The standard approximate three phase voltage drop formula used for wire sizing in North America is VD = (√3 × K × I × D) / CM. This equation calculates the line-to-line voltage lost as heat across a balanced three-phase circuit, ensuring your conductors are thick enough to deliver adequate voltage to the load without violating NEC-style guidance (which recommends a maximum 3% drop for branch circuits and 5% for the total feeder plus branch).
The Core Formula and Symbol Definitions
Before running calculations, you must understand the exact assumptions behind this formula. This specific iteration assumes a balanced three-phase load, steady-state current, and unity power factor (or a power factor close enough to 1.0 that reactance is negligible for smaller wire sizes). It calculates the line-to-line voltage drop.
The formula is expressed as:
VD = (√3 × K × I × D) / CM
| Symbol | Definition | Standard US Unit | Notes & Bench Realities |
|---|---|---|---|
| VD | Voltage Drop | Volts (V) | The actual line-to-line voltage lost across the conductors. |
| √3 | Square Root of 3 | Dimensionless | Approximately 1.732. Accounts for the 120° phase shift in 3-phase systems. |
| K | Specific Resistance Constant | Ω·cmil/ft | Use 12.9 for Copper and 21.2 for Aluminum at 75°C operating temp. |
| I | Current (Load) | Amperes (A) | The continuous or full-load ampacity of the equipment. |
| D | One-Way Distance | Feet (ft) | Distance from source to load. (Do not double this for 3-phase). |
| CM | Circular Mils | cmil | Cross-sectional area of the wire. Found in NEC Chapter 9, Table 8. |
Rearranged Forms for Practical Wire Sizing
On the jobsite, you rarely need to find the voltage drop of an existing wire; you usually need to find the right wire size for a new run. By algebraically rearranging the core formula, we can solve for any missing variable.
- Solving for CM (Required Wire Area):
CM = (√3 × K × I × D) / VD
Use this to find the minimum circular mils required, then round UP to the next standard AWG size. - Solving for I (Maximum Allowable Current):
I = (VD × CM) / (√3 × K × D)
Use this to see how much load an existing buried feeder can handle before exceeding a 3% drop. - Solving for D (Maximum Run Distance):
D = (VD × CM) / (√3 × K × I)
Use this to determine how far you can push a specific wire size before needing to step up to a larger gauge.
Worked Example 1: Calculating Voltage Drop
Scenario: You are feeding a 480V, 3-phase, 50A industrial motor located 200 feet from the panel. You plan to use 4 AWG copper THHN. What is the exact voltage drop, and does it meet the 3% guideline?
- Identify and list the variables:
- System Voltage = 480V
- I = 50 A
- D = 200 ft
- K = 12.9 (Copper at 75°C)
- CM = 41,740 cmil (from NEC Chapter 9, Table 8 for 4 AWG)
- Substitute values into the formula:
VD = (1.732 × 12.9 × 50 × 200) / 41,740 - Multiply the numerator:
1.732 × 12.9 = 22.342822.3428 × 50 = 1,117.141,117.14 × 200 = 223,428 - Divide by the denominator (CM):
VD = 223,428 / 41,740VD = 5.35 Volts - Calculate the percentage drop:
Percentage = (5.35V / 480V) × 100Percentage = 1.11%
Result: The voltage drop is 5.35V, which is 1.11%. This is well below the 3% NEC-style recommendation. 4 AWG copper is electrically sufficient for voltage drop, provided it also meets the ampacity requirements for a 50A motor circuit after applying NEC Article 430 multipliers.
Worked Example 2: Sizing Wire for a Maximum Drop
Scenario: You are installing a new 208V, 3-phase subpanel feeder. The continuous load is 100A, and the distance is 150 feet. You want to limit the voltage drop to exactly 3%. What size copper wire do you need?
