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Three-phase circuit analysis — Y, Δ, and the power triangle

Line vs phase, Y ↔ Δ conversion, P = √3·V_LL·I_L·cos θ, complex power S = V·I*, and PF correction with capacitors.

Sophomore ~11 min

Step 1 — Balanced 3-phase set: V_LL = √3 · V_phase, with a 30° phase shift

Animation speed 0.55×

P — kW Q — kVAR S — kVA PF —

Reference notes

Use Next → on the narrator above to walk through balanced three-phase circuit analysis: line vs phase quantities, Y and Δ connections, Y ↔ Δ conversion, and the power triangle that every load-flow study lives on.

The balanced 3-phase set

A balanced three-phase set has three phasors of equal magnitude, exactly 120° apart. Two voltages must be kept straight:

V_phase (also V_LN, line-to-neutral) — voltage from one phase conductor to the neutral.
V_LL (line-to-line) — voltage between two phase conductors.

Their relationship comes from phasor subtraction: V_ab = V_a − V_b, where V_a and V_b are 120° apart and equal in magnitude. The phasor algebra gives:

V_LL = √3 · V_phase ∠ +30°

On a 480 V Y system, V_phase = 480 / √3 ≈ 277 V. Nameplate voltages, transformer ratings, and exam problem statements default to V_LL. Never read a 13.8 kV machine as 13.8 kV per-phase.

Y (star) connection

In a Y connection each phase impedance Z sits between a line and a common neutral point. The voltage across each impedance is V_phase; the current through each impedance is the line current.

V_phase = V_LL / √3 · I_phase = V_phase / Z · I_line = I_phase

Y is the standard for HV side of step-up transformers (each phase winding sees only V_LL/√3, easing insulation) and for generators (where a neutral is needed for grounding through a resistor or reactor).

Δ (delta) connection

In a Δ connection each phase impedance sits between two lines — no neutral exists. The voltage across each impedance is V_LL; the line current is the phasor sum of two phase currents meeting at the corner.

V_phase = V_LL · I_phase = V_LL / Z · I_line = √3 · I_phase ∠ −30°

The dual symmetry between Y and Δ: where Y has √3 in voltages, Δ has √3 in currents.

Δ has no neutral. It's common on the LV side of distribution transformers (the closed delta loop captures triplen harmonics) and on motor windings. Step 3 of the animation shows the impedance loop and the phase / line current relationship.

Y ↔ Δ conversion

For a balanced load, converting between the two connections obeys:

Z_Δ = 3 · Z_Y

Derivation: write the line current for each and force them to match when fed by the same source. The line current for Y is V_phase/Z_Y = V_LL/(√3·Z_Y). The line current for Δ is √3·V_LL/Z_Δ. Setting them equal gives Z_Δ = 3·Z_Y.

For unbalanced loads the general Y ↔ Δ conversion formulas are more involved (Kennelly's theorem). On the PE exam you'll mostly use the balanced form.

Three-phase power

The total three-phase real, reactive, and apparent power for a balanced load are:

P_3φ = √3 · V_LL · I_L · cos θ (kW)

Q_3φ = √3 · V_LL · I_L · sin θ (kVAR)

S_3φ = √3 · V_LL · I_L (kVA)

where θ is the power-factor angle (between phase voltage and phase current). Power factor PF = cos θ = P / |S|.

These three quantities form the power triangle: P horizontal, Q vertical, S as the hypotenuse, θ as the angle.

Lagging PF (Q > 0) — current lags voltage; inductive load (motors, transformers, fluorescent ballasts).
Leading PF (Q < 0) — current leads voltage; capacitive load (capacitor banks, over-excited synchronous motors).

Worked example

A 500 kW load at 0.8 PF lagging on a 480 V (V_LL) system:

S = P / cos θ = 500 / 0.8 = 625 kVA
Q = √(S² − P²) = √(625² − 500²) = 375 kVAR
I_L = S / (√3 · V_LL) = 625 000 / (1.732 · 480) ≈ 752 A

Complex power S = V · I*

The complex form lets you keep P and Q in a single number. Define S = V · I* (current is conjugated). If V = |V|∠α and I = |I|∠β, then:

S = |V|·|I| ∠ (α − β) = |V|·|I|·cos θ + j |V|·|I|·sin θ = P + jQ

The conjugate is essential — without it, Q would have the wrong sign for inductive loads. The convention "Q > 0 for lagging" only works with S = V·I*.

Power factor correction

An inductive load has lagging PF and a positive Q. A capacitor bank in parallel supplies leading Q (negative Q). The capacitor's reactive power per phase is Q_cap = ω·C·V² (where ω = 2πf is the angular frequency and C is the capacitance). Net Q drops. P is unchanged. S = √(P² + Q²) shrinks, so line current I = S / (√3 · V_LL) drops in proportion. Two consequences:

Lower copper losses on the feeder: I²R losses fall as I².
Lower utility demand charge: most large-customer rate schedules penalize PF below 0.90 or 0.95.

Target PF for industrial sites is typically 0.95 lagging (some utilities require 0.98). Over-correction past unity creates a leading PF, which utilities also penalize.

Take-away. A balanced three-phase circuit reduces to single-phase math with two conversions. Y connections have V_LL = √3·V_phase; Δ connections have I_line = √3·I_phase. Power is always P = √3·V_LL·I_L·cos θ. The power triangle (P, Q, S, θ) is the single picture you need on every load-flow, transformer-sizing, and PF-correction problem on the PE exam.

Keyboard shortcuts

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On step 6, click the + / − buttons on the canvas to add/remove capacitor kVAR and watch the triangle shrink.