6.1 Power Fundamentals & Three-Phase

Why power fundamentals matter

The Power knowledge area is one of the highest-weight topics on the NCEES FE Electrical and Computer exam at roughly 8 to 12 of the 110 questions (tied with Circuit Analysis as a top-five area). Almost every power problem reduces to applying the power triangle and the balanced three-phase relations, all of which are printed in the searchable NCEES FE Reference Handbook.

Because the exam is open to that Handbook, the tested skill is not memorizing derivations but locating the right relation and keeping voltage, current, and angle references consistent. A wrong formula or a peak-vs-RMS slip produces a confidently wrong answer, so verification beats speed.

Start every power problem with a three-part classification: (1) single-phase or three-phase; (2) wye or delta; and (3) which quantity is asked — real power (W), reactive power (VAR), apparent power (VA), or power factor. Misclassifying the connection is the single most common reason a correct equation yields a wrong number. Write the governing relation explicitly before substituting; the few seconds spent are cheaper than a re-work.

The power triangle

For an AC load whose voltage and current phasors differ by angle theta, three quantities describe power flow:

• Real (active) power P = V·I·cos(theta), in watts (W) — the average power doing useful work.
• Reactive power Q = V·I·sin(theta), in volt-amperes reactive (VAR) — energy that sloshes between source and reactive elements, doing no net work.
• Apparent power |S| = V·I, in volt-amperes (VA), where complex power S = P + jQ.

Here V and I are RMS magnitudes. The three form a right triangle: |S|^2 = P^2 + Q^2. The power factor is pf = cos(theta) = P/|S|. Sign of Q encodes the load type.

Trap: "0.8 lagging" and "0.8 leading" both give cos(theta)=0.8 but opposite Q signs. Always carry the lagging/leading word.

Power-factor correction

Utilities (and the exam) care about power-factor correction because a low pf forces more line current for the same real power, raising I^2R losses and voltage drop. To raise a lagging pf, add a shunt capacitor that locally supplies reactive power, cutting the net Q the source must deliver.

The capacitor's required reactive power is Q_C = P·[tan(theta_1) - tan(theta_2)], where theta_1 is the original angle and theta_2 the target angle. Real power P is unchanged by correction; only Q and |S| drop, so line current falls. The needed capacitance follows from Q_C = V^2/X_C = V^2·(2·pi·f·C), giving C = Q_C/(2·pi·f·V^2) for a single-phase load.

Worked example: power-factor correction

A single-phase load draws P = 12 kW at pf = 0.70 lagging, 240 V, 60 Hz. Correct it to 0.95 lagging.

1. theta_1 = arccos(0.70) = 45.57 deg, tan(theta_1) = 1.020.
2. theta_2 = arccos(0.95) = 18.19 deg, tan(theta_2) = 0.329.
3. Q_C = 12·(1.020 - 0.329) = 12·0.691 = 8.29 kVAR.
4. C = 8290/(2·pi·60·240^2) = 8290/(5.43e6) = 1.53e-3 F = 1530 uF. The load still consumes 12 kW; only the reactive burden on the source shrank.

Three-phase: wye and delta

Three-phase systems use three sources spaced 120 degrees apart and deliver constant instantaneous power, which is why they dominate generation, transmission, and large motors. The two connection schemes are wye (Y) and delta.

For any balanced load, total real power is P = sqrt(3)·V_L·I_L·cos(theta), with V_L and I_L the line quantities and theta the per-phase impedance angle. Replace cos with sin for Q, and drop the trig factor for |S| = sqrt(3)·V_L·I_L. The sqrt(3) is the most-forgotten factor on the exam; write it before substituting.

Worked example: three-phase power

A balanced load draws line current I_L = 25 A at V_L = 480 V, pf = 0.85 lagging.

• P = sqrt(3)·480·25·0.85 = 1.732·480·25·0.85 = 17,667 W ≈ 17.7 kW.
• |S| = sqrt(3)·480·25 = 20,785 VA ≈ 20.8 kVA.
• Q = sqrt(3)·480·25·sin(31.79 deg) = 20,785·0.527 = 10.95 kVAR (positive, lagging).

Per-unit basics

The per-unit (pu) system expresses each quantity as a fraction of a chosen base: pu value = actual value / base value. Engineers pick a base apparent power (often the three-phase MVA) and a base voltage per zone; base impedance follows as Z_base = (V_base)^2 / S_base and base current as I_base = S_base/(sqrt(3)·V_base) for three-phase.

The payoff: when bases are scaled by the transformer turns ratio, the turns ratios disappear and a multi-voltage network collapses into one connected per-unit circuit. Per-unit values also cluster near 1.0, so a result of 0.05 pu or 12 pu instantly flags a likely error. To shift an impedance to a new base, use Z_pu,new = Z_pu,old·(V_base,old/V_base,new)^2·(S_base,new/S_base,old). Manufacturers quote machine and transformer impedances in pu on the device rating precisely because the value is then ratio-independent.