If you need the direct answer: the resistance of a wire equation is R = ρ(L/A). Resistance (R) equals the material's resistivity (ρ) multiplied by the wire's length (L), divided by its cross-sectional area (A). On the bench or the jobsite, this single formula dictates whether your 240V well pump will run smoothly or overheat and burn out its windings due to severe voltage drop.

Abstract theory won't keep a breaker from tripping. To use this formula effectively, you need to track your units ruthlessly, understand the difference between 20°C laboratory assumptions and 75°C attic realities, and know how to rearrange the math when you're solving for wire gauge instead of voltage drop. Below is the complete derivation, symbol mapping, and step-by-step worked examples for both metric and imperial (AWG) systems.

The Core Equation, Symbols, and Assumptions

The fundamental physics of wire resistance relies on the resistivity model defined by standard physics references. Before plugging in numbers, you must understand what each symbol represents and the physical assumptions baked into the math.

Symbol Variable Name Metric Unit (SI) Imperial Unit (US) Practical Context
R Resistance Ohms (Ω) Ohms (Ω) The opposition to current flow. Causes voltage drop and heat.
ρ (rho) Resistivity Ω·m Ω·cmil/ft Copper at 20°C: 1.68×10⁻⁸ Ω·m or 10.37 Ω·cmil/ft.
L Length Meters (m) Feet (ft) The one-way physical length of the conductor.
A Cross-Sectional Area Square meters (m²) Circular mils (cmil) The slice of the wire. Larger area = lower resistance.
When the Formula Applies (and its Assumptions):
  • Temperature: Standard resistivity values (like 10.37 for copper) assume a conductor temperature of 20°C (68°F). In a hot attic under full load, copper resistivity increases by roughly 20%. Always use temperature-corrected ρ for precise voltage drop calculations on long runs.
  • Current Type: This formula applies perfectly to DC and low-frequency (50/60Hz) AC. At high frequencies, the skin effect forces current to the outer edge of the wire, effectively reducing 'A' and increasing 'R' beyond what this equation predicts.
  • Material Homogeneity: It assumes a solid, uniform conductor. Stranded wire has a slightly larger overall diameter for the same AWG due to air gaps between strands, but the effective conductive area remains the same as solid wire of the same AWG.

Rearranged Forms for Field Calculations

You rarely solve for R in isolation. Usually, you know your acceptable resistance (derived from a maximum 3% voltage drop limit) and need to find the required wire size, or you are troubleshooting an unknown cable length. Here are the algebraic rearrangements:

  • Solve for Area (Sizing Wire): A = ρ(L / R)
  • Solve for Length (Finding Distance): L = R(A / ρ)
  • Solve for Resistivity (Identifying Material): ρ = R(A / L)

Worked Problem 1: Metric Solar Array DC Run

Scenario: You are wiring a 48V DC solar array to a charge controller. You have a spool of solid copper wire with a 2.0 mm diameter. The one-way run from the panels to the controller is 50 meters. What is the resistance of this single conductor?

Step 1: Identify and convert knowns to base SI units.

  • ρ (Copper) = 1.68 × 10⁻⁸ Ω·m
  • L = 50 m
  • Diameter (d) = 2.0 mm = 0.002 m. Therefore, radius (r) = 0.001 m.

Step 2: Calculate Cross-Sectional Area (A).

The formula for the area of a circle is A = π × r². A common trap is using the diameter instead of the radius, or forgetting to square the unit conversion.

  • A = π × (0.001 m)²
  • A = 3.14159 × 0.000001 m²
  • A = 3.14159 × 10⁻⁶ m²

Step 3: Apply the resistance of a wire equation with unit tracking.

  • R = ρ × (L / A)
  • R = (1.68 × 10⁻⁸ Ω·m × 50 m) / (3.14159 × 10⁻⁶ m²)
  • R = (8.4 × 10⁻⁷ Ω·m²) / (3.14159 × 10⁻⁶ m²)
  • Notice how the m² units cancel out, leaving only Ohms (Ω).
  • R = 0.267 Ω

Worked Problem 2: Imperial 12 AWG THHN Branch Circuit

Scenario: You are pulling 12 AWG THHN copper wire through conduit for a 120V receptacle circuit. The one-way distance from the subpanel to the outlet is 200 feet. What is the resistance of the hot conductor?

In the US, we rely on standardized wire gauge tables rather than measuring physical diameters with calipers. According to NEC Chapter 9, Table 8, the cross-sectional area of 12 AWG solid copper is 6,530 circular mils (cmil).

Step 1: Identify knowns in Imperial units.

  • ρ (Copper) = 10.37 Ω·cmil/ft (at 20°C)
  • L = 200 ft
  • A = 6,530 cmil

Step 2: Apply the equation.

