Sinusoidal Steady State, Phasors and AC Power
RMS and average values, phasors and impedance, series and parallel AC circuits, real, reactive, apparent and complex power, power factor correction and maximum power transfer in AC.
In sinusoidal steady state, every voltage and current is a sine wave at the source frequency. Phasors turn the differential equations into complex algebra.
RMS and average values
| Waveform | RMS | Average (half cycle) | Form factor |
|---|---|---|---|
| Sine, peak V_m | V_m/√2 | 2V_m/π | 1.11 |
| Square, peak V_m | V_m | V_m | 1 |
| Triangle, peak V_m | V_m/√3 | V_m/2 | 1.155 |
Form factor = RMS/average; crest (peak) factor = peak/RMS (√2 for a sine).
Phasors and impedance
- v(t) = V_m cos(ωt + φ) ↔ phasor V = V∠φ (using RMS magnitudes in power work).
- Impedances: Z_R = R, Z_L = jωL, Z_C = 1/(jωC) = −j/(ωC).
- In an inductor current lags voltage by 90°; in a capacitor it leads by 90° (ELI the ICE man).
- Impedances combine like resistances. Admittance Y = 1/Z = G + jB.
AC power
With RMS phasors V and I and the angle φ between them:
- Real power P = VI cos φ (W).
- Reactive power Q = VI sin φ (VAR); positive for inductive loads.
- Apparent power S = VI (VA), with S² = P² + Q².
- Complex power S = V I* = P + jQ.
- Power factor = cos φ = P/S (lagging for inductive loads).
Power factor correction: a capacitor in parallel supplies reactive power Q_C = P(tan φ₁ − tan φ₂), with C = Q_C/(ωV²).
Maximum power transfer (AC): Z_L = Z_th* (complex conjugate), giving P_max = |V_th|²/(4R_th).
Worked example
A load takes 10 kW at 0.8 pf lagging from 230 V, 50 Hz. Find the capacitor to raise the pf to 1.
- tan φ₁ = 0.75, so Q = 10 × 0.75 = 7.5 kVAR.
- C = 7500/(2π × 50 × 230²) = 451 µF.
Practice
A sine wave has a peak of 325 V. What is its RMS value in V? (Give 1 decimal place.)
Show answer and explanation
Answer: 229.8V_rms = 325/√2 = 229.8 V.
A load takes 3 kW and 4 kVAR. What is its apparent power in kVA?
Show answer and explanation
Answer: 5S = √(P² + Q²) = √(9 + 16) = 5 kVA (pf 0.6).
What is the reactance (Ω) of a 10 µF capacitor at 50 Hz? (Give 2 decimal places.)
Show answer and explanation
Answer: 318.31X_C = 1/(2πfC) = 1/(2π × 50 × 10⁻⁵) = 318.31 Ω.
In a purely capacitive circuit, the current:
Show answer and explanation
Answer: Ai = C dv/dt, so the current peaks a quarter cycle before the voltage.