- Calculate the maximum allowable voltage drop (VD):
Max VD = 208V × 0.03Max VD = 6.24 Volts - Identify the known variables:
- VD = 6.24 V
- I = 100 A
- D = 150 ft
- K = 12.9 (Copper at 75°C)
- Substitute values into the rearranged CM formula:
CM = (1.732 × 12.9 × 100 × 150) / 6.24 - Multiply the numerator:
1.732 × 12.9 = 22.342822.3428 × 100 = 2,234.282,234.28 × 150 = 335,142 - Divide by the maximum VD:
CM = 335,142 / 6.24CM = 53,708.6 cmil - Select the standard AWG size:
Looking at NEC Chapter 9, Table 8, 3 AWG copper is 52,620 cmil (too small). 2 AWG copper is 66,360 cmil.
Result: You must use a minimum of 2 AWG copper to maintain a 3% or less voltage drop on this 150-foot run. Never round down to a smaller wire size when the calculated CM falls between standard AWG tiers.
Unit Mistakes That Break the Math
When troubleshooting a failed inspection or a motor that hums but won't start, I usually find one of these three unit errors in the original design math:
- Confusing Line-to-Line vs. Line-to-Neutral Voltage: The formula calculates the line-to-line voltage drop (e.g., 480V or 208V). When calculating your percentage, you must divide by the line-to-line voltage. Dividing a 480V system drop by 277V (line-to-neutral) will artificially inflate your percentage and cause you to overspend on copper.
- Mixing Metric and Imperial: The constant K (12.9) and the Circular Mil (CM) area are strictly Imperial. If your distance is in meters, or your wire area is in square millimeters (mm²), this specific formula will yield garbage data. For metric calculations, use the formula:
VD = √3 × I × (R cosθ + X sinθ) × L, where L is in kilometers and R/X are in Ω/km. See Electrical Technology's metric voltage drop guide for IEC standard conversions. - Doubling the Distance (D): In single-phase calculations, we often multiply the distance by 2 to account for the hot and the neutral return path. In a balanced three-phase system, the √3 multiplier mathematically accounts for the return paths and phase angles. Use the strict one-way physical distance for D. Doubling it will result in a calculated drop twice as high as reality.
Frequently Asked Questions
How does the three phase voltage drop formula differ from single phase?
The primary difference is the multiplier. Single-phase voltage drop uses a multiplier of 2 (representing the out-and-back length of the hot and neutral conductors). Three-phase uses √3 (1.732). Because 1.732 is less than 2, a three-phase system inherently experiences less voltage drop than a single-phase system carrying the same current over the same distance and wire size, making it highly efficient for industrial power distribution.
What is a realistic voltage drop percentage for a 3-phase feeder?
According to NFPA 70 (NEC) Informational Notes, a realistic and recommended target is a maximum 2% drop for the feeder, and a maximum 3% drop for the branch circuit, resulting in a total system drop of no more than 5%. In heavy industrial environments with large motors, designers often aim for an even tighter 1.5% total drop to prevent starting torque issues and excessive heat generation in motor windings.
Does the three phase voltage drop formula account for power factor?
The standard approximate formula (using the K-factor) assumes a power factor of 1.0 (unity), meaning it only calculates the resistive voltage drop. For highly inductive loads like large, uncorrected motors, or for very large wire sizes (typically 1/0 AWG and larger) where AC reactance (X) becomes significant, this formula will slightly under-calculate the true drop. For precise engineering on large feeders, you must use the exact impedance formula: VD = √3 × I × (R cosθ + X sinθ) × L, where θ is the power factor angle.
Why do we use 1.732 instead of 2 in the three phase formula?
The value 1.732 is the square root of 3 (√3). In a wye-connected three-phase system, the line-to-line voltage is √3 times the line-to-neutral voltage (e.g., 208V is √3 × 120V). When calculating the voltage drop across the lines, the vector mathematics of the 120-degree phase separation between the three conductors dictates that the effective drop measured line-to-line is scaled by √3 relative to the drop on a single conductor.
