← All courses KANSAS STATE UNIVERSITY / ECE 685

L08 / Three-phase AC circuits

Three-Phase AC Circuits I

Balanced sets, phase sequence, symmetrical components, and wye/delta voltage and current relations.

Available52 slides
Sequence and Y/Δ line/phase relations

01 / UNDERSTAND & PREDICT

Understand the model, then predict the result

Finalized lecture slides

Open / download original PDF ↗

Follow the original explanations, diagrams, derivations, and examples in slide order, then use the companion experiment below.

L08 original slide 1 of 52
L08 · 1 / 52
Slide text
Loading slide text.

Figures and page order follow the student PDF for this lecture.

At 400 V line voltage with three identical 20+j15 Ω impedances, does changing Y to Δ make line current √3 or 3 times larger?
  • Distinguish phase/line voltage and branch/line current.
  • Check √3 and 30° using voltage differences and terminal KCL.
  • Check three-phase power and zero neutral current in a balanced set.
Three-phase AC circuits: concept and calculation route
Course-authored concept route; the numerical experiment follows below.

Sequence and reference

This module uses source VAN = Vphase∠0° relative to a reference neutral. In abc, B and C are −120° and +120°. Reversing sequence changes angles while retaining balanced-load total power.

Wye Y

V_{LL}=\sqrt3 V_{phase},\qquad I_{line}=I_{phase}

A Y branch sees VLL/√3, and branch current equals line current. For positive sequence VAB = VAN−VBN leads VAN by 30°.

Delta Δ

I_A=I_{AB}-I_{CA},\qquad Z_Y=Z_\Delta/3

A Δ branch sees line voltage. Find IAB, IBC, ICA, then IA = IAB−ICA. Balanced line-current magnitude is √3 times branch current. Equivalent Y impedance is ZΔ/3.

Baseline example: check each step

  1. Y branch voltage = 400/√3 ≈ 230.940 V.
  2. Branch and line current ≈ 9.238 A; impedance angle ≈ 36.870°.
  3. P=5120 W, Q=3840 var, |S|=6400 VA.
  4. With the same branch impedance in Δ: branch current=16 A, line current≈27.713 A, P=15360 W.
Original slide headings for this lecture46
  1. 1Lecture outline
  2. 2Why three phases: constant total power
  3. 3Why three phases: economical conductors
  4. 4Why three phases: a rotating magnetic field
  5. 5Balanced source
  6. 6Phasor-to-waveform mapping
  7. 7Sequence rule
  8. 8Positive abc geometry
  9. 9Negative acb geometry
  10. 10Phase angles and peak times
  11. 11Time-order test
  12. 12Three forms of a complex number
  13. 13Balanced phasor sum
  14. 14Example: positive-sequence phasors
  15. 15Example: negative-sequence phasors
  16. 16Example: instantaneous voltage checks
  17. 17Shifted reference
  18. 18Three sequence basis vectors
  19. 19Why do the weighted sums isolate one sequence?
  20. 20Finding the three sequence coefficients
  21. 21Example: symmetrical components
  22. 22Wye source
  23. 23Orientation before arithmetic
  24. 24Wye voltage: complex subtraction
  25. 25Component triangle
  26. 26Other two line voltages
  27. 27Line-to-line rating
  28. 28Example: two voltage reference choices
  29. 29Current relation
  30. 30Balanced neutral current
  31. 31Delta phase voltage
  32. 32Current notation in delta
  33. 33Delta branch currents from voltage
  34. 34From voltage to line current at terminal A
  35. 35All three line currents from voltage
  36. 36A-line current: algebra from voltage
  37. 37A-line current: voltage geometry
  38. 38Balanced delta current result
  39. 39Positive-sequence delta current phasors
  40. 40Negative-sequence connection shifts
  41. 41Wye and delta: one reference table
  42. 42Wye and delta: visual summary
  43. 43Example: balanced 13.8-kV delta
  44. 44Example: delta branch-current calculation
  45. 45Example: line current from rectangular KCL
  46. 46Example: branch and line current sets
Cross-check the original slides

02 / EXPLORE

Change one input and explain the response

Switch to Δ and multiply both impedance components by 3. Verify the original Y current and power return. Reverse sequence and inspect VAB phase.

Advanced parameters / test readings

Preparing the model.

Load branch voltage—
Branch current—
Line current—
Total real power—
Total reactive power—
Balanced current sum—

Three phase voltages

Three line currents

Current intermediate values and numerical checks

The model uses an ideal balanced source, three identical impedances, and zero line impedance. The source neutral provides a phase reference; a delta load has no neutral conductor.

03 / EDIT & COMPUTE

Edit code to reproduce the model independently

Reproduce the baseline, then modify the parameter scan. The source contains reusable independent model functions; edit the current function and inspect numerical checks.

case is a snapshot of the controls when you press Run. Call solve(case) and assign the final solution to result to plot it.

Download teaching models

The first run needs internet access to download Python. Computation stays in your browser; the solver uses only the standard library.

Ready to run.

Output appears here.
Inspect and edit the model source (advanced)

Edit this module's function and run again. case.module selects the module; solve(case) returns values, plots, and checks. The parameter experiment keeps the original JavaScript reference for comparison.

04 / CHECK & EXPLAIN

Companion experiment practice and feedback

Fixed practice inputs

VLL=400 V RMS, abc sequence, Y branch Z=20+j15 Ω, VAN angle 0°.

Practice uses fixed baseline inputs independently of the controls. Each field displays its tolerance.

±0.05 V
±0.05 A
±0.05 W

With fixed line voltage and identical branch impedance, how much does line current increase from Y to Δ?

Finally, explain in your own words

  1. What are the inputs, references, and main assumptions?
  2. Switch to Δ and multiply both impedance components by 3. Verify the original Y current and power return. Reverse sequence and inspect VAB phase.
  3. Did your code edit change physical parameters, the method, or representation bases? Which check helps identify that?

Passing numerical and understanding checks records this lecture’s companion practice as “practice checks passed.”