Editing Three-phase