Journeyman refresher Refresher 2 of 10 · about 9 min
Rusty on three-phase math?
What you’ll learn
- Which of voltage or current carries the √3 in a wye and in a delta.
- The power formula and the current formula, both directions, with the right voltage.
- The power triangle — kW, kVA, kVAR — without drawing it.
- Quick sanity checks that catch a slipped √3 before it reaches a panel.
Before this: Wye and delta
Line and phase, in one table
| Wye (star) | Delta | |
|---|---|---|
| Voltage | V line = √3 × V phase | V line = V phase |
| Current | I line = I phase | I line = √3 × I phase |
| Neutral? | Yes — the centre point | No (a high-leg delta taps one winding) |
| Everyday pairs | 208Y/120, 480Y/277, 600Y/347 | 240 V, 480 V, 600 V delta |
The formulas you actually use
- kVA = √3 × V × I ÷ 1000line-to-line V, line I
- kW = kVA × PFreal power
- I = kW × 1000 ÷ (√3 × V × PF)current from kW
- I = kVA × 1000 ÷ (√3 × V)current from kVA
- kVAR = √(kVA² − kW²)reactive power
- √3 × V LL = 3 × V LNwhy both forms agree
Worked check: both directions
- A 208 V three-phase load draws 50 A per line at PF 0.85. kVA = 1.732 × 208 × 50 ÷ 1000 = 18.0 kVA.
- kW = 18.0 × 0.85 = 15.3 kW. kVAR = √(18.0² − 15.3²) = 9.5 kVAR.
- Now back: I = 15,300 ÷ (1.732 × 208 × 0.85) = 15,300 ÷ 306.2 = 50.0 A. It closes — that is the check.
- Same load worked with phase values: 3 × 120 × 50 × 0.85 = 15,300 W. Same answer, as it should be.
Sanity checks
- Line voltage on a wye is always the bigger number. If you divided to get it, you went the wrong way.
- At the same kW, three-phase current is lower than single-phase current at the same voltage by 1.732.
- Rough rule on 600 V: about 1 A per kVA (1000 ÷ (1.732 × 600) = 0.96). On 480 V: about 1.2 A per kVA. On 208 V: about 2.8 A per kVA.
- A lower power factor always means more current for the same kW, never less.
On the job
Tips
- Carry the amps-per-kVA figure for the voltages you work on — it checks a nameplate in your head.
- When the numbers do not close, work them backwards. A formula applied twice in opposite directions should return what you started with.
- Write the unit on every line. kVA, kW and kVAR mixed up are the usual slip.
Common mistakes
- Using line-to-neutral voltage with √3 (or line-to-line with 3).
- Putting the √3 on current in a wye or on voltage in a delta.
- Adding kW and kVAR to get kVA.
- Forgetting the power factor when going from kW to amps.
Check yourself
-
1.A balanced three-phase heater draws 40 A per line at 600 V. Roughly how many kW is it?
kW = √3 × V × I × PF ÷ 1000 = 1.732 × 600 × 40 × 1.0 ÷ 1000 = 41.6 kW. 24 kW forgets the √3; 72 kW uses 3 instead of √3 with line values.
-
2.A 30 kW three-phase load at 0.9 PF on 480 V. What is the line current?
I = kW × 1000 ÷ (√3 × V × PF) = 30,000 ÷ (1.732 × 480 × 0.9) = 30,000 ÷ 748.2 = 40.1 A. 36.1 A would be the answer at PF 1.0 — a poor power factor always raises the current for the same kW.
-
3.On a 480Y/277 system, what is the voltage from any line to neutral?
Line to neutral on a wye is the line voltage ÷ √3: 480 ÷ 1.732 = 277 V.
-
4.A balanced delta load has 20 A in each winding. What is the line current?
In a delta the line current is √3 × the winding (phase) current: 20 × 1.732 = 34.6 A.
-
5.The three-phase formula uses √3 × V × I. What V and I does it expect?
The √3 is there precisely because you are using line values — what you can measure. If you use line-to-neutral volts instead, the formula becomes 3 × V phase × I.
-
6.A load is 100 kVA at 0.8 PF. What are its kW and kVAR?
kW = kVA × PF = 80. kVAR = √(100² − 80²) = √3600 = 60. The power triangle is a 3-4-5 right triangle here — kW and kVAR do not simply add up to kVA.
Questions people ask
When do I use √3 and when do I use 3?
Use √3 with line-to-line voltage and line current: P = √3 × V LL × I × PF. Use 3 with line-to-neutral (phase) voltage: P = 3 × V LN × I × PF. They give the same answer on a balanced load, because V LL = √3 × V LN.
Is 1.732 accurate enough?
Yes. √3 is 1.73205…; 1.732 gets you within a hundredth of a percent, which is far better than any meter reading you will put into the formula.
What if the load is unbalanced?
For a small imbalance, average the three line currents and use the balanced formula. For a large one, work each phase on its own (V LN × I × PF) and add the three. The neutral current on a wye is a separate question — see the unbalanced-loads lesson.
Related
- Wye and delta: line vs phase — the full lesson
- Three-phase power and power factor — the full lesson
- Unbalanced loads and neutral current
- Three-phase power calculator
- Journeyman formula card — every formula, printable
- Single & three phase formula sheet
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