The resistance of a wire dictates voltage drop, heat generation, and fault current magnitude in any electrical circuit. Whether you are sizing a feeder for a subpanel or calculating the trace width on a custom PCB, the underlying physics remains identical. This guide breaks down the resistance of wire equation, provides real-world material data, and walks through exact jobsite calculations with full unit tracking.

The Core Resistance of Wire Equation and Symbol Definitions

In electrical theory, the DC resistance of a uniform conductor is directly proportional to its length and inversely proportional to its cross-sectional area. The fundamental metric formula is:

R = ρ × (L / A)

For US-based electricians working with American Wire Gauge (AWG) and circular mils, the practical equivalent used in NEC voltage drop calculations is:

R = (K × L) / Acmil

Symbol Definitions and Standard Units
Symbol Definition Metric Unit US / Imperial Unit
R Total electrical resistance of the conductor Ohms (Ω) Ohms (Ω)
ρ (rho) Resistivity of the material (intrinsic property) Ohm-meters (Ω·m) Ohm-circular mils per foot
K Specific resistance constant (equivalent to ρ in US units) N/A Ω·cmil / ft
L Length of the conductor Meters (m) Feet (ft)
A Cross-sectional area of the conductor Square meters (m²) or mm² Circular mils (cmil)

Real-World Resistivity Data for Common Conductors

A massive trap for DIYers and junior engineers is using textbook resistivity values measured at 20°C (68°F) for wires that will actually operate at 75°C (167°F) inside a conduit. Resistivity increases with temperature. According to NFPA 70 (NEC) Chapter 9, Table 8, the K-factor for copper shifts from roughly 10.4 at 20°C to 12.9 at 75°C. Always use the 75°C K-factor for standard THHN/THWN branch circuit and feeder voltage drop calculations.

Conductor Resistivity and K-Factors at Operating Temperatures
Material ρ at 20°C (Ω·m × 10-8) K-Factor at 20°C K-Factor at 75°C (NEC Practice) Common Application
Copper (Annealed) 1.724 10.4 12.9 NM-B, THHN, branch circuits
Aluminum (EC Grade) 2.820 17.0 21.2 Service entrance feeders, SER cable
Silver 1.590 9.6 ~11.8 High-end audio contacts, RF plating
Gold 2.440 14.7 ~18.1 Corrosion-resistant PCB edge connectors
Iron 10.00 60.2 N/A Structural grounding rods (not conductors)

For deeper physics on how lattice vibrations scatter electrons to create this resistance, the Georgia State University HyperPhysics database provides excellent foundational models.

Rearranged Forms and Practical Unit Traps

On the bench or jobsite, you rarely solve for R directly. You usually know the acceptable resistance (or voltage drop) and need to find the maximum length or the required wire gauge. Here are the algebraically rearranged forms of the metric equation:

  • Solve for Length (L): L = (R × A) / ρ (Use to find maximum run length before exceeding a voltage drop threshold).
  • Solve for Area (A): A = (ρ × L) / R (Use to size a wire for a specific load and distance).
  • Solve for Resistivity (ρ): ρ = (R × A) / L (Use to identify an unknown alloy or verify material purity).

Unit Mistakes That Break the Math

The most common reason a calculated wire resistance yields a wildly incorrect answer is a unit mismatch. Watch out for these specific traps:

Trap 1: Mixing mm² with Circular Mils. A 10 AWG wire has an area of 10,380 cmil, but roughly 5.26 mm². If you plug 10,380 into the metric formula as if it were square meters or mm², your resistance will be off by a factor of thousands.
Trap 2: One-Way Length vs. Loop Length. The formula calculates the resistance of a single physical wire. For a 120V circuit, current travels out on the hot wire and returns on the neutral. If your load is 100 feet away, the electrical loop length is 200 feet. You must multiply L by 2 when calculating total circuit voltage drop.
Trap 3: Diameter vs. Area. The denominator is Area, not diameter. If you measure a wire with calipers and get 2mm, the area is π × r² (3.14 mm²), not 2. Furthermore, in the US system, Area in cmil is simply the diameter in mils (thousandths of an inch) squared.

Worked Examples: From Theory to the Jobsite

Let’s apply the math to two distinct scenarios, tracking every unit to ensure the dimensional analysis holds up.

