Lesson Three-Phase Power · Power and power factor
Three-phase power
Three-phase true power for a balanced system is P = √3 × ELine × ILine × PF (watts). Apparent power is VA = √3 × ELine × ILine. Reactive power uses the same structure with sin θ or measured VARs. Watts, VARs, and VA still form a power triangle — now on three-phase quantities.
Why √3 appears
Each phase delivers power. For a balanced wye or delta, summing the three phases and expressing everything in line quantities produces the universal field formulas with √3.
VA (apparent) = √3 × ELine × ILine P (true) = √3 × ELine × ILine × PF VARs = √3 × ELine × ILine × sin θ (or from the power triangle)
Watts and VARs
True power (watts) does work and makes heat in resistive parts. VARs come from inductors and capacitors. Apparent power (VA) sizes supply equipment. PF = P / VA still holds.
Motors consume lagging VARs; capacitor banks supply leading VARs. Net VARs shrink → PF rises → line current falls for the same kW.
Balanced example pattern
Suppose ELine = 480 V, ILine = 10 A, PF = 0.85 lagging:
VA = 1.732 × 480 × 10 ≈ 8314 VA P = 8314 × 0.85 ≈ 7067 W
Measuring approach
- Two-wattmeter method (classic) or digital three-phase power analyzers
- Clamp each line; measure L-L voltages; read PF on the analyzer
- For rough checks: nameplate HP and efficiency estimate kW; compare to √3 EI PF
Same formulas for wye and delta
Once you use line voltage and line current, you do not switch formulas between wye and delta. The √3 identities already absorbed the topology when you converted to line values.
Horsepower link (field)
Mechanical output relates to electrical input by efficiency:
P_out = P_in × efficiency
Do not expect input watts to equal 746 × HP without including efficiency and PF.
Numbers you should be able to work cold
480 V, 25 A, PF 0.8:
VA = 1.732×480×25 ≈ 20.8 kVA P ≈ 16.6 kW VARs ≈ √(VA²−P²) ≈ 12.5 kVAR
Double-check: P/VA = 0.8.
Per-phase method (sanity check)
For balanced wye: EPhase=ELine/√3, IPhase=ILine, P = 3 × EPhase × IPhase × PF — algebraically same as √3 ELine ILine PF.
If your per-phase and √3 methods disagree, you mixed line/phase quantities.
Metering reminder
A single clamp × one L-L voltage × √3 × PF works for balanced estimates. Unbalanced systems need per-phase power sums or a proper analyzer.
Field case
Situation. A 480 V motor clamps at 30 A. Someone claims “power is 480×30×3.” Another claims “480×30.” Both wrong for three-phase watts without PF and √3.
What happened. Correct apparent power uses √3: VA = 1.732×480×30. True power still needs PF.
Applied lesson. Write the formula before punching numbers. Missing √3 or missing PF are the two most common power errors.
### Teaching pause — say this out loud
Before you leave this lesson, explain the main idea to an imaginary first-month helper in under one minute. If you need the book open to do it, reread How it works once more. Field diagnosis only helps after the concept is yours.
Also sketch the key diagram from memory (triangle, wye/delta, filter shape, or charge curve — whichever this lesson used). Labels beat artistic skill.
### Why this lesson matters on Monday morning
Three-phase power is not trivia. You will meet it when a meter reading looks “impossible,” when a replacement part is almost right, or when a helper asks why the book uses √3 or lead/lag. Master the model here so the next call is pattern recognition, not panic.
Common Monday uses: verify a nameplate against clamps, explain a PF or capacitor change to a customer, or catch a miswired series/parallel or wye/delta assumption before energizing.
In the field
Symptom
Bill vs clamp mismatch; undersized feeder heat; wrong kW estimates
Where to look
PF, balance, true RMS meters vs average-sensing
Likely causes
- Omitted √3
- assumed PF=1
- unbalanced phases
What to measure
- EL, IL (all phases), PF or kW directly
What not to do
- Use single-phase P=EI on a three-phase feeder
Checklist
- I write VA = √3 × EL × IL
- I write P = √3 × EL × IL × PF
- I keep the power triangle clear
- I use line quantities for both wye and delta
- I include efficiency when relating HP to electrical input
- I measure PF instead of assuming 1.0