Three-phase power
Three-phase apparent power is line-to-line voltage times line current times 1.732: VA = E × I × 1.732. True power adds the power factor, W = E × I × 1.732 × PF, and line current is I = VA ÷ (E × 1.732). The 1.732 (the square root of 3) appears because the three phases are 120 degrees apart, so phase and line values do not simply add.
What three-phase power is
Three-phase power is three AC voltages of the same size, each shifted 120 degrees from the others. Because one phase is always near its peak, power delivery is smooth, motors start and run without extra parts, and the same load needs smaller conductors than it would on single phase. Almost every commercial and industrial building runs on it, so the exam tests it constantly.
The formulas
VA = E × I × 1.732W = E × I × 1.732 × PF
I = VA ÷ (E × 1.732)
I = W ÷ (E × 1.732 × PF)
E is always the line-to-line voltage and I is the line current. For kVA or kW, divide the result by 1,000. If the load is resistive, such as a heater, PF is 1.0 and drops out.
The voltage and current relationships inside wye and delta connections also use 1.732:
Wye: E(line) = E(phase) × 1.732, I(line) = I(phase)Delta: E(line) = E(phase), I(line) = I(phase) × 1.732
Line vs phase values in wye and delta
"Phase" means one winding or one load element. "Line" means the conductors running between the source and the load.
| Connection | Voltage | Current |
|---|---|---|
| Wye (star) | Line voltage is 1.732 × phase voltage | Line current equals phase current |
| Delta | Line voltage equals phase voltage | Line current is 1.732 × phase current |
An easy way to remember it: in a wye, each line connects in series with one winding, so the current is the same and the voltage differs. In a delta, each winding sits directly across two lines, so the voltage is the same and the current differs.
208Y/120 and 480Y/277 systems
Both are 4-wire wye systems. The first number is line to line, and the second is line to neutral.
- 208Y/120 V: 208 ÷ 1.732 = 120 V to neutral. Used for receptacles, lighting and small equipment. Two-pole loads get 208 V, not 240 V.
- 480Y/277 V: 480 ÷ 1.732 = 277 V to neutral. Used for commercial lighting at 277 V and motors and HVAC at 480 V.
A 120/240-volt 4-wire delta system is different. Two legs read 120 V to neutral, but the high leg reads about 120 × 1.732 = 208 V to neutral. NEC 110.15 requires that leg to be identified with orange or another effective means.
Balanced loads and the neutral
A balanced load draws the same current on all three lines. On a wye system with balanced linear loads, the three currents cancel at the neutral, so neutral current is zero. When loads are unbalanced, the neutral carries the imbalance, but it is not a simple subtraction because the currents are 120 degrees apart. Nonlinear loads such as LED drivers and computers add triplen harmonics that stack up in the neutral instead of canceling.
Worked examples
Example 1: true power with power factor
A balanced three-phase load draws 60 A per line at 208 V with a 0.85 power factor.
- VA = 208 × 60 × 1.732 = 21,615 VA (21.6 kVA)
- W = 21,615 × 0.85 = 18,373 W (18.4 kW)
Example 2: line current of a heater
A 50 kW three-phase resistive heater runs at 480 V.
- I = 50,000 ÷ (480 × 1.732) = 50,000 ÷ 831.4 = 60.1 A
Example 3: line-to-neutral voltage
On a 480Y/277 system, 480 ÷ 1.732 = 277 V to neutral. On a 208Y/120 system, 208 ÷ 1.732 = 120 V. Going the other way, a 277-volt wye winding gives 277 × 1.732 = 480 V line to line.
Example 4: delta winding current
A delta-connected motor draws 52 A on each line. The current in each winding is 52 ÷ 1.732 = 30.0 A.
Example 5: single-phase load on a three-phase system
A 5,000-watt single-phase water heater is connected line to line on a 208Y/120 system. It uses only two lines, so there is no 1.732: 5,000 ÷ 208 = 24.0 A.
Example 6: why higher voltage helps
A 100 kVA balanced load draws 100,000 ÷ (480 × 1.732) = 120.3 A at 480 V, but 100,000 ÷ (208 × 1.732) = 277.6 A at 208 V. Same power, less than half the current at 480 V, which means smaller conductors and less voltage drop. Try the numbers in the voltage drop calculator.
Common exam mistakes
- Leaving out 1.732. 75 kVA at 208 V is 208 A. Dividing by 208 alone gives 361 A, which is a common wrong choice.
- Using 1.732 on a single-phase load. A load connected across two lines of a three-phase system is still single phase.
- Using line-to-neutral voltage in the formula. Use 208 V or 480 V, not 120 V or 277 V.
- Mixing up wye and delta. Wye changes voltage by 1.732. Delta changes current by 1.732.
- Forgetting power factor when the question gives it, or applying it when asked for VA.
- Using 3 instead of 1.732. Multiplying single-phase power by 3 only works with phase (line-to-neutral) voltage and phase current, not line values.
Practice questions
Frequently asked questions
Why is 1.732 used in three-phase calculations?
1.732 is the square root of 3. In a three-phase system the voltages are 120 degrees apart, so the voltage between two lines is 1.732 times the voltage of one phase, not two times. The same factor ties line current to phase current in a delta.
How do you convert three-phase kVA to amps?
Multiply kVA by 1,000 and divide by the line-to-line voltage times 1.732. For example, 45 kVA at 480 V is 45,000 ÷ (480 × 1.732), about 54.1 A. The transformers page applies the same formula to full-load current.
What is the difference between 208Y/120 and 480Y/277?
Both are 4-wire wye systems. 208Y/120 gives 208 V line to line and 120 V to neutral for general receptacles and lighting. 480Y/277 gives 480 V for motors and large equipment and 277 V for commercial lighting.
Is line current the same as phase current?
In a wye, yes. In a delta, line current is 1.732 times the phase (winding) current. Voltage works the opposite way: equal in a delta, 1.732 times larger on the lines in a wye.
How can I get faster at three-phase problems?
Memorize the four formulas, then practice deciding whether each problem is single or three phase before you touch the calculator. The Ohm's law calculator, the power factor guide and the practice test are good places to drill.