3.2 AC Steady-State & Phasors

Phasors turn AC into algebra

A sinusoid v(t) = V_m cos(omegat + phi) is represented by a phasor V = V_m angle phi, where omega = 2pi*f (rad/s). Once every element is an impedance, KVL, KCL, voltage dividers, Thevenin, and node/mesh methods all apply exactly as in DC, but with complex arithmetic. The NCEES on-screen calculator handles rectangular-to-polar conversion, so keep numbers in whichever form simplifies the current step: rectangular (a + jb) for adding impedances, polar (M angle theta) for multiplying or dividing.

FE problems usually quote root-mean-square (RMS) values. For a sinusoid, V_rms = V_m / sqrt(2) = 0.707 V_m. All average-power calculations use RMS values; a common trap is plugging peak amplitudes into a power formula, which overstates power by a factor of 2.

Impedance and reactance

Each passive element has an impedance Z (ohms):

• Resistor: Z_R = R (no phase shift).
• Inductor: Z_L = jX_L where X_L = omegaL = 2pif*L. Voltage leads current by 90 degrees.
• Capacitor: Z_C = -jX_C where X_C = 1/(omegaC) = 1/(2pif*C). Current leads voltage by 90 degrees.

The mnemonic ELI the ICE man captures phase: in an inductor (L), voltage E leads current I (ELI); in a capacitor (C), current I leads voltage E (ICE).

Total series impedance is Z = R + j(X_L - X_C). Its magnitude is |Z| = sqrt(R^2 + (X_L - X_C)^2) and angle theta = arctan((X_L - X_C)/R). The current phasor is I = V / Z.

Worked impedance example: A 50 ohm resistor in series with a 0.1 H inductor at f = 60 Hz. X_L = 2pi60*0.1 = 37.7 ohm, so Z = 50 + j37.7 ohm. |Z| = sqrt(50^2 + 37.7^2) = sqrt(2500 + 1421) = 62.6 ohm; theta = arctan(37.7/50) = 37.0 degrees. Driven by 120 V RMS, the current magnitude is 120/62.6 = 1.92 A, lagging the voltage by 37 degrees.

Complex power: P, Q, and S

The complex power S = V_rms * I_rms* (conjugate of current) resolves into:

• Real (average) power P = V_rms I_rms cos(theta), in watts (W) — the power converted to work/heat.
• Reactive power Q = V_rms I_rms sin(theta), in volt-amperes reactive (VAR) — oscillating between source and reactive elements.
• Apparent power S = V_rms I_rms = |P + jQ|, in volt-amperes (VA).

The power triangle gives S^2 = P^2 + Q^2, with theta the voltage-to-current angle and pf = cos(theta) = P/S.

Inductive loads absorb positive Q (lagging pf); capacitive loads supply Q (leading pf).

Worked power example: For the 62.6 ohm circuit above, I_rms = 1.92 A, S = 120 * 1.92 = 230 VA, P = 230 * cos(37) = 184 W, Q = 230 * sin(37) = 138 VAR (inductive, lagging).

Power factor correction and resonance

Low lagging power factor forces high line current for a given real power, raising I^2 R losses. Adding shunt capacitance supplies reactive power locally and raises pf toward unity. The capacitive reactive power required to move from angle theta_1 to theta_2 is Q_C = P*(tan(theta_1) - tan(theta_2)).

In a series RLC circuit, resonance occurs when X_L = X_C, so 2pif_0L = 1/(2pif_0C), giving f_0 = 1 / (2pisqrt(L*C)). At resonance the reactances cancel, impedance is purely resistive and minimum (Z = R), and current is maximum. Quality factor Q = (1/R)sqrt(L/C) = omega_0L/R, and bandwidth BW = f_0 / Q. A parallel RLC (tank) circuit resonates at the same f_0 but presents maximum impedance, so line current is minimum.

Balanced three-phase power

Three-phase systems use wye (Y) or delta connections. In wye, line voltage leads phase voltage and is larger by sqrt(3): V_L = sqrt(3)*V_phase, while line current equals phase current. In delta, line voltage equals phase voltage, while line current is sqrt(3) times phase current: I_L = sqrt(3)*I_phase.

For either balanced connection the total real power is P = sqrt(3) * V_L * I_L * cos(theta), reactive Q = sqrt(3) V_L I_L sin(theta), and apparent S = sqrt(3) V_L I_L. Note that the angle theta here is still the per-phase load angle (between phase voltage and phase current), not 30 degrees — the sqrt(3) already accounts for the line-to-phase geometry. Worked example: a balanced load on a 480 V (line) three-phase supply draws 20 A line current at pf = 0.85: P = sqrt(3)48020*0.85 = 14,130 W = 14.1 kW. A common trap is forgetting the sqrt(3) and computing only single-phase power, which understates the total by 1.73.

Three-phase delivers constant instantaneous power (the three pulsating single-phase contributions sum to a steady value), which is why motors run smoothly, and it transmits a given power with less conductor than single-phase. Per-phase analysis is the standard solving technique: reduce a balanced three-phase problem to one phase of an equivalent wye, solve it as a single-phase circuit, then scale by 3 for total power.