduit where ambient temperatures hit 50°C, a copper feeder's resistance will be nearly 12% higher than the baseline formula predicts, directly increasing your voltage drop.
Why does the resistance of a conductor formula fail at high frequencies?
At DC or standard 50/60Hz mains frequencies in small wires, current flows uniformly through the entire cross-section (Area). However, as frequency increases (such as in RF circuits, variable frequency drives, or high-speed data lines), alternating magnetic fields induce eddy currents that push the electron flow toward the outer skin of the conductor. This is called the skin effect. Because the center of the wire carries almost no current, the effective cross-sectional area (A) shrinks. Since A is in the denominator of the formula, a smaller effective Area results in a much higher AC resistance than the DC formula calculates.
What is the resistance of a conductor formula for a non-uniform wire?
If a conductor tapers (like a wedge or a conical shape) or varies in thickness along its length, you cannot use a single value for A. Instead, you must treat the wire as an infinite series of infinitesimally thin slices and use integral calculus. The formula becomes R = ∫ [ρ / A(x)] dx, integrated from 0 to L, where A(x) is the mathematical function describing how the cross-sectional area changes at any given point x along the length. For standard solid or stranded electrical wiring, however, the cross-section is uniform, and the basic algebraic formula is perfectly accurate.
