3.3 Three-Phase Systems and Power Factor

Why This Matters for the Texas Journeyman Exam

The TDLR/PSI journeyman exam is built around commercial and industrial installations, and almost every commercial service in Texas is three-phase. The Calculations part tests three-phase power formulas directly (motor load calcs, feeder sizing, service demand calculations), and the Knowledge part expects you to recognize standard voltage configurations (208Y/120, 480Y/277) and understand how power factor affects conductor sizing. A candidate who cannot distinguish wye from delta, or who forgets the 1.732 factor, will lose points on a large cluster of questions.

Three-Phase System Configurations

Wye (Y) Systems

A wye (or "star") system has three phase windings connected at a common neutral point. Four wires are brought out: three ungrounded (phase) conductors and one grounded (neutral) conductor.

The defining relationship in a wye system is:

E_line-to-line = 1.732 × E_line-to-neutral

Common wye voltages on the exam:

The "208Y/120" notation tells you both voltages at once: 208 V phase-to-phase, 120 V phase-to-neutral. Verify: 120 × 1.732 = 207.8 ≈ 208 V. Likewise 277 × 1.732 = 480.1 ≈ 480 V.

Delta (Δ) Systems

A delta system has three windings connected end-to-end in a closed loop. Only three wires are brought out (no neutral), and the line-to-line voltage equals the winding (phase) voltage:

E_line-to-line = E_phase

Common delta voltages: 240 V, 480 V, 600 V. Some deltas are corner-grounded or center-tapped (240/120 V high-leg delta, NEC 110.15), but the exam primarily treats delta as a 3-wire, no-neutral system.

Quick Comparison

The 1.732 Factor (√3)

The number 1.732 is the square root of 3. It appears in every three-phase formula because at any instant, a balanced three-phase load draws current through two conductors, and the vector sum of two phase currents 120° apart is √3 times one phase current. You will use 1.732 in:

• Voltage relationships (wye): E_LL = 1.732 × E_LN
• Power: P = 1.732 × E_LL × I_line × PF
• Voltage drop: VD = 1.732 × I × R × L ÷ 1,000

Memorize 1.732 — you will use it dozens of times on the exam.

Three-Phase Power Formula

For a balanced three-phase load, the real power is:

P = 1.732 × E_LL × I_line × PF

Where:

• P = real power in watts (W)
• E_LL = line-to-line voltage in volts (V)
• I_line = line current in amperes (A)
• PF = power factor (decimal, 0 to 1)

If the load is purely resistive (PF = 1.0), this simplifies to P = 1.732 × E × I.

Worked Example 1 — Resistive Three-Phase Load

A 480 V three-phase electric heater (resistive, PF = 1.0) draws 60 A per phase. What is the power?

P = 1.732 × 480 × 60 × 1.0 = 49,882 W (≈ 49.9 kW)

Worked Example 2 — Inductive Load with Power Factor

A 208Y/120 V three-phase motor draws 30 A line current at a power factor of 0.85. What real power does it consume?

P = 1.732 × 208 × 30 × 0.85

P = 1.732 × 208 × 30 = 10,806 VA (apparent power)

P = 10,806 × 0.85 = 9,185 W (≈ 9.2 kW)

The apparent power is 10.8 kVA; the real power is 9.2 kW. The difference is reactive power (VARs), which the motor uses to magnetize its windings but does not convert to mechanical work.

Power Factor (PF)

Power factor is the ratio of real power (watts) to apparent power (volt-amperes):

PF = W ÷ VA = cos θ

Where θ is the phase angle between voltage and current.

Why Power Factor Matters on the Exam

For the same real power, a lower power factor means more current:

I = P ÷ (1.732 × E × PF)

As PF drops, I rises. Higher current means:

• Larger conductors (NEC 210.19, 215.2)
• Larger overcurrent protection (NEC 240.4)
• More voltage drop (Section 3.2 formulas)
• Higher transformer kVA rating required

Worked Example 3 — Power Factor and Current

A 480 V three-phase load consumes 30 kW of real power.

• At PF = 1.0: I = 30,000 ÷ (1.732 × 480 × 1.0) = 30,000 ÷ 831.4 = 36.1 A
• At PF = 0.7: I = 30,000 ÷ (1.732 × 480 × 0.7) = 30,000 ÷ 582.0 = 51.5 A

The 0.7 PF load draws 43% more current for the same 30 kW. Every conductor, breaker, and feeder in that circuit must be sized for 51.5 A, not 36.1 A.

Worked Example 4 — Computing Power Factor

A load draws 40 A at 480 V three-phase. A wattmeter reads 25,000 W. What is the power factor?

Apparent power = 1.732 × 480 × 40 = 33,254 VA

PF = W ÷ VA = 25,000 ÷ 33,254 = 0.752 (≈ 0.75 lagging)

NEC Treatment of Power Factor

The NEC does not mandate a specific power factor for most installations, but it accounts for motor PF implicitly through its tables:

• NEC Table 430.250 — Full-load currents for 3-phase AC motors. The table values already incorporate typical motor PF and efficiency. You do not compute motor current from P = 1.732 × E × I × PF using nameplate horsepower; you use the table value directly.
• NEC 430.22 — Conductors serving a single continuous-duty motor must be sized at 125% of the FLC from Table 430.250 (or the nameplate FLC if higher).
• NEC 220.18(A) — Branch-circuit motor loads are computed using the Table 430.250 current, not a calculated current from nameplate kW.
• NEC Article 460 — Power-factor correction capacitors: rules for installation, discharge means, and conductor sizing. The exam may test the 460.12 rating requirement and the 460.28 discharge-circuit rules.

Worked Example 5 — Motor Conductor Sizing

A 50 hp, 460 V, 3-phase motor is supplied by a feeder. Size the conductor.

1. From NEC Table 430.250: 50 hp at 460 V → 77 A FLC (this value embeds the motor's typical PF and efficiency — do not recompute).
2. From NEC 430.22: Conductor ampacity = 125% × 77 A = 96.25 A.
3. From NEC Table 310.16 (75 °C Cu): #2 AWG Cu = 115 A — meets 96.25 A.

So the minimum conductor is #2 AWG copper (before any voltage-drop check). Note that we never used the power formula directly — the NEC table bypasses it. The formula matters when you are asked to verify nameplate data, compute power factor from wattmeter readings, or size a capacitor bank.

Practical Effects of Low Power Factor

Correction is typically achieved by installing shunt capacitors (NEC Article 460) or synchronous condensers that supply leading reactive power to cancel the lagging reactive power of inductive loads. A corrected system approaches PF = 1.0, reducing line current and voltage drop.

Key Formulas for Three-Phase and Power Factor