Short answer: For a balanced three-phase load, apparent power is kVA = (1.732 × line voltage × line current) ÷ 1,000, so current is I = VA ÷ (1.732 × V). Real power adds the power factor: W = 1.732 × V × I × PF. The 1.732 is the square root of 3. In a wye system the line voltage is 1.732 times the phase voltage, which is why 120 V becomes 208 V. In a delta system the line current is 1.732 times the phase current.
PSI does not list three-phase power by name. Its Calculations outline includes Services, Service Equipment and Separately Derived Systems (4 scored questions, covering load calculations and transformers) and Calculations and Theory (2 scored questions). These counts cover the full subject areas, not three-phase power alone. The Texas Journeyman Practice Pack has 1,000 original questions and 10 timed mock exams in the real two-part format, with step-by-step solutions for every calculation. Try free practice questions without signing up. Get the full pack for US$39 with one payment.
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The formulas
| Find | Single-phase | Balanced three-phase |
|---|---|---|
| Current from VA | I = VA ÷ V | I = VA ÷ (1.732 × V) |
| VA from current | VA = V × I | VA = 1.732 × V × I |
| Watts | W = V × I × PF | W = 1.732 × V × I × PF |
| kVA from kW | kVA = kW ÷ PF | kVA = kW ÷ PF |
In the three-phase formulas, V is the line-to-line voltage and I is the line current. In the single-phase formulas, V is the voltage across the load and I is the current through it. PF is the power factor, a number from 0 to 1. For a purely resistive load such as a heater, PF = 1 and watts equal volt-amperes.
Some books write 1.73 or the square root of 3 instead of 1.732. On an exam, the small difference rarely changes which answer you pick, but use the factor the question shows if it gives one.
Where 1.732 comes from
The three phases are 120 electrical degrees apart. In a balanced wye system, when you measure between two phases, the two phase-to-neutral voltages do not simply add. The result is the square root of 3, or about 1.732, times the phase-to-neutral voltage.
- 120/208 V wye: 120 × 1.732 = about 208 V line to line.
- 277/480 V wye: 277 × 1.732 = about 480 V line to line.
Wye vs delta
| Wye (star) | Delta | |
|---|---|---|
| Line voltage | 1.732 × phase voltage | equal to phase voltage |
| Line current | equal to phase current | 1.732 × phase current |
| Neutral | common point of the three windings | none from the center; a 240 V delta may have a center-tapped winding |
| Common systems | 120/208 V, 277/480 V | 240 V, 480 V; 120/240 V with a high leg |
On a 120/240 V four-wire delta with a center-tapped winding, one line conductor measures about 208 V to the neutral, while the other two measure 120 V. That high leg must be identified, usually with orange, and must never feed a 120 V line-to-neutral load.
Worked example 1: transformer kVA to amps
A 45 kVA three-phase transformer has a 480 V primary and a 120/208 V secondary. Find the full-load current on each side.
- Secondary: 45,000 ÷ (1.732 × 208) = 45,000 ÷ 360.3 = 124.9 A.
- Primary: 45,000 ÷ (1.732 × 480) = 45,000 ÷ 831.4 = 54.1 A.
- Check: the voltage ratio is 480 ÷ 208 = 2.31, and 124.9 ÷ 54.1 = 2.31. Current goes up by the same ratio that voltage goes down.
Worked example 2: a motor load with power factor
A balanced 480 V three-phase load uses 50 kW at a power factor of 0.85.
- Current: 50,000 ÷ (1.732 × 480 × 0.85) = 50,000 ÷ 706.7 = 70.8 A.
- Apparent power: 50 ÷ 0.85 = 58.8 kVA.
A lower power factor means more current for the same real power, so the conductors and the transformer must carry more. For motor branch circuits, the NEC uses its own full-load current tables, not nameplate kW, for most sizing. That is the next lesson.
Worked example 3: a single-phase load on a three-phase panel
A 5,000 W, 208 V water heater is connected between two phases of a 120/208 V panel. It is a single-phase load.
- I = 5,000 ÷ 208 = 24.0 A.
- Do not use 1.732 here. The 1.732 formula is for a balanced three-phase load. A single-phase load connected between two conductors uses its actual applied voltage.
Using 1.732 on a single-phase load, or leaving it out on a three-phase load, is one of the most common calculation mistakes.
Check yourself
- What is the full-load current of a 30 kVA, 208 V three-phase load? 30,000 ÷ 360.3 = 83.3 A.
- In a wye system, is the line current or the line voltage 1.732 times the phase value? The line voltage.
- A 7,200 W, 240 V single-phase heater is connected to a three-phase panel. Do you use 1.732? No. I = 7,200 ÷ 240 = 30 A.
To check your numbers, use our three-phase power calculator and transformer kVA calculator.
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Practice calculations with step-by-step lessons
Texas Journeyman Electrician Exam Prep 2026 NEC (paperback, 327 pages). 486 practice questions: 316 chapter questions across 20 chapters plus two full mock exams of 85 questions each, 59 NEC Knowledge in 130 minutes and 26 Calculations in 110 minutes. Includes calculation lessons and code lookup paths, and NEC questions come with article, section or table references.
FAQ
Why is 1.732 used in three-phase calculations?
1.732 is the square root of 3. The phases are 120 degrees apart, so the voltage between two phases is 1.732 times the voltage of one phase in a wye system.
How do I convert kVA to amps for three-phase?
Divide the volt-amperes by 1.732 times the line-to-line voltage. A 45 kVA load at 208 V draws 45,000 ÷ 360.3 = about 124.9 A.
What is the difference between kW and kVA?
kVA is apparent power and kW is real power. For a balanced three-phase load, kVA = 1.732 × line voltage × line current ÷ 1,000, and kW = kVA × power factor. For a purely resistive load the power factor is 1, so they are equal.
Is three-phase power on the Texas Journeyman exam?
PSI does not list three-phase power as a separate topic or say how many questions use the formula. The three-phase current formula is used in load, transformer and motor calculations.
Sources checked October 2026
- PSI Candidate Information Bulletin, Texas electrician examinations (updated September 15, 2026): https://test-takers.psiexams.com/api/content/bulletin/1682
