AC Circuits and Three-Phase Power

Key Takeaways

  • 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: June 2026

AC Fundamentals and RMS

Alternating current varies sinusoidally with time: v(t)=Vmsin(ωt+ϕ)v(t) = V_m \sin(\omega t + \phi) 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. 707,V_m, \qquad I_{rms} = \frac{I_m}{\sqrt{2}}$$ Every AC voltage rating you see — 120 V outlets, 480 V industrial feeders — is an RMS value; the 120 V outlet actually peaks near 170 V. ** A frequent FE error is plugging peak values into P = VI or treating a quoted line voltage as a peak.

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+jX,Z=R2+X2,θ=arctan ⁣XRZ = R + jX, \qquad |Z| = \sqrt{R^2 + X^2}, \qquad \theta = \arctan\!\frac{X}{R}

ElementImpedanceReactancePhase relationship
ResistorZ = RX = 0V and I in phase
InductorZ = jωLX_L = ωLcurrent LAGS voltage 90°
CapacitorZ = −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 they are 90° out of phase with each other (opposite signs): Z=R+j(ωL1ωC),Z=R2+(XLXC)2Z = R + j\left(\omega L - \frac{1}{\omega C}\right), \qquad |Z| = \sqrt{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: f0=12πLCf_0 = \frac{1}{2\pi\sqrt{LC}}

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:

QuantitySymbolFormulaUnit
Real (active) powerPVI cos φ = I²Rwatts (W)
Reactive powerQVI sin φ = I²XVAR
Apparent powerSVI = 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: PF=cosϕ=PS=RZPF = \cos\phi = \frac{P}{S} = \frac{R}{|Z|} 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):

ConfigurationVoltage relationCurrent relation
Wye (Y)V_L = √3 V_φI_L = I_φ
Delta (Δ)V_L = V_φI_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: Ptotal=3VLILcosϕ,Stotal=3VLILP_{total} = \sqrt{3}\,V_L I_L \cos\phi, \qquad S_{total} = \sqrt{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.

Test Your Knowledge

A series RLC circuit has R = 30 Ω, X_L = 40 Ω, and X_C = 10 Ω. The impedance magnitude is:

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B
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D
Test Your Knowledge

A load draws 5 kW of real power and 4 kVAR of reactive power. The power factor is closest to:

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B
C
D
Test Your Knowledge

In a balanced wye-connected three-phase system with a 480 V line-to-line voltage, the phase (line-to-neutral) voltage is:

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B
C
D
Test Your Knowledge

At resonance in a series RLC circuit, which statement is true?

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B
C
D