When sizing conductors for a subpanel feeder or a low-voltage solar array, guessing the voltage drop is a fast track to overheated insulation and tripped breakers. The foundation of voltage drop analysis is calculating the resistance of a wire based on its material, length, and cross-sectional area. While a multimeter can measure resistance directly on a bench, you cannot clip leads onto a 150-foot underground feeder to check it before pulling. You must calculate it mathematically to ensure your circuit complies with NEC-style voltage drop recommendations (typically under 3% for branch circuits and 5% overall).
This guide provides the exact formulas, material constant tables, and step-by-step worked examples you need to calculate wire resistance in both metric (mm²) and US customary (AWG/Circular Mils) systems.
The Core Formula and Symbol Definitions
The fundamental physics equation for DC resistance (and low-frequency AC resistance) of a uniform conductor is derived from the material's inherent opposition to current flow. In the US electrical trade, we frequently use a simplified variant that swaps metric area for Circular Mils (CM) to align with standard AWG wire tables.
Standard Metric Formula:
R = ρ × (L / A)
US Customary (AWG) Formula:
R = (K × L) / CM
| Symbol | Definition | Metric Unit | US Customary Unit |
|---|---|---|---|
| R | Total DC Resistance of the conductor | Ohms (Ω) | Ohms (Ω) |
| ρ (rho) | Electrical resistivity of the material | Ω·m (Ohm-meters) | N/A (Use K) |
| K | Specific resistance constant (approximate) | N/A | Ω·cmil/ft |
| L | One-way length of the conductor | Meters (m) | Feet (ft) |
| A | Cross-sectional area of the conductor | Square meters (m²) | N/A (Use CM) |
| CM | Cross-sectional area in Circular Mils | N/A | Circular Mils (cmil) |
Rearranged Forms
On the jobsite, you often know the maximum allowable resistance (based on your voltage drop limit) and need to find the maximum run length or the required wire size. Here are the algebraically rearranged forms:
- Solving for Length (Metric):
L = (R × A) / ρ - Solving for Area (Metric):
A = (ρ × L) / R - Solving for Resistivity:
ρ = (R × A) / L - Solving for Length (US):
L = (R × CM) / K - Solving for Wire Size (US):
CM = (K × L) / R
Material Resistivity and Temperature Constants
Resistivity is not a static number; it scales with temperature. A copper wire at 20°C (68°F) has significantly lower resistance than the same wire operating at 75°C (167°F) inside a hot attic or bundled in a conduit with other loaded conductors. According to engineering material databases and NEC Chapter 9, Table 8, the constants shift based on thermal load.
Below is the critical data table for common wiring materials. Bookmark these values for your voltage drop calculations.
| Material | ρ at 20°C (Ω·m) | K at 20°C (Ω·cmil/ft) | K at 75°C (Ω·cmil/ft) | Common Application |
|---|---|---|---|---|
| Copper (Annealed) | 1.724 × 10-8 | 10.4 | 12.9 | THHN, NM-B, Standard Branch Circuits |
| Aluminum (EC Grade) | 2.820 × 10-8 | 17.0 | 21.2 | Service Entrance, Large Feeders (XHHW) |
| Silver | 1.590 × 10-8 | 9.6 | 11.9 | High-end audio, specialized contacts |
| Gold | 2.440 × 10-8 | 14.7 | 18.2 | Corrosion-resistant PCB plating |
| Nichrome (80/20) | 1.100 × 10-6 | 660.0 | ~675.0 | Heating elements, dummy loads |
Note: The K-factor is derived by converting the metric resistivity into circular mils and feet. For standard US copper wiring calculations where temperature is unknown or assumed ambient, K=10.4 is the baseline. For loaded conductors operating near their ampacity limit, use K=12.9.
Worked Examples: Calculating the Resistance of a Wire
Let's apply the math to two real-world scenarios. Notice how intermediate steps and unit conversions are tracked explicitly—this is where most DIYers and students lose points (or burn up wires).
Problem 1: Metric Solar Array Run (Finding Resistance)
Scenario: You are wiring a 48V DC solar array to a charge controller. The one-way physical distance is 30 meters. You are using 6 mm² copper wire. The ambient temperature in the conduit is roughly 20°C. What is the total one-way resistance of the wire?
Step 1: Identify knowns and select formula.
- Material: Copper at 20°C → ρ = 1.724 × 10-8 Ω·m
- Length (L): 30 m
- Area (A): 6 mm². Crucial conversion: 1 mm² = 1 × 10-6 m². Therefore, A = 6 × 10-6 m².
- Formula:
R = ρ × (L / A)
Step 2: Substitute and solve.