Example 1: Metric Calculation for a Solar DC Run

Scenario: You are wiring a 24V DC solar array to a charge controller using a single run of 4 mm² copper cable. The physical distance is 15 meters. What is the resistance of the positive wire?

  1. Identify Knowns:
    ρ (copper at 20°C) = 1.724 × 10-8 Ω·m
    L = 15 m
    A = 4 mm² = 4 × 10-6 m² (Converting mm² to base meters squared)
  2. Setup Equation:
    R = ρ × (L / A)
  3. Substitute Values:
    R = (1.724 × 10-8 Ω·m × 15 m) / (4 × 10-6 m²)
  4. Calculate Numerator:
    1.724 × 10-8 × 15 = 2.586 × 10-7 Ω·m²
  5. Divide by Denominator:
    R = (2.586 × 10-7) / (4 × 10-6)
    R = 0.06465 Ω

Result: The single wire has a resistance of roughly 64.6 milliohms. For the full DC loop (out and back), the total resistance would be 0.129 Ω.

Example 2: US Imperial Calculation for a 240V Feeder

Scenario: You are running a 240V subpanel feeder using 2 AWG aluminum SER cable. The one-way physical distance is 120 feet. The terminations are rated for 75°C. Find the total loop resistance and the voltage drop at a 70A load.

  1. Identify Knowns:
    K (aluminum at 75°C) = 21.2 Ω·cmil/ft
    Lloop = 120 ft × 2 = 240 ft
    Acmil for 2 AWG = 66,360 cmil (from NEC Chapter 9, Table 8)
  2. Setup Equation:
    R = (K × Lloop) / Acmil
  3. Substitute Values:
    R = (21.2 × 240) / 66,360
  4. Calculate Numerator:
    21.2 × 240 = 5,088
  5. Divide by Denominator:
    R = 5,088 / 66,360 = 0.07667 Ω
  6. Calculate Voltage Drop (Vdrop = I × R):
    Vdrop = 70A × 0.07667 Ω = 5.36V

Result: The total loop resistance is 0.076 Ω. The voltage drop is 5.36V, which is 2.2% of 240V. This is well under the NEC recommended 3% maximum for feeders, confirming 2 AWG aluminum is correctly sized.

Assumptions, Limitations, and Realistic Magnitudes

The resistance of wire equation is a linear DC model. To use it correctly in AC home wiring, you must understand its boundaries.

When the Formula Applies (and When It Doesn’t)

  • Uniform Cross-Section: The formula assumes the wire is a perfect cylinder. It fails if the wire is stranded and you mistakenly use the outer jacket diameter instead of the sum of the strand areas, or if the wire is kinked and stretched.
  • DC vs. AC (Skin Effect): At 60 Hz AC, current pushes slightly toward the outer edge of the conductor (skin effect). For wire sizes smaller than 1/0 AWG, this effect is negligible (less than 1% error) and the DC formula holds true. For massive conductors like 500 kcmil or busbars, AC resistance (Rac) is higher than DC resistance, and you must apply a skin-effect multiplier from All About Circuits or IEEE standards.
  • Steady-State Temperature: Resistivity changes by roughly 0.4% per degree Celsius for copper. If a wire is heavily loaded in a hot attic (ambient 45°C + conductor heating), the actual operating resistance will be 15-20% higher than the 20°C textbook value.

What a Realistic Answer Magnitude Looks Like

Developing an intuitive sense for the answer prevents catastrophic decimal errors. In standard residential wiring (14 AWG to 4/0 AWG), wire resistance is measured in milliohms per foot or fractions of an ohm for whole runs.

  • 14 AWG Copper: ~3.14 Ω per 1,000 ft (or 0.00314 Ω/ft).
  • 10 AWG Copper: ~1.24 Ω per 1,000 ft.
  • 4/0 AWG Aluminum: ~0.25 Ω per 1,000 ft.

The Sanity Check: If you calculate the resistance of a 100-foot run of 12 AWG copper wire and your math yields 45 Ω, you have made a unit error (likely failing to convert mm² to m², or dropping a zero in the circular mil area). A 100-foot run of 12 AWG should yield approximately 0.2 Ω. Always compare your final calculated number against these baseline magnitudes before cutting wire or energizing a panel.