2.2 Single-Phase & Three-Phase AC Fundamentals and Power Factor
Key Takeaways
- AC sine wave characteristics: Root-Mean-Square (RMS) voltage represents the equivalent DC thermal heating value, where V_RMS = 0.707 * V_peak and V_peak = 1.414 * V_RMS. Multimeters and NEC ratings always express AC quantities in RMS.
- AC impedance (Z) is the vector sum of pure resistance (R) and net reactance (X = X_L - X_C): Z = √(R² + (X_L - X_C)²). Inductive reactance (X_L = 2πfL) causes current to lag voltage, while capacitive reactance (X_C = 1 / (2πfC)) causes current to lead voltage.
- Power Factor (PF) is the ratio of True Power (P in watts) to Apparent Power (S in VA): PF = W / VA = cos(θ). Inductive motor loads create lagging power factors that draw excess line current without producing mechanical work.
- In balanced 3-phase Wye systems, Line-to-Line voltage is √3 (1.732) times Line-to-Neutral voltage (V_LL = 1.732 * V_LN), while Line current equals Phase current (I_Line = I_Phase). Standard configurations include 208Y/120V and 480Y/277V.
- In balanced 3-phase Delta systems, Line voltage equals Phase voltage (V_LL = V_Phase), while Line current is √3 times Phase current (I_Line = 1.732 * I_Phase). Three-phase apparent power is always calculated as VA = √3 * E_LL * I_Line.
2.2 Single-Phase & Three-Phase AC Fundamentals and Power Factor
Alternating current (AC) is the universal medium for electrical power transmission and distribution. Unlike direct current, AC periodically reverses its direction of flow and continuously changes its magnitude. This sinusoidal behavior introduces reactive components—inductance and capacitance—that shift the timing relationship between voltage and current waveforms.
For the Idaho Journeyman exam, you must master the mathematical relationships of single-phase and three-phase circuits, understand the square root of three (1.732 ≈ 1.732) conversion factor, and solve complex power factor problems.
1. AC Sine Wave Metrics & Principles
A pure AC voltage wave is generated by a conductor rotating through a uniform magnetic field, producing a sinusoidal waveform described by the function e(t) = V_peak sin(2π f t).
+-----------------------------------------------------------------------------+
| AC SINE WAVE CHARACTERISTICS |
| |
| +V_peak --+ * * * |
| | * * |
| +V_RMS --+ - - * - - - - - * - - - - - - - - - - - - - |
| | * * |
| 0 Volts --+---*---------------+---------------*----------> Time |
| | * * |
| -V_RMS --+ - - - - - - - - - - * - - - - - * - - - - - |
| | * * |
| -V_peak --+ * * * |
| |<------------- 1 Cycle (360°) ------------->| |
+-----------------------------------------------------------------------------+
Core Waveform Definitions
- Frequency (f): The number of complete cycles (360^°) executed per second, measured in Hertz (Hz). North American electrical systems operate at standard 60 Hz.
- Period (T): The time required to complete one full cycle: T = 1 / f. At 60 Hz, T = 1 / 60 = 0.01667 seconds = 16.67 ms.
- Peak Voltage (V_peak): The maximum instantaneous voltage reached in either the positive or negative half-cycle.
- Peak-to-Peak Voltage (V_pk-pk): The total amplitude between positive and negative peaks: V_pk-pk = 2 x V_peak.
- RMS (Root-Mean-Square) Effective Voltage (V_RMS): The effective value of an AC voltage that delivers the exact same heating power to a resistive load as an equivalent DC voltage.
V_RMS = V_peak x 0.7071 = V_peak / sqrt(2) V_peak = V_RMS x 1.4142 = V_RMS x sqrt(2) V_average = V_peak x 0.637
[!NOTE] Unless explicitly stated otherwise, every voltage and current specified on nameplates, test instruments, blueprints, and the NEC is an RMS value. A nominal 120 V receptacle actually swings between +169.7 V and -169.7 V peak (120 x 1.4142).
2. Reactance & Impedance in AC Circuits
When AC flows through coils (motors, transformers, ballasts) or capacitors, the changing electromagnetic and electrostatic fields create an opposition to current flow known as reactance (X).
