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AC Circuits and Three-Phase Power

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AC Circuits and Three-Phase Power

Key Takeaways Explain This Topic AI

An AC sinusoid v(t) = V_m sin(ωt + φ) has angular frequency ω = 2πf; US line frequency is 60 Hz.

RMS values do the work of power: V_rms = V_m/√2 ≈ 0.707 V_m; "120 V" household power is RMS (peak ≈ 170 V).

Impedance Z = R + jX; inductive reactance X_L = ωL, capacitive reactance X_C = 1/(ωC); |Z| = √(R² + X²).

ELI the ICE man: in an inductor voltage (E) leads current (I); in a capacitor current (I) leads voltage (E).

Power factor PF = cos φ = P/S; real power P (W), reactive Q (VAR), apparent S (VA), with S² = P² + Q².

Balanced three-phase power: P = √3 V_L I_L cos φ; in wye V_L = √3 V_φ, in delta I_L = √3 I_φ.

Last updated: September 2026

AC Fundamentals and RMS

Alternating current varies sinusoidally with time: v ( t ) = V m sin ⁡ ( ω t + ϕ ) v(t) = V_m \sin(\omega t + \phi) v ( t ) = V m ​ sin ( ω t + ϕ )

where V_m is the peak (amplitude), ω = 2πf is the angular frequency in rad/s, f is the frequency in hertz (60 Hz in North America, 50 Hz in much of the world), and φ is the phase angle. For a sinusoid, the RMS values are: V r m s = V m 2 ≈ 0.707 V m , I r m s = I m 2 V_{rms} = \frac{V_m}{\sqrt{2}} \approx 0.707\,V_m, \qquad I_{rms} = \frac{I_m}{\sqrt{2}} V r m s ​ = 2 ​ V m ​ ​ ≈ 0.707 V m ​ , I r m s ​ = 2 ​ I m ​ ​

Nominal utility voltages such as 120 V and 480 V are RMS values; a sinusoidal 120 V supply peaks near 170 V. Use RMS values for AC power calculations: Real power is P = VᵣₘₛIᵣₘₛ cos φ, where φ is the voltage–current phase difference. The simpler P = VᵣₘₛIᵣₘₛ applies to unity power factor. Do not substitute peak values into these RMS formulas.

Impedance and Reactance

In AC circuits, capacitors and inductors oppose current in a frequency-dependent, phase-shifting way described by impedance Z, the AC generalization of resistance, expressed as a complex number: Z = R + j X , ∣ Z ∣ = R 2 + X 2 , θ = arctan ⁡ ⁣ X R Z = R + jX, \qquad |Z| = \sqrt{R^2 + X^2}, \qquad \theta = \arctan\!\frac{X}{R} Z = R + j X , ∣ Z ∣ = R 2 + X 2 ​ , θ = arctan R X ​

  • Element Impedance Reactance Phase relationship
  • Resistor Z = R X = 0 V and I in phase
  • Inductor Z = jωL X_L = ωL current LAGS voltage 90°
  • Capacitor Z = −j/(ωC) X_C = 1/(ωC) current LEADS voltage 90°

The mnemonic ELI the ICE man captures the phasing: in an inductor (L), voltage E leads current I (ELI); in a capacitor (C), current I leads voltage E (ICE). Reactance grows with frequency for an inductor (X_L = ωL) but shrinks for a capacitor (X_C = 1/ωC) — at very high frequency an inductor blocks and a capacitor passes current.

Series RLC, Resonance, and the Power Triangle

For a series RLC circuit the reactances of L and C subtract because their imaginary components have opposite signs (the inductor and capacitor voltages are 180° apart for the same series current): Z = R + j ( ω L − 1 ω C ) , ∣ Z ∣ = R 2 + ( X L − X C ) 2 Z = R + j\left(\omega L - \frac{1}{\omega C}\right), \qquad |Z| = \sqrt{R^2 + (X_L - X_C)^2} Z = R + j ( ω L − ω C 1 ​ ) , ∣ Z ∣ = R 2 + ( X L ​ − X C ​ ) 2 ​

At the resonant frequency the two reactances are equal (X_L = X_C) and cancel, leaving Z = R (purely resistive) and current at its maximum: f 0 = 1 2 π L C f_0 = \frac{1}{2\pi\sqrt{LC}} f 0 ​ = 2 π L C ​ 1 ​

Worked impedance example. A series circuit has R = 30 Ω, X_L = 40 Ω, X_C = 10 Ω. Net reactance = 40 − 10 = 30 Ω (inductive). |Z| = √(30² + 30²) = √1,800 = 42.4 Ω , with phase angle arctan(30/30) = 45° (current lags). At 120 V_rms, I = 120/42.4 = 2.83 A.

