Editing Three-phase electric power (section)

===Wye (or, star; Y)===
[[File:3 Phase Power Connected to Wye Load.svg|thumb|Three-phase AC generator connected as a wye or star source to a wye- or star-connected load. In the circuit shown, unbalanced currents will flow between the source and load through the ground, creating undesired [[Stray voltage#Neutral return currents through the ground|stray ground voltages]].<ref>{{cite web |title=What is "Stray Voltage"? |date=August 10, 2015 |publisher=Utility Technology Engineers-Consultants (UTEC) |url=https://www.utilitytec.com/uploads/1/3/0/9/130926103/2015_stray_voltage_analysis.pdf |access-date=December 10, 2023}}</ref>]]

The voltage seen by the load will depend on the load connection; for the wye case, connecting each load to a phase (line-to-neutral) voltages gives<ref name="GloverSarma2011" />

:<math>\begin{align}
  I_1 &= \frac{V_1}{|Z_\text{total}|}\angle (-\theta), \\
  I_2 &= \frac{V_2}{|Z_\text{total}|}\angle (-120^\circ - \theta), \\
  I_3 &= \frac{V_3}{|Z_\text{total}|}\angle ( 120^\circ - \theta),
\end{align}</math>

where ''Z''<sub>total</sub> is the sum of line and load impedances (''Z''<sub>total</sub> = ''Z''<sub>LN</sub> + ''Z''<sub>Y</sub>), and ''θ'' is the phase of the total impedance (''Z''<sub>total</sub>).

The phase angle difference between voltage and current of each phase is not necessarily 0 and depends on the type of load impedance, ''Z''<sub>y</sub>. Inductive and capacitive loads will cause current to either lag or lead the voltage. However, the relative phase angle between each pair of lines (1 to 2, 2 to 3, and 3 to 1) will still be −120°.

[[File:Wye connection line voltages.png|thumb|A phasor diagram for a wye configuration, in which ''V''<sub>ab</sub> represents a line voltage, and ''V''<sub>an</sub> represents a phase voltage. Voltages are balanced as{{ubl
| ''V''<sub>ab</sub> {{=}} (1∠α − 1∠α + 120°) {{sqrt|3}}{{nnbsp}}{{pipe}}''V''{{pipe}}∠α + 30°,
| ''V''<sub>bc</sub> {{=}} {{sqrt|3}}{{nnbsp}}{{pipe}}''V''{{pipe}}∠α − 90°,
| ''V''<sub>ca</sub> {{=}} {{sqrt|3}}{{nnbsp}}{{pipe}}''V''{{pipe}}∠α + 150°
}}{{paragraph}}
(α = 0 in this case).
]]
By applying [[Kirchhoff's current law]] (KCL) to the neutral node, the three phase currents sum to the total current in the neutral line. In the balanced case:
: <math>I_1 + I_2 + I_3 = I_\text{N} = 0.</math>