The standard single-phase voltage drop formula calculation is VD = (2 × K × I × L) / CM. For a 120V, 20A circuit running 150 feet using 12 AWG copper wire, the calculated drop is 11.85V (9.8%). Because this vastly exceeds the NEC-recommended 3% maximum for branch circuits, the concrete fix is to upgrade to 6 AWG copper, which drops the loss to 2.95V (2.4%).
The Core Voltage Drop Formula Calculation
To calculate voltage drop accurately, you must use the circular mil (CM) area of the conductor rather than the AWG number directly. The foundational formula for single-phase AC (with a power factor near 1.0) and DC circuits is:
VD = (2 × K × I × L) / CM
| Symbol | Definition | Standard Units / Values |
|---|---|---|
| VD | Voltage Drop | Volts (V) |
| 2 | Constant for single-phase round-trip path | Dimensionless (Use 1.732 for 3-phase) |
| K | Specific resistance of the conductor material | 12.9 for Copper, 21.2 for Aluminum (at 75°C) |
| I | Current (Load) | Amperes (A) |
| L | One-way length of the circuit | Feet (ft) |
| CM | Cross-sectional area in Circular Mils | CM (e.g., 12 AWG = 6,530 CM) |
The 'K' value represents the ohms-cmil per foot of the material. While many textbooks use 10.8 for copper (which assumes a 20°C ambient room temperature), practical jobsite wiring operates hotter. Using K = 12.9 accounts for the 75°C operating temperature column found in NEC Table 310.16, providing a safer, real-world worst-case calculation.
Rearranged Forms for Circuit Design
You rarely use the formula just to find the voltage drop. Usually, you know your allowable drop and need to find the right wire. Here are the rearranged forms solving for each critical variable:
- Solving for Wire Size (CM): CM = (2 × K × I × L) / VD
- Solving for Maximum Length (L): L = (VD × CM) / (2 × K × I)
- Solving for Maximum Current (I): I = (VD × CM) / (2 × K × L)
Assumptions, Limits, and Realistic Magnitudes
This formula applies strictly to steady-state DC or single-phase AC circuits where the Power Factor (PF) is approximately 1.0 (typical for resistive loads like heaters and incandescent lighting). If you are sizing wire for heavy inductive loads (large motors, transformers) where PF is 0.8 or lower, this simplified formula will under-calculate the drop; you must use the full impedance formula (VD = I × Z) incorporating both resistance and reactance.
What does a realistic answer magnitude look like?
For a 120V branch circuit, a 3% drop is 3.6V. For a 240V feeder, a 3% drop is 7.2V. If your calculation spits out a drop of 45V on a standard residential branch, you have either made a unit error (like using meters instead of feet) or you are attempting to push 100A through 14 AWG wire. Total system drop (feeder + branch combined) should never exceed 5%.
Worked Examples with Unit Tracking
Let's run two real-world scenarios, tracking every unit to ensure the math holds up.
Problem 1: Finding the Voltage Drop on an Existing Run
Scenario: You are wiring a 120V, 20A receptacle circuit to a detached workshop. The one-way distance is 150 feet. You plan to use 12 AWG THHN copper wire. What is the voltage drop and percentage?
- Identify Knowns: K = 12.9 (Cu at 75°C), I = 20A, L = 150 ft. From NEC Chapter 9 Table 8, 12 AWG = 6,530 CM.
- Set up the equation: VD = (2 × 12.9 × 20 × 150) / 6530
- Multiply the numerator: 2 × 12.9 = 25.8. Then 25.8 × 20 = 516. Then 516 × 150 = 77,400.
- Divide by CM: 77,400 / 6,530 = 11.85V.
- Calculate Percentage: (11.85V / 120V) × 100 = 9.87%.
Verdict: A 9.87% drop is unacceptable. The workshop tools will experience severe voltage sag under load.
Problem 2: Sizing Wire for a 240V Feeder
Scenario: You need to run a 240V, 40A subpanel feeder. The one-way distance is 200 feet. You want to limit the voltage drop to exactly 3%. What size copper wire do you need?
- Identify Knowns: K = 12.9, I = 40A, L = 200 ft. Max VD = 240V × 0.03 = 7.2V.
- Use the rearranged formula for CM: CM = (2 × K × I × L) / VD
- Set up the equation: CM = (2 × 12.9 × 40 × 200) / 7.2
- Multiply the numerator: 2 × 12.9 × 40 × 200 = 206,400.
- Divide by VD: 206,400 / 7.2 = 28,666 CM.
- Select the AWG: Looking at NEC Chapter 9 Table 8, 6 AWG is 26,240 CM (too small). 4 AWG is 41,740 CM.
Verdict: You must pull 4 AWG Copper to maintain a 3% or lower drop on this feeder.
Common Unit Mistakes That Break the Math
When the math yields absurd results, it is almost always due to one of these three unit errors:
- Using Round-Trip Length for 'L': The '2' in the numerator of the formula already accounts for the hot and neutral/ground return path. 'L' must be the strict one-way physical distance from panel to load. If you double 'L' and keep the '2', you will quadruple your calculated drop.
- Mixing Metric and Imperial: The K constant (12.9) is calibrated for feet and circular mils. If you measure your run in meters, the formula breaks. Convert meters to feet (multiply by 3.281) before plugging into 'L'.
- Confusing mm² with CM: International wire sizes use square millimeters. You cannot plug mm² into the 'CM' slot. To convert, use the multiplier: CM = mm² × 1,973.5. For example, a 4mm² wire is roughly 7,894 CM.
Decision Path: Final Wire Sizing Pick
Use this decision matrix to finalize your wire gauge after running the voltage drop formula calculation. This aligns with standard NFPA 70 National Electrical Code efficiency guidelines and practical Southwire engineering recommendations.
| Calculated % Voltage Drop | Circuit Type | Action Required |
|---|---|---|
| ≤ 3.0% | Branch or Feeder | Keep calculated AWG. Optimal efficiency. |
| 3.1% to 5.0% | Feeder Only | Acceptable. Feeders can tolerate up to 5% if the subsequent branch circuit is short and stays under 3%. |
| 3.1% to 5.0% | Branch Circuit | Bump up one AWG size. Lighting will flicker on motor startups; sensitive electronics may brownout. |
| > 5.0% | Any Circuit | Mandatory upgrade. Recalculate CM targeting a 2% drop to find your new minimum AWG. |