+-----------------------------------------------------------------------------+
| REACTANCE & PHASE ANGLE MNEMONICS |
| |
| " E L I " " I C E " |
| In an Inductive Circuit: In a Capacitive Circuit: |
| Voltage (E) Leads Current (I) Current (I) Leads Voltage (E) |
| Current (I) LAGS Voltage (E) Voltage (E) LAGS Current (I) |
+-----------------------------------------------------------------------------+
Inductive Reactance (X_L)
Inductors (wire coils) oppose changes in current by generating a counter-electromotive force (CEMF). Inductive reactance is directly proportional to frequency and inductance (L in Henrys):
X_L = 2 π f L
Capacitive Reactance (X_C)
Capacitors store electrostatic charge and oppose changes in voltage. Capacitive reactance is inversely proportional to frequency and capacitance (C in Farads):
X_C = (1) / (2 π f C)
Total Impedance (Z)
Impedance (Z) is the total opposition to AC current flow, combining pure DC resistance (R) and net reactance (X = X_L - X_C). Because resistance and reactance are 90^° out of phase, they must be added vectorially using the Pythagorean theorem:
Z = sqrt(R^2 + (X_L - X_C)^2)
Impedance Triangle (Vector Addition)
/|
/ |
/ |
Impedance / | Net Reactance
(Z)/ | (X = X_L - X_C)
/ |
/ θ |
+-------+
Resistance (R)
Worked Example: AC Impedance and Current
An industrial motor coil has a winding resistance of 8 Ω and an inductance of 0.0159 H. It is connected to a 120 V, 60 Hz supply. Calculate the inductive reactance, circuit impedance, and current.
- Calculate Inductive Reactance (X_L): X_L = 2 x π x 60 Hz x 0.0159 H = 377 x 0.0159 ≈ 6.0 Ω
- Calculate Total Impedance (Z): Z = sqrt(R^2 + X_L^2) = sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10.0 Ω
- Calculate Circuit Current (I): I = E / Z = (120 V) / (10.0 Ω) = 12.0 A
3. The Power Triangle & Power Factor
In AC circuits with reactive components, power is divided into three distinct vector components forming the Power Triangle:
+-----------------------------------------------------------------------------+
| THE POWER TRIANGLE |
| |
| Apparent Power (S) in Volt-Amperes (VA) |
| /| |
| / | |
| / | Reactive Power (Q) |
| / | in VARs (kVAR) |
| / | (Magnetizing Current) |
| / θ | |
| +------+ |
| True Power (P) in Watts (kW) |
| (Actual Work / Thermal Heat) |
+-----------------------------------------------------------------------------+
Power Component Definitions:
- True Power (P): Measured in Watts (W) or Kilowatts (kW). Represents the actual energy consumed to perform mechanical work or produce heat: P = E x I x cos(θ) = I^2 x R.
- Reactive Power (Q): Measured in Volt-Amperes Reactive (VAR) or kVAR. Power alternately stored and returned to the circuit by inductive magnetic fields: Q = E x I x sin(θ) = I^2 x X.
- Apparent Power (S): Measured in Volt-Amperes (VA) or kVA. The total vector combination of True and Reactive power supplied by the utility: S = E x I = sqrt(P^2 + Q^2).
Power Factor (PF)
Power factor is the mathematical ratio of True Power to Apparent Power:
PF = (True Power (W)) / (Apparent Power (VA)) = cos(θ) True Power (Watts) = Apparent Power (VA) x PF = E x I x PF
[!WARNING] Why Low Power Factor Matters: An inductive motor operating at a 0.70 power factor requires 1.43 times more line current (1 / 0.70) than a unity-PF load doing the same mechanical work. This excess current increases I^2R thermal line losses, drops branch-circuit voltage, and causes utilities to levy steep power factor surcharge penalties.
4. Three-Phase AC Principles: Wye vs. Delta Systems
Three-phase AC power is generated by three separate coils spaced 120^° apart mechanically within the alternator stator, producing three sinusoidal voltages of equal magnitude shifted by 120^° in phase.