  • R = ρ × (L / A)
  • R = 10.37 Ω·cmil/ft × (200 ft / 6,530 cmil)
  • The 'ft' and 'cmil' units cancel out, leaving Ohms.
  • R = 2,074 / 6,530
  • R = 0.317 Ω

Bench Note: For a complete single-phase AC circuit, current must travel to the load and return. The total loop resistance for voltage drop calculations would be double this value (0.634 Ω), assuming the neutral wire is the same gauge and length.

Real-World Scenario: The 240V Well Pump Failure

Equations don't just live on paper; they dictate equipment lifespan. Here is a teardown of a real-world jobsite failure where ignoring the resistance equation destroyed a $1,200 submersible well pump.

The Setup:
A homeowner needed to run a 240V, 30-amp deep well pump. The trench from the main panel to the wellhead was 250 feet long. The electrician sized the wire based purely on ampacity (the ability of the wire to handle heat without melting the insulation). According to NEC Table 310.16, 10 AWG THHN copper is rated for 30A at 60°C. They pulled two 10 AWG conductors (plus a ground) and energized the system.

The Numbers:
Let's run the resistance of a wire equation for this setup to see the hidden problem.

  • ρ = 10.37 Ω·cmil/ft (Using 20°C baseline; actual operating temp in the trench would make this worse).
  • L = 250 ft.
  • A = 10,380 cmil (NEC Chapter 9, Table 8 for 10 AWG).
  • R (one-way) = (10.37 × 250) / 10,380 = 0.249 Ω.
  • R (total loop) = 0.249 Ω × 2 = 0.498 Ω.

The Outcome:
Using Ohm's Law (V = I × R), the voltage drop across the wire loop at full 30A load is:
V_drop = 30A × 0.498 Ω = 14.94 Volts.
The pump only received 225V instead of 240V. This represents a 6.2% voltage drop (NEC recommends a maximum of 3% for branch circuits, 5% total).

What Went Wrong:
Induction motors are constant-power devices. When voltage drops, the motor draws more current to produce the same mechanical horsepower. The pump drew closer to 34A, which further increased the I²R heating in the 10 AWG wire and the motor windings. Within three months, the pump's internal thermal overload tripped repeatedly, the insulation on the motor windings degraded from sustained heat, and the pump seized. The fix required pulling new 8 AWG wire (Area = 16,510 cmil), dropping the loop resistance to 0.31 Ω and the voltage drop to an acceptable 3.8% under actual operating load.

Unit Traps and Realistic Answer Magnitudes

When your multimeter reads a value that doesn't match your math, you've likely fallen into one of these unit traps:

  1. The Millimeter Squared Trap: 1 mm² is NOT 0.001 m². Because area is a square dimension, 1 mm² = (10⁻³ m)² = 10⁻⁶ m². Forgetting to square the conversion factor will throw your metric calculation off by a factor of 1,000.
  2. The Diameter vs. Radius Trap: The area formula requires radius (A = πr²). If you are given a wire diameter of 4mm, you must divide by 2 to get a 2mm radius before squaring.
  3. The Circular Mil Confusion: A circular mil is the area of a circle with a diameter of one mil (1/1000th of an inch). It is calculated as A = d² (where d is in mils). Do not mix standard square inches with circular mils; they are entirely different unit systems.
  4. Temperature Blindness: Using the 20°C resistivity constant (10.37) for a wire running at 75°C inside a hot conduit will understate your resistance by roughly 20%. For precise voltage drop on long feeders, use the 75°C resistivity constant for copper: 12.9 Ω·cmil/ft.

What Does a Realistic Answer Look Like?
If your math yields 50 Ω for a home wiring run, you've made a decimal error. Here is a baseline for realistic magnitudes:

  • Short Bench Jumper (2 ft, 12 AWG): ~0.003 Ω (Barely measurable on a standard multimeter; requires a milliohm meter or Kelvin clamp to verify accurately).
  • Standard Branch Circuit (100 ft loop, 14 AWG): ~0.25 Ω.
  • Long Feeder (400 ft loop, 2/0 AWG Aluminum): ~0.08 Ω (Aluminum has a higher ρ than copper, roughly 17.0 Ω·cmil/ft, requiring a larger 'A' to compensate).
  • High Voltage Transmission Line: Milliohms per mile, achieved by bundling massive conductors to maximize 'A' and minimize 'R'.

Mastering the resistance of a wire equation means moving beyond memorizing R = ρ(L/A). It means internalizing the units, respecting the temperature coefficients, and recognizing that every fraction of an ohm dictates the physical reality of the electrons moving through your copper.