The power triangle. In AC, current and voltage may be out of phase, splitting power into three parts:

  • Quantity Symbol Formula Unit
  • Real (active) power P VI cos φ = I²R watts (W)
  • Reactive power Q VI sin φ = I²X VAR
  • Apparent power S VI = I²·|Z| VA

They relate by S² = P² + Q², i.e. S = √(P² + Q²). Only real power P does useful work or shows as heat; reactive power Q sloshes back and forth into inductors/capacitors each cycle. The power factor is: P F = cos ⁡ ϕ = P S = R ∣ Z ∣ PF = \cos\phi = \frac{P}{S} = \frac{R}{|Z|} PF = cos ϕ = S P ​ = ∣ Z ∣ R ​

Inductive (lagging) loads — motors, transformers — dominate industry. Utilities penalize low power factor because it raises line current for the same useful power; adding capacitors corrects a lagging PF toward unity , cutting current and losses.

Three-Phase Power

Large power systems use three-phase AC: three voltages 120° apart, delivering constant total power and using conductors efficiently. Loads connect in wye (Y / star) or delta (Δ) :

  • Configuration Voltage relation Current relation
  • Wye (Y) V_L = √3 V_φ I_L = I_φ
  • Delta (Δ) V_L = V_φ I_L = √3 I_φ

where V_L is line-to-line voltage, V_φ phase voltage, I_L line current, I_φ phase current. For a balanced load, total real and apparent power are: P t o t a l = 3 V L I L cos ⁡ ϕ , S t o t a l = 3 V L I L P_{total} = \sqrt{3}\,V_L I_L \cos\phi, \qquad S_{total} = \sqrt{3}\,V_L I_L P t o t a l ​ = 3 ​ V L ​ I L ​ cos ϕ , S t o t a l ​ = 3 ​ V L ​ I L ​

Using line quantities, the √3 already accounts for all three phases — do not multiply by 3 again. The common 480/277 V building system is wye: 480 V between lines, 277 V (= 480/√3) line-to-neutral.

Worked power-factor example. A load draws P = 5 kW of real power and Q = 4 kVAR of reactive power. Apparent power S = √(P² + Q²) = √(5² + 4²) = √41 = 6.40 kVA, so PF = P/S = 5/6.40 = 0.78 lagging . To raise the PF to unity, a capacitor bank supplying 4 kVAR of leading reactive power is added, canceling the load's 4 kVAR and dropping line current.

Worked three-phase example. A balanced three-phase motor runs at V_L = 480 V, draws I_L = 10 A, PF = 0.85, with 90% efficiency. Input power P_in = √3 × 480 × 10 × 0.85 = 7,064 W. Output power P_out = 0.90 × 7,064 = 6,358 W ≈ 8.5 hp (÷ 745.7 W/hp).

Measuring devices. A voltmeter (high internal resistance) connects in parallel; an ammeter (low resistance) connects in series; a wattmeter uses a parallel voltage coil and a series current coil; an ohmmeter is used only on a de-energized circuit. Connecting an ammeter in parallel (a near-short) is a classic destructive mistake.

The Power Triangle and Power-Factor Correction

AC power splits into three quantities related by the power triangle : real power P (watts, does work), reactive power Q (VAR, shuttles energy in reactances), and apparent power S (VA). They satisfy S = √(P²+Q²) and power factor pf = cosθ = P/S.

  • Quantity Symbol/unit Formula
  • Real power P (W) S·cosθ
  • Reactive power Q (VAR) S·sinθ
  • Apparent power S (VA) V·I (rms)
  • Power factor pf P/S = cosθ

Inductive loads (motors) draw lagging reactive power, lowering pf and forcing larger currents for the same real power. Power-factor correction adds capacitors to supply that reactive power locally, raising pf toward 1, cutting line current and losses.

Worked Example

A load draws 8 kW at pf = 0.8 lagging: S = 8/0.8 = 10 kVA, Q = √(10²−8²) = 6 kVAR — the reactive demand a correction capacitor must offset.