The standard approximate voltage drop formula for single-phase AC or DC circuits is VD = (2 × K × I × L) / CM. This calculates the total voltage lost across both the ungrounded (hot) and grounded (neutral/return) conductors. If you are sizing wire for a branch circuit or feeder, this is the foundational equation you need to ensure your equipment receives adequate voltage under load.
The Single-Phase Voltage Drop Formula & Symbol Definitions
The formula is a direct derivation of Ohm’s Law (V = I × R). Because resistance (R) of a wire is defined by its material resistivity (K), length (L), and cross-sectional area (CM), we substitute R with (K × L / CM). We multiply by 2 because single-phase current must travel out to the load and return to the source, traversing two wire lengths.
VD = (2 × K × I × L) / CM
| Symbol | Definition | Standard Unit |
|---|---|---|
| VD | Voltage Drop (total line-to-line loss) | Volts (V) |
| 2 | Multiplier for the out-and-back current path (single-phase) | Dimensionless |
| K | Specific resistance (resistivity) of the conductor material | Ω·cmil/ft |
| I | Current flowing through the circuit (load amperage) | Amperes (A) |
| L | One-way distance from source to load (NOT total wire length) | Feet (ft) |
| CM | Cross-sectional area of the wire in Circular Mils | cmil |
Standard Wire Parameters & Constants
To use the formula, you cannot plug in the AWG number directly. You must use the Circular Mil (CM) area. Furthermore, the 'K' constant changes based on the conductor material and its operating temperature. The NFPA 70 (NEC) Chapter 9, Table 8 provides the baseline CM values, while Note 4 provides the K constants.
Below is a data-dense reference table for common residential and commercial wire sizes. We use the 75°C column, as most modern breakers and terminations are rated for 75°C, making it the standard baseline for voltage drop calculations under load.
| AWG Size | Circular Mils (CM) | Copper K (75°C) | Aluminum K (75°C) | Copper K (20°C / Ambient) |
|---|---|---|---|---|
| 14 AWG | 4,110 | 12.9 | 21.2 | 10.4 |
| 12 AWG | 6,530 | 12.9 | 21.2 | 10.4 |
| 10 AWG | 10,380 | 12.9 | 21.2 | 10.4 |
| 8 AWG | 16,510 | 12.9 | 21.2 | 10.4 |
| 6 AWG | 26,240 | 12.9 | 21.2 | 10.4 |
| 4 AWG | 41,740 | 12.9 | 21.2 | 10.4 |
Rearranged Forms for Circuit Design
On the jobsite, you rarely solve for VD in isolation. Usually, you know your allowable voltage drop (e.g., 3% of 120V = 3.6V) and need to find the required wire size, or you have an existing wire and need to know the maximum distance you can run it. Here are the algebraically rearranged forms of the single-phase voltage drop formula:
- Solve for Wire Size (CM):
CM = (2 × K × I × L) / VD
Use this to find the minimum Circular Mils required, then round UP to the next standard AWG size. - Solve for Maximum Distance (L):
L = (VD × CM) / (2 × K × I)
Use this to find the maximum one-way run length before you exceed your VD limit. - Solve for Maximum Current (I):
I = (VD × CM) / (2 × K × L)
Use this to determine the absolute maximum load a specific wire run can support within VD limits (do not exceed the wire's ampacity rating). - Solve for Material Constant (K):
K = (VD × CM) / (2 × I × L)
Rarely used in design, but useful for forensic troubleshooting to determine if a conductor is suffering from high-resistance corrosion or faulty terminations.
Worked Examples with Unit Tracking
Let’s apply the formula to two real-world scenarios. Tracking units through the calculation is the best way to catch errors before you pull wire.
Example 1: 120V Branch Circuit for a Receptacle
Scenario: You are running a 120V, 15A dedicated circuit for a shop tool using 12 AWG solid copper THHN. The one-way distance from the panel to the outlet is 60 feet. Is the voltage drop within the NEC recommended 3% limit?
