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L05 / Single-phase AC circuits

Single-Phase AC Circuits I

Map sinusoidal waveforms to RMS phasors; convert and add polar and rectangular representations.

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Finalized lecture slides

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

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

01 / UNDERSTAND

From one waveform to one phasor

Why can the same voltage be described as both 120 V and 169.7 V? Start with the slide deck's opening question.

v(t) = 120√2 cos(377t + 30°) V
What is the RMS value of this voltage?

By the end, distinguish peak from RMS, convert a sinusoid to polar and rectangular phasors, identify lead/lag, and add compatible voltage phasors before reconstructing the waveform.

Three representations of one voltage: a peak-amplitude waveform, an RMS polar phasor, and an RMS rectangular phasor
All three describe the same signal. The time coefficient is peak; the phasor magnitude is RMS. The waveform illustration is schematic.

1. Declare references and units

v(t) = √2 Vᵣₘₛ cos(ωt + φ), ω = 2πf

Use a cosine reference, RMS phasors, and degree inputs. Convert phase angles to radians in calculation. Angular frequency ω is in rad/s; f is in Hz; T = 1/f.

2. Understand RMS

Vᵣₘₛ² = (1/T) ∫ v²(t) dt = V̂²/2

With cos²θ = (1 + cos2θ)/2, the oscillating term averages to zero over a period. A sinusoid therefore has V̂ = √2 Vᵣₘₛ. RMS gives the DC voltage producing the same average heating in the same resistor.

3. Choose the calculation form

V = Vᵣₘₛ ∠φ = a + jb; a = Vᵣₘₛ cosφ; b = Vᵣₘₛ sinφ

For addition, add real and imaginary parts separately. Recover magnitude with √(a² + b²) and angle with atan2(b, a) to retain the quadrant. Magnitudes add directly only when the two phasors align.

4. Identify lead and lag

Δφ = φV − φI

At the same frequency, voltage at +30° leads current at −20° by 50°. Positive phase shifts a waveform left of the zero-phase cosine. Use the principal difference from −180° to 180°; a 180° separation is opposition without a unique lead/lag direction.

Before adding: match frequency, physical quantity and units, RMS/peak convention, cosine reference, and angle origin. Compare voltage and current by phase; do not add them into a voltage phasor.

Slide example CE-2A: convert → add → reconstruct

Given vₛ(t) above, i(t) = 10√2 cos(377t − 20°) A, and Vₓ = 40∠−60° V, follow the slide sequence. Displayed results use three decimals; calculations retain full precision.

  1. Use RMS and retain phaseVₛ = 120∠30° V; I = 10∠−20° A; φV − φI = 50°.
  2. Convert to rectangular formVₛ = 103.923 + j60.000 V; Vₓ = 20.000 − j34.641 V.
  3. Add matching componentsVₜ = 123.923 + j25.359 V = 126.491∠11.565° V.
  4. Reconstruct and checkvₜ(t) = 126.491√2 cos(377t + 11.565°) V; V̂ₜ = 178.885 V.

The slides use ω = 377 rad/s, giving f ≈ 60.001 Hz, displayed as 60.0 Hz. The experiment retains 377 rather than replacing it with a rounded frequency.

Cross-check the original slide headings32
  1. 1Lecture outline
  2. 2Starting problem
  3. 3Starting-problem solution
  4. 4Sinusoid
  5. 5Waveform magnitudes
  6. 6Frequency and period
  7. 7What phase means
  8. 8Read phase from the time curve
  9. 9Phase difference
  10. 10RMS
  11. 11RMS derivation
  12. 12Three equivalent representations
  13. 13Time domain to polar form
  14. 14Polar form to time domain
  15. 15Polar form to rectangular form
  16. 16Rectangular form to polar form
  17. 17Choose polar or rectangular form
  18. 18Before adding, confirm compatibility
  19. 19Phasor addition is component addition
  20. 20Phasor addition as vectors
  21. 21CE-2A setup
  22. 22CE-2A Step 1
  23. 23CE-2A Step 2
  24. 24CE-2A Step 3
  25. 25CE-2A Step 4
  26. 26CE-2A Step 5
  27. 27CE-2A Step 6
  28. 28Practice problem
  29. 29Rectangular addition
  30. 30Polar reconstruction
  31. 31RMS waveform reconstruction
  32. 32Phasor summary

02 / PREDICT & EXPLORE

Make the phasors and waveforms move together

Predict first: if Vₓ aligns with Vₛ, will the resultant RMS be 120 + 40? If equal magnitudes oppose, is the resultant angle still defined? Then test your prediction.

Resultant RMS126.491 V
Resultant angle11.565°
Resultant peak178.885 V
f / T60.001 Hz / 16.666 ms

Preparing the experiment.

RMS phasors: head-to-tail addition

Translate Vₓ to the tip of Vₛ. Phasors retain their t = 0 reference angles; moving the time cursor does not rotate them.

Time domain: same-frequency voltage sum

Vₛ / vₛVₓ / vₓVₜ / vₜ

Current calculation: components → phasor → waveform
Compare voltage and current by phase
Each curve is divided by its own peak to compare phase only; they do not share physical units or amplitudes. Current RMS remains 10 A.
  1. Change only ω. How does the time scale change? Do phasor magnitudes or angles change?
  2. Move Vₓ from −60° to +30°. Explain why the magnitude changes instead of always equaling the sum of magnitudes.
  3. Select Cancellation. Inspect voltage, RMS, and angle, and explain why a zero vector has no direction.
03 / EDIT & COMPUTE
Edit Python to reproduce and test the result
The code receives the slider inputs. Run the baseline to check the waveform and RMS, then edit a phase or the model. Code results are plotted separately against the current reference.
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 model
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)
The phasor conversion, complex addition, and waveform reconstruction functions execute on the next run. Edit the calculation and compare numerically integrated RMS with phasor magnitude. The parameter experiment retains the original reference model.
Your turn: before solve(case), add case["vx_angle_deg"] = case["vs_angle_deg"]. Predict the RMS, then run it. Next load the phase sweep from −180° to 180° and explain its maximum and minimum magnitudes.
Differences may come from edited inputs or calculations. Check the case before interpreting every difference as an error.
04 / CHECK & EXPLAIN
Solve slide practice CE-2D independently
vₐ(t) = 80√2 cos(377t − 25°) V; vᵦ(t) = 60√2 cos(377t − 70°) V
Find V꜀ = Vₐ + Vᵦ and v꜀(t). This exercise uses the fixed inputs above, independent of the sliders. Find rectangular components, then RMS, angle, and peak.
Voltage tolerance: ±0.05 V; angle tolerance: ±0.05°. Angles are compared modulo 360°.
 
Which pair can be added directly using this lecture's phasor method?
Explain in your own words
  1. Why does a phasor use RMS, while waveform reconstruction needs √2?
  2. Why does changing ω leave the phasor result unchanged but change the waveform's time scale?
  3. Did your Python edit change the inputs, representation, or physical model? How can you check?
L06 connects these voltage and current phasors to real, reactive, and complex power.
Source：L05_single_phase_ac_i/L05_single_phase_ac_i.pdf · Opening problem, CE-2A, and CE-2D follow the current slides. Web diagrams and experiments are course-authored. The model covers sinusoidal quantities at a shared frequency.