+-----------------------------------------------------------------------------+
| THREE-PHASE SYSTEM ARCHITECTURES |
| |
| WYE (STAR) CONNECTION DELTA CONNECTION |
| |
| Phase A Phase A |
| o o |
| \ / \ |
| \ Phase B / \ |
| Neutral o--o---o / \ |
| / o-------o |
| / Phase B Phase C |
| o |
| Phase C |
+-----------------------------------------------------------------------------+
The 1.732 (1.732) Geometric Relationship
Because phases are separated by 120^° rather than 180^°, line-to-line voltages do not equal 2 x V_phase. Instead, vector subtraction yields the fundamental multiplier:
1.732 = 2 x cos(30^°) = 2 x 0.866025 = 1.73205
Summary of 3-Phase Rules
| Property | Wye (Y) 4-Wire System | Delta (Δ) 3-Wire System | High-Leg Delta 4-Wire |
|---|---|---|---|
| Voltage Relation | V_Line-Line = 1.732 x V_Line-Neutral | V_Line-Line = V_Phase | Phase A & C to N = 120 V;<br>Phase B (High-Leg) to N = 208 V |
| Current Relation | I_Line = I_Phase | I_Line = 1.732 x I_Phase | Standard branch rules apply |
| Standard Voltages | 208Y/120V, 480Y/277V | 240V, 480V | 240/120V (NEC 110.15 High-Leg) |
| Common Use | Commercial lighting, receptacles, motors | Heavy industrial plants, ungrounded/corner-grounded | Light commercial with heavy 240V 3-phase motor loads |
[!CAUTION] NEC 110.15 High-Leg Identification: On a 4-wire, delta-connected system where the midpoint of one phase winding is grounded to supply 120V lighting loads, the phase conductor with the higher voltage-to-ground (120V x 1.732 = 208V) must be identified by an outer finish that is ORANGE in color (or by tagging) and must be connected to Phase B in panelboards and switchboards.
5. Three-Phase Power Formulas & Worked Calculations
When calculating power in balanced three-phase systems, the formulas always incorporate the 1.732 ≈ 1.732 constant:
VA_Three-Phase = 1.732 x E_Line-Line x I_Line = 1.732 x E_LL x I_L P_Watts = 1.732 x E_Line-Line x I_Line x PF = 1.732 x E_LL x I_L x PF I_Line = (VA_Three-Phase) / (1.732 x E_Line-Line) = (P_Watts) / (1.732 x E_Line-Line x PF)
Comprehensive Step-by-Step 3-Phase Problems
Problem 2.1: Sizing a 3-Phase Transformer Secondary Feeder
A 75 kVA, 3-phase transformer has a 480V primary and a 208Y/120V secondary. Calculate the rated full-load line current output on the secondary side.
Given: VA = 75,000 VA, E_Line-Line = 208 V I_Line = (VA) / (1.732 x E_LL) = (75,000) / (1.732 x 208) = 75,000 / 360.256 = 208.19 Amperes (A standard 208Y/120V 75 kVA transformer supplies approximately 208 A per phase at full load).
Problem 2.2: Sizing Feeders for a 3-Phase Industrial Motor
A 480 V, 3-phase, 50 HP induction motor operates at an efficiency of 91% and a lagging power factor of 0.86. Calculate the actual full-load line current drawn by the motor.
- Step 1: Convert Output Mechanical Horsepower to Electrical Watts Input: P_output = 50 HP x 746 W/HP = 37,300 Watts P_input = P_output / Efficiency = (37,300 W) / (0.91) = 40,989.0 Watts
- Step 2: Calculate Apparent Power (VA): VA = P_input / PF = 40,989.0 / 0.86 = 47,661.6 VA
- Step 3: Calculate Line Current (I_Line): I_Line = (VA) / (1.732 x E_LL) = (47,661.6) / (1.732 x 480) = 47,661.6 / 831.36 = 57.33 Amperes (Note: For NEC branch circuit sizing, electricians refer to NEC Table 430.250, which lists a standardized FLC of 65A for a 50HP 460V motor to ensure safety margins).
What is the peak-to-peak voltage of a nominal 120V AC RMS branch circuit?
A 480V 3-phase balanced load draws 45 Amperes of line current with a lagging power factor of 0.80. What is the true power in kilowatts consumed by this load?
In an ungrounded 3-phase 4-wire 240/120V high-leg delta panelboard, what is the nominal voltage measured from the high-leg (Phase B) conductor to the neutral conductor?