Three Phase Voltage Drop Equation: Derivation, Solved Examples & FAQ

The standard simplified three phase voltage drop equation used for sizing feeders and branch circuits under NEC guidelines is VD = (√3 × K × I × L) / CM. This formula allows electricians and engineers to calculate the expected voltage loss across a balanced three-phase AC circuit using the DC resistance approximation of the conductor. While it is highly accurate for conductors sized 1/0 AWG and smaller, it assumes a balanced load, steady-state sinusoidal current, and ignores AC reactance. For precise engineering on large feeders, the exact impedance formula must be used, which we will cover below.

The Core Three Phase Voltage Drop Equation

The formula relies on the geometric relationship between phase-to-neutral and phase-to-phase voltages in a three-phase system. The √3 factor (approximately 1.732) bridges this gap, replacing the multiplier of '2' used in single-phase DC or split-phase calculations.

VD = (√3 × K × I × L) / CM

Rearranged Forms for Practical Sizing

On the jobsite or in the design office, you rarely solve for VD directly. Usually, you know the maximum allowable voltage drop and need to find the required wire size or the maximum distance you can run a specific cable. Here are the algebraically rearranged forms of the three phase voltage drop equation:

• Solving for Wire Size (CM):
  CM = (√3 × K × I × L) / VD
  Use this to find the minimum Circular Mils required, then match it to the next largest AWG size in NEC Table 8.
• Solving for Maximum Distance (L):
  L = (VD × CM) / (√3 × K × I)
  Use this to determine how far you can run a specific wire gauge before exceeding your 3% drop threshold.
• Solving for Maximum Current (I):
  I = (VD × CM) / (√3 × K × L)
  Use this to find the maximum load an existing buried 3-phase feeder can handle without excessive voltage sag.

Step-by-Step Worked Examples

Theory is useless without application. Below are two common scenarios demonstrating how to use the formula with strict unit tracking to prevent calculation errors.

Example 1: Calculating Voltage Drop for an Existing Motor Feeder

Scenario: You are commissioning a 480V, 100A three-phase HVAC compressor. The feeder is 1/0 AWG copper (THHN), and the one-way physical distance from the panel to the disconnect is 250 feet. What is the voltage drop and percentage?

1. Identify Knowns:
   √3 = 1.732
K = 12.9 (Copper at 75°C)
   I = 100 A
L = 250 ft
CM = 105,600 (1/0 AWG from NEC Table 8)
2. Substitute into Equation:
   VD = (1.732 × 12.9 × 100 × 250) / 105,600
3. Calculate Numerator (Unit Tracking):
   1.732 × (12.9 Ω·cmil/ft) × (100 A) × (250 ft) = 558,570 Ω·A·cmil
   Note: The 'ft' unit cancels out.
4. Divide by Denominator:
   558,570 Ω·A·cmil / 105,600 cmil = 5.289 Ω·A
   Since Ω × A = Volts, VD = 5.29 V
5. Calculate Percentage:
   (5.29 V / 480 V) × 100 = 1.1%

Verdict: A 1.1% drop is well within the recommended 3% limit. The 1/0 AWG copper wire is perfectly adequate for this run.

Example 2: Sizing Wire for a New 208V Panelboard

Scenario: You need to feed a new 208Y/120V, 60A three-phase subpanel located 150 feet from the main switchgear. You want to limit the voltage drop to exactly 3%. What is the minimum AWG copper wire size required?

1. Determine Maximum Allowable VD:
   Max VD = 208 V × 0.03 = 6.24 V
2. Identify Knowns:
   √3 = 1.732, K = 12.9, I = 60 A, L = 150 ft, VD = 6.24 V
3. Use the Rearranged Formula for CM:
   CM = (1.732 × 12.9 × 60 × 150) / 6.24
4. Calculate Numerator:
   1.732 × 12.9 × 60 × 150 = 201,085.2
5. Divide by VD:
   201,085.2 / 6.24 = 32,225.19 CM
6. Select Wire Size:
   Look up 32,225 CM in NEC Chapter 9, Table 8.
   6 AWG = 26,240 CM (Too small)
   4 AWG = 41,740 CM (Sufficient)

Verdict: You must pull 4 AWG copper conductors to maintain a 3% or lower voltage drop on this 208V feeder.

