The three-phase formula uses √3 × V × I. What V and I does it expect?

A. Line-to-line voltage and line current

B. Line-to-neutral voltage and line current

C. Line-to-line voltage and winding current

D. Any voltage, as long as you use the average current

Repaso para oficiales 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
Tensión	V line = √3 × V phase	V line = V phase
Corriente	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 power triangle. kVA is the long side, so kW and kVAR never add up to it directly — kVA² = kW² + kVAR².

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.

En la obra

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

0 right of 0 answered 6 questions · pick one answer each

1.A balanced three-phase heater draws 40 A per line at 600 V. Roughly how many kW is it?
2.A 30 kW three-phase load at 0.9 PF on 480 V. What is the line current?
3.On a 480Y/277 system, what is the voltage from any line to neutral?
4.A balanced delta load has 20 A in each winding. What is the line current?
5.The three-phase formula uses √3 × V × I. What V and I does it expect?
6.A load is 100 kVA at 0.8 PF. What are its kW and kVAR?

Answers

B. 41.6 kW — 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.
B. 40.1 A — 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.
B. 277 V — Line to neutral on a wye is the line voltage ÷ √3: 480 ÷ 1.732 = 277 V.
C. 34.6 A — In a delta the line current is √3 × the winding (phase) current: 20 × 1.732 = 34.6 A.
A. Line-to-line voltage and line current — 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.
B. 80 kW, 60 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.

Preguntas frecuentes

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.

Ver también

Wye and delta: line vs phase — the full lesson
Three-phase power and power factor — the full lesson
Unbalanced loads and neutral current
Calculadora de potencia trifásica
Journeyman formula card — every formula, printable
Single & three phase formula sheet

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