- R = (1.724 × 10-8 Ω·m × 30 m) / (6 × 10-6 m²)
- R = (5.172 × 10-7 Ω·m²) / (6 × 10-6 m²)
- R = 0.0862 Ω (or 86.2 milliohms)
Problem 2: US AWG Subpanel Feeder (Finding Wire Size)
Scenario: You are running a 240V aluminum feeder to a detached garage subpanel. The one-way trench distance is 150 feet. The load is 40A continuous. You want to limit the one-way resistance to a maximum of 0.25 Ω to keep voltage drop manageable. The wire will be operating in a warm environment (assume 75°C). What minimum AWG size (in Circular Mils) do you need?
Step 1: Identify knowns and select rearranged formula.
- Material: Aluminum at 75°C → K = 21.2 Ω·cmil/ft
- Length (L): 150 ft
- Max Resistance (R): 0.25 Ω
- Formula:
CM = (K × L) / R
Step 2: Substitute and solve.
- CM = (21.2 Ω·cmil/ft × 150 ft) / 0.25 Ω
- CM = 3180 Ω·cmil / 0.25 Ω
- CM = 12,720 cmil
Step 3: Map to standard AWG.
Looking at standard conductor property tables, 8 AWG aluminum is 16,510 cmil, while 10 AWG is 10,380 cmil. Since 12,720 cmil exceeds the capacity of 10 AWG, you must step up to 8 AWG aluminum (or larger) to meet your resistance target. (Note: Always verify the final AWG against NEC ampacity tables for 40A continuous loads, which may force you to 6 AWG or 4 AWG depending on terminal temperature ratings).
Assumptions, Unit Traps, and Realistic Magnitudes
The formulas above are elegant, but they rely on strict physical assumptions. If you violate these assumptions, your math will lie to you.
When the Formula Applies (and When It Doesn't)
- DC and Low-Frequency AC: These formulas calculate DC resistance. For standard 50/60Hz AC mains power, the DC resistance is practically identical to the AC impedance for wire sizes up to roughly 2/0 AWG. Above that, or at higher frequencies (like VFD outputs or RF), skin effect and proximity effect force current to the outer edges of the conductor, effectively reducing the cross-sectional area (A) and increasing the real-world resistance.
- Uniform Temperature: The formula assumes the entire length of the wire is at the temperature used to select ρ or K. If half your run is in a freezing crawlspace and the other half is in a 120°F attic, your actual resistance will be a weighted average of the two.
- Homogeneous Material: It assumes pure copper or EC-grade aluminum. Copper-clad aluminum (CCA) wire, often found in cheap extension cords or fraudulent Amazon wiring, has a completely different resistivity profile and will yield higher resistance than pure copper calculations suggest.
Unit Mistakes That Break the Math
If your calculator spits out an answer that looks like 1.724e-14 or 45,000, you have fallen into a unit trap:
- The mm² to m² Trap: In the metric formula, Area (A) must be in square meters. Wire is sold in square millimeters. Forgetting to multiply your mm² value by 10-6 will result in a resistance calculation that is one million times too small.
- Circular Mils vs. Square Mils: In the US system, CM (Circular Mils) is calculated by squaring the diameter of the wire in mils (thousandths of an inch). It is not the same as square mils. 1 Circular Mil = 0.7854 Square Mils. Never plug a square mil area into the K-factor formula.
- One-Way vs. Round-Trip Length: The formulas calculate the resistance of a single conductor. If you are calculating voltage drop for a complete circuit, you must either double the length (L) before plugging it into the formula, or double the final resistance result. Forgetting this is the #1 reason DIY solar builds suffer from massive voltage sag.
What a Realistic Answer Magnitude Looks Like
When calculating the resistance of a wire for home electrical or low-voltage projects, your answers should almost always be in the milliohm (mΩ) to low single-digit ohm range.
A standard 100-foot coil of 12 AWG solid copper wire has a calculated resistance of roughly 0.193 Ω at 20°C. If you put your digital multimeter (DMM) across this coil and it reads 4.5 Ω, your math isn't wrong—your wire is broken, or your probe contacts are dirty. Conversely, if you are testing a 2-foot jumper wire and your DMT reads 0.0 Ω, that is normal; standard multimeters lack the resolution to read sub-10 milliohm values. You need a dedicated milliohm meter or a 4-wire Kelvin measurement setup to verify short jumper resistances.
By anchoring your calculations to the correct temperature constants and rigorously tracking your unit conversions, you can confidently size conductors, predict voltage drop, and ensure your wiring projects operate safely and efficiently.