Assumptions, Limitations, and Unit Traps

The simplified three phase voltage drop equation is a workhorse, but it is not a universal law of physics. It relies on specific assumptions and is highly vulnerable to unit-entry errors.

When the Formula Applies (and When It Doesn't)

This formula assumes a balanced three-phase load with steady-state sinusoidal current. It uses the DC resistance of the conductor (the 'K' constant) and ignores AC skin effect and proximity effect. For conductors 1/0 AWG and smaller, AC reactance is negligible, making this formula highly accurate. However, for large parallel feeders (e.g., 500 kcmil or larger) running in steel conduit, inductive reactance (X) becomes significant. In those cases, you must use the exact impedance equation: VD = √3 × I × L × (R cosθ + X sinθ), where θ is the power factor angle of the load.

Fatal Unit Mistakes

• The Round-Trip Trap: In single-phase DC math, we multiply the distance by 2 to account for the hot and neutral return paths. Do not do this here. The √3 constant inherently accounts for the three-phase geometry and phase angles. 'L' must strictly be the one-way physical tape-measure distance from source to load.
• AWG vs. CM: Plugging the AWG number (e.g., '4') into the CM slot instead of the Circular Mils value (e.g., '41,740') will result in a calculated voltage drop thousands of times higher than reality. Always use NEC Table 8 to convert AWG to CM.

Realistic Answer Magnitudes

What should your final number look like? A realistic, well-designed answer magnitude is between 1% and 3% of the nominal system voltage. On a 480V system, expect 4.8V to 14.4V of drop. If your calculation yields a drop greater than 5%, the wire is severely undersized for the distance. Operating at >5% drop causes three-phase induction motors to draw excessive current to compensate for the lost voltage, leading to rapid insulation degradation, contactor chatter, and nuisance breaker tripping.

Frequently Asked Questions

How does the three phase voltage drop equation differ from single phase?

The primary difference is the multiplier. The single-phase AC voltage drop equation uses a multiplier of 2 (representing the out-and-back physical length of the hot and neutral wires). The three-phase equation uses √3 (1.732). This is because, in a balanced three-phase system, the return current is distributed across the other two phases, and the phase-to-phase voltage is √3 times the phase-to-neutral voltage. Consequently, for the same current, distance, and wire size, a three-phase circuit will experience roughly 13% less voltage drop than an equivalent single-phase circuit (1.732 / 2 = 0.866).

What is the exact AC impedance three phase voltage drop equation?

When dealing with large conductors (typically larger than 1/0 AWG) or circuits with a poor power factor, the DC resistance approximation breaks down. The exact equation recognized by IEEE and advanced engineering texts is:
VD = √3 × I × L × (R cosθ + X sinθ)
In this formula, R is the AC resistance per 1,000 feet, X is the inductive reactance per 1,000 feet (found in NEC Chapter 9, Table 9), and θ is the power factor angle of the load. This accounts for the voltage drop caused by the magnetic fields surrounding large conductors in steel or PVC conduit.

Does the three phase voltage drop equation apply to unbalanced loads?

No. The standard √3 formula strictly assumes a perfectly balanced load across all three phases (L1, L2, and L3). If you are calculating voltage drop for a heavily unbalanced load—such as a panelboard where L1 is carrying 80A, L2 is carrying 20A, and L3 is carrying 40A—the neutral will carry significant current, and the phase-to-neutral voltages will shift. To calculate voltage drop in unbalanced scenarios, you must abandon the simplified three-phase formula and instead calculate the voltage drop on each individual phase-to-neutral conductor using single-phase math, factoring in the neutral current return vectorially.