GEK-3 — AC Systems and Power

Read AC quantities correctly

Alternating current changes magnitude and direction periodically. A common power-system waveform is sinusoidal. Frequency (f) is cycles per second, measured in hertz (Hz). At 60 Hz, the period of one cycle is T = 1/f = 1 ÷ 60 s⁻¹ = 0.0167 s, or 16.7 ms.

Ordinary AC nameplate voltage and current values are root-mean-square (RMS) values. RMS lets AC heating effect be compared with DC: a 120 V RMS sine wave produces the same resistive heating as 120 V DC in the same resistance. Do not substitute peak voltage into a calculation that gives RMS voltage. For a sine wave, peak voltage is about √2 × V_RMS, but most LLE calculations use the stated RMS value directly.

Resistance, reactance, and impedance

Resistance dissipates real power. Inductors and capacitors also oppose AC, but their opposition depends on frequency and is called reactance, measured in ohms. Ideal inductive reactance is X_L = 2πfL, where L is henries. Ideal capacitive reactance is X_C = 1/(2πfC), where C is farads.

At higher frequency, inductive reactance increases and capacitive reactance decreases. In an ideal inductor, current lags voltage; in an ideal capacitor, current leads voltage. A purely resistive load has voltage and current in phase. A memory aid is useful, but the physical distinction matters: magnetic fields store energy in inductors, electric fields store energy in capacitors, and the stored energy can return to the circuit.

Example: assume an ideal 0.100 H inductor connected to a 120 V RMS, 60 Hz source. X_L = 2π × 60 Hz × 0.100 H = 37.7 Ω. With winding resistance neglected, I = V/X_L = 120 V ÷ 37.7 Ω = 3.18 A RMS. A real coil also has resistance and losses, so its current and real power require impedance information.

Impedance (Z) is total opposition to AC and is measured in ohms. In a series resistance-inductance circuit, resistance and inductive reactance are perpendicular components, so magnitude is Z = √(R² + X_L²), not simply R + X_L. For a series resistance-capacitance circuit, magnitude similarly uses √(R² + X_C²).

Assume a series load has R = 30 Ω, X_L = 40 Ω, and is connected to 120 V RMS at the stated frequency. Z = √[(30 Ω)² + (40 Ω)²] = 50 Ω. Current is I = 120 V ÷ 50 Ω = 2.40 A. The impedance must be at least as large as either perpendicular component, so 50 Ω is reasonable.

Real, reactive, and apparent power

Real power (P) performs work or produces heat and is measured in watts. Apparent power (S) is the RMS volt-ampere product and is measured in volt-amperes: for single phase, S = VI. Reactive power (Q) represents energy exchanged with magnetic or electric fields and is measured in reactive volt-amperes (var).

Power factor (PF) is P/S. For single phase, P = VI × PF. Power factor ranges from 0 to 1 in magnitude. A purely resistive load has PF = 1.00; many motors have a lagging power factor below 1.00. Low power factor means more current is required to deliver the same real power at the same voltage.

For the 30 Ω and 40 Ω series RL example, apparent power is S = 120 V × 2.40 A = 288 VA. Real power is P = I²R = (2.40 A)² × 30 Ω = 172.8 W. Thus PF = 172.8 W ÷ 288 VA = 0.600, which also equals R/Z = 30 Ω ÷ 50 Ω.

For a separate 240 V single-phase load drawing 30 A at PF = 0.80, assume RMS values and steady operation. Apparent power is S = 240 V × 30 A = 7,200 VA = 7.20 kVA. Real power is P = 7,200 VA × 0.80 = 5,760 W = 5.76 kW. Omitting power factor would incorrectly report apparent power as real power.

Balanced three-phase relationships

For a balanced three-phase load using line-to-line voltage V_L and line current I_L:

• S = √3 × V_L × I_L
• P = √3 × V_L × I_L × PF

Assume a balanced 480 V three-phase load draws 20 A at PF = 0.90. Apparent power is S = 1.732 × 480 V × 20 A = 16,627 VA, or 16.6 kVA. Real power is P = 16,627 VA × 0.90 = 14,964 W, or 15.0 kW after appropriate rounding. The square-root-of-three factor is used once.

Connection relationships help explain it. In a balanced wye system, V_L = √3 × V_phase and I_L = I_phase. In balanced delta, V_L = V_phase and I_L = √3 × I_phase. Do not apply a wye line-to-neutral relationship to a delta load without evidence.

Exam checks

Identify whether the question asks for watts, VA, current, or power factor. State single-phase or balanced three-phase assumptions. Use line-to-line voltage with the three-phase formula. At fixed real power and voltage, lowering power factor must increase current; a result showing the opposite is suspect. Finally, a theory result is not automatically an NEC ampacity or overcurrent selection. Later chapters apply continuous-load, conductor, terminal, motor, and equipment rules from the 2017 NEC.

Information checkpoint

When a problem omits power factor or phase configuration, do not silently invent either value. Identify whether the requested quantity can be calculated from the supplied data and state every assumption required by the selected formula.