3.1 Three-Phase Power & Wye/Delta Relationships

The Three-Phase Power Formula

Quick Answer: For a balanced three-phase load, total real power P = √3 × VLL × IL × PF, apparent power S = √3 × VLL × IL, and reactive power Q = √3 × VLL × IL × sin θ. Use line-to-line voltage and line current — never mix in a phase voltage by mistake.

The single-phase formula P = V × I × PF extends to three phases by multiplying by 3 if you use per-phase quantities: P = 3 × Vphase × Iphase × PF. In a wye connection Iphase = IL and Vphase = VLN, so P = 3 × VLN × IL × PF. Because VLN = VLL / √3, this simplifies to P = √3 × VLL × IL × PF. In a delta connection Vphase = VLL and Iphase = IL / √3, which produces the same √3 result. The formula is identical for both connections when you use line quantities.

θ is the angle between the phase voltage and phase current. Power factor PF = cos θ. A 0.85 PF means θ ≈ 31.8° and sin θ ≈ 0.527.

Where the √3 Comes From

The √3 factor is not a fudge number — it falls directly out of the 120° phase displacement. In a wye system the line-to-line voltage is the vector difference of two phase voltages that are each VLN in magnitude and 120° apart. Using the law of cosines, the magnitude of that difference is VLL = √(V LN² + V LN² − 2·V LN²·cos 120°) = √(2·V LN²·(1 − (−0.5))) = √(3·V LN²) = √3 · V LN. The same geometry, applied to currents in a delta, gives IL = √3 · Iphase. Remembering the geometry makes the wye/delta tables below intuitive instead of memorized: voltage converts by √3 in wye, current converts by √3 in delta, and the conversion always goes the same way because the underlying 120° relationship is identical.

Wye vs. Delta Voltage and Current Relationships

A 480Y/277 V service is a wye system: 480 V between any two lines, 277 V from any line to neutral, and the neutral is brought out as a fourth conductor. A 480 V delta service has 480 V between lines and no neutral.

Phase Rotation: ABC vs ACB

Positive sequence (ABC) means phase A reaches its positive peak first, then B lags A by 120°, then C lags B by 120°. Negative sequence (ACB) reverses the order — C leads B leads A. Phase rotation matters for two big reasons on the NETA exam:

• Motor rotation: A three-phase induction motor's shaft direction follows the phase rotation. Swapping any two leads reverses rotation. Before commissioning a motor, verify rotation with a phase-rotation tester — reversing a 1,500 hp pump motor can shear a coupling or wreck a pump.
• Directional relays: ANSI 67 directional overcurrent and 87 transformer differential relays use phase rotation to establish torque direction. Wrong rotation makes the relay trip for through-faults or fail to trip for internal faults. Always confirm rotation during commissioning with a phase-sequence indicator before energizing protection.

Balanced vs. Unbalanced Systems

A balanced three-phase system has equal magnitudes on all three phases and 120° displacement between them. In a balanced wye system the vector sum of the three line currents is zero, so a neutral conductor (if present) carries zero current at fundamental frequency. In an unbalanced system the magnitudes or angles differ, the neutral carries the imbalance, and negative-sequence currents appear (covered in 3.2).

Worked Example

A 480 V three-phase motor draws 100 A per phase at 0.85 power factor. Find real, apparent, and reactive power.

1. Apparent power: S = √3 × VLL × IL = 1.732 × 480 × 100 = 82,944 VA ≈ 82.9 kVA.
2. Real power: P = S × PF = 82.9 × 0.85 = 70,466 W ≈ 70.5 kW. (Direct: √3 × 480 × 100 × 0.85 = 70,656 W — the small difference is rounding of √3.)
3. Reactive power: Q = S × sin θ = 82.9 × sin(31.8°) = 82.9 × 0.527 = 43.7 kVAR ≈ 43.7 kVAR.
4. Line-to-neutral voltage: VLN = 480 / √3 = 277 V.

If PF were unity (1.0), P would equal S at 82.9 kW and Q would be zero. As PF drops, more current flows for the same real power — which is why utilities and NETA test reports track PF.

Delta Current Example

A 480 V delta-connected load draws 50 A through each phase winding. What is the line current and the total apparent power?

1. Line current: IL = √3 × Iphase = 1.732 × 50 = 86.6 A. In a delta the line current splits between two windings, so the line current is larger than the winding current by √3.
2. Apparent power: S = √3 × VLL × IL = 1.732 × 480 × 86.6 = 72,000 VA ≈ 72 kVA. Cross-check per phase: each winding sees Vphase = VLL = 480 V at Iphase = 50 A, so 3 × 480 × 50 = 72,000 VA. The two methods agree, which confirms you used the correct √3 relationship for each quantity.

Delta-Wye Transformer Phase Shift

Most NETA acceptance tests involve at least one delta-wye transformer, and these introduce a 30° phase shift between the primary (delta) and secondary (wye) voltages. The high-voltage delta winding's line-to-line voltage leads the low-voltage wye's line-to-neutral voltage by 30° for a standard Dy1 connection. This shift matters because transformer differential relays (ANSI 87T) compare currents on both sides; the relay's compensation windings or numerical phase-shifting must cancel the 30° shift, or the relay sees a false differential current and trips on through-load. When commissioning a delta-wye bank, verify the vector group nameplate (e.g., Dy1, Dy11) and confirm the relay's phase compensation matches it — a common NETA Level 2 commissioning task.

Exam Trap

Do not use VLN in the √3 formula. The √3 factor already converts line-to-line voltage to the per-phase quantity. Mixing VLN into P = √3 × VLL × IL × PF double-counts the √3 and gives an answer 1.732× too high. If a question gives 277 V and 100 A in a wye system, either convert to 480 V line-to-line and use the √3 formula, or use P = 3 × 277 × 100 × PF directly.