The 3-Phase Voltage Drop Formula & Symbol Definitions
When sizing conductors for 3-phase AC power, calculating voltage drop ensures your equipment receives adequate voltage at the far end of the run. Using a TI 30XS calculator online emulator is a standard workflow for electrical engineers and journeyman exam candidates who need the MultiView fraction display without carrying physical hardware. The standard approximate formula, derived from NFPA 70 (NEC) Chapter 9 principles, is:
| Symbol | Definition | Standard Unit / Value |
|---|---|---|
| VD | Voltage Drop (Line-to-Line) | Volts (V) |
| √3 | 3-Phase Phasor Constant | ~1.732 (unitless) |
| K | Conductor DC Resistivity Constant | 12.9 Ω·cmil/ft (Copper at 75°C) 21.2 Ω·cmil/ft (Aluminum at 75°C) |
| I | Load Current | Amperes (A) |
| D | One-Way Circuit Distance | Feet (ft) |
| CM | Conductor Cross-Sectional Area | Circular Mils (cmil) |
Rearranged Forms: Solving for Every Variable
On the jobsite or in the exam room, you rarely just solve for VD. You usually know your allowable drop and need to find the wire size or maximum run length. Here are the algebraic rearrangements:
- Solving for Wire Size (CM): CM = (√3 × K × I × D) / VD
- Solving for Max Distance (D): D = (VD × CM) / (√3 × K × I)
- Solving for Max Current (I): I = (VD × CM) / (√3 × K × D)
Deriving the √3: Phasor Math vs. Single-Phase
Why does the single-phase voltage drop formula use a multiplier of 2 (representing the out-and-back 180° path of the current), while the 3-phase formula uses √3? This isn't an arbitrary code rule; it's rooted in vector math.
In a 3-phase Wye system, the line-to-line voltage is the vector difference between two phase voltages that are 120° apart. When you calculate the voltage drop across a 3-phase feeder, you are looking at the line-to-line drop. Using the law of cosines for two equal vectors separated by 120°, the resultant magnitude is exactly √3 (1.732) times the per-phase drop. The Copper Development Association (CDA) and standard engineering texts confirm that this √3 multiplier accounts for the phase angle geometry, effectively reducing the calculated line-to-line drop compared to a naive "times three" assumption.
Assumptions, Unit Traps, and Realistic Magnitudes
When this formula applies: This approximation assumes a balanced 3-phase AC circuit, sinusoidal waveforms, and a standard power factor (0.85 to 1.0). For highly inductive loads with a low power factor (e.g., large unloaded motors), the exact formula incorporating reactance (X) and power factor angles is required.
Unit mistakes that will break your math:
- Mixing Meters and Feet: The
Kconstant of 12.9 is mathematically bound to feet. If your blueprint is in meters, convert distance to feet before plugging it into the TI-30XS. - Using mm² instead of CM: The formula demands Circular Mils. 1 mm² ≈ 1973.5 CM. Never plug metric cross-sectional area directly into the CM variable.
- Using Line-to-Neutral Voltage: VD here represents the line-to-line drop. Do not compare your result against a line-to-neutral limit without dividing by √3 first.
Worked Examples: Step-by-Step on the TI 30XS Calculator Online
The TI-30XS MultiView interface allows you to see the entire fraction before hitting enter, which prevents parenthesis errors. Below are two common scenarios with exact unit tracking and keystrokes.
Problem 1: Finding Voltage Drop
Scenario: A 480V 3-phase motor draws 60A. The one-way run is 250 feet using 4 AWG Copper wire. Find the voltage drop.
Knowns: K = 12.9, I = 60, D = 250, CM = 41,740 (from NEC Ch. 9 Table 8).
Unit Tracking: (unitless × [Ω·cmil/ft] × [A] × [ft]) / [cmil] = Volts.
- Press
2ndthenx^2(this inputs the √ symbol). The screen shows√(. - Type
3, then press)to close the radical. (Critical Trap: If you forget this parenthesis, the calculator will apply the square root to the entire rest of your equation!) - Press
×12.9×60×250. - Press
÷41740. - Press
enter.
Result: 8.01V. This is a 1.67% drop on a 480V system, which is well within the 3% NEC recommendation.
Problem 2: Sizing the Wire (Finding CM)
Scenario: You are designing a 208V 3-phase feeder for a 100A panelboard. The distance is 150 feet. The maximum allowable voltage drop is 3% (6.24V). What size copper wire do you need?
Knowns: K = 12.9, I = 100, D = 150, VD = 6.24.
Equation: CM = (√3 × K × I × D) / VD
- Press
2ndx^23)×12.9×100×150. - Press
÷6.24. - Press
enter.
Result: 53,711.53 CM. You now need to map this raw number to a physical wire size.
Decision Path: Translating Math to Physical Wire Sizes
Once your TI-30XS spits out a Circular Mil requirement, you must cross-reference it with NEC Chapter 9, Table 8. You can never round down; the physical wire must have at least the CM area your math demands.
| Calculated CM Requirement | Standard NEC AWG Size | Actual Circular Mils | Concrete Pick (Copper THHN) |
|---|---|---|---|
| ≤ 41,740 | 4 AWG | 41,740 | 4 AWG THHN |
| 41,741 – 52,620 | 3 AWG | 52,620 | 3 AWG THHN |
| 52,621 – 66,360 | 2 AWG | 66,360 | 2 AWG THHN (Pick for Prob 2) |
| 66,361 – 83,690 | 1 AWG | 83,690 | 1 AWG THHN |
| 83,691 – 105,600 | 1/0 AWG | 105,600 | 1/0 AWG THHN |
Default Recommendation: Always round up to the next standard AWG size based on your calculated CM, and then verify that final pick against the ampacity limits in NEC Table 310.16. For the 53,711 CM calculation in Problem 2, 3 AWG (52,620 CM) is too small. The mandatory, code-compliant pick is 2 AWG THHN Copper, which also safely handles the 100A load at 75°C terminations.