- Identify Variables:
- I = 15 A
- L = 60 ft
- CM = 6,530 cmil (from 12 AWG table above)
- K = 12.9 Ω·cmil/ft (Copper at 75°C)
- Substitute into Formula:
VD = (2 × 12.9 [Ω·cmil/ft] × 15 [A] × 60 [ft]) / 6,530 [cmil] - Calculate Numerator:
2 × 12.9 × 15 × 60 = 23,220(Units: Ω·A·ft, which simplifies to V·ft when considering the cmil cancellation) - Divide by CM:
23,220 / 6,530 = 3.555... V - Calculate Percentage:
%VD = (3.56 V / 120 V) × 100 = 2.96%
Verdict: At 2.96%, this run passes the NEC Informational Note recommendation of 3% for branch circuits. 12 AWG is acceptable.
Example 2: 240V Feeder to a Detached Garage Subpanel
Scenario: You are feeding a 240V single-phase subpanel with a continuous load of 40A. You plan to use 8 AWG copper wire. The trench distance (one-way) is 120 feet. Calculate the voltage drop.
- Identify Variables:
- I = 40 A
- L = 120 ft
- CM = 16,510 cmil (8 AWG)
- K = 12.9 Ω·cmil/ft
- Substitute into Formula:
VD = (2 × 12.9 × 40 × 120) / 16,510 - Calculate Numerator:
2 × 12.9 × 40 × 120 = 123,840 - Divide by CM:
123,840 / 16,510 = 7.50 V - Calculate Percentage:
%VD = (7.50 V / 240 V) × 100 = 3.12%
Verdict: A 3.12% drop slightly exceeds the 3% branch/feeder individual recommendation, though it is well within the 5% total system limit. If the subpanel will run sensitive electronics or motor loads that require strict voltage regulation, you should upsizing to 6 AWG (CM = 26,240), which would drop the loss to 1.96%. For general lighting and standard receptacles, 8 AWG is functionally fine.
Application Limits, Assumptions, and Common Unit Traps
The VD = (2 × K × I × L) / CM formula is an approximation. It is highly accurate for standard residential and light commercial wiring, but you must understand its boundaries to avoid dangerous miscalculations.
When the Formula Applies (and When It Doesn't)
This formula assumes a single-phase, 2-wire circuit (or a perfectly balanced 3-wire split-phase circuit where the neutral carries zero current). It also assumes a power factor (PF) of 1.0, meaning the load is purely resistive (like incandescent lighting or strip heaters).
If you are calculating voltage drop for a large inductive load (like a 5HP air compressor motor) where the power factor might be 0.80, the wire's AC reactance (X) comes into play. In those cases, electrical engineers use the exact AC formula: VD = 2 × I × L × (R cosθ + X sinθ) / 1000 (where R and X are resistance and reactance per 1000 ft, and θ is the phase angle). For runs under 100 feet carrying loads under 50A, the simpler formula is universally accepted by inspectors.
Common Unit Mistakes That Break the Math
Warning: If your calculator outputs a voltage drop of 45V on a 120V circuit, stop. You didn't discover a massive energy leak; you made a unit error. Here are the three most common traps:
- Using AWG instead of CM: Plugging '12' into the denominator instead of '6530' will inflate your result by a factor of 500.
- Using Total Wire Length instead of One-Way Distance: If you pull 120 feet of physical wire to reach a load 60 feet away, L is 60. The '2' in the numerator already accounts for the return path. Doubling L and keeping the '2' double-counts the return trip.
- Mixing Metric and Imperial: This formula is strictly Imperial (feet, circular mils). If your distance is in meters, convert to feet first (multiply by 3.281), or switch entirely to the metric formula:
VD = (2 × ρ × I × L) / A(where ρ is resistivity in Ω·m, L is meters, and A is cross-sectional area in mm²).
What a Realistic Answer Magnitude Looks Like
According to Southwire's engineering guidelines and NEC Informational Notes, a properly designed circuit should exhibit a voltage drop of 2% to 5% under full load.
- On a 120V circuit, a realistic VD is between 2.4V and 6.0V.
- On a 240V circuit, a realistic VD is between 4.8V and 12.0V.
If your calculated magnitude falls outside this range—especially if it exceeds 10%—you either have a math error, or you are about to install a wire that will overheat, cause lights to flicker, and prematurely burn out motor windings. Always verify your math against a trusted voltage drop reference chart before purchasing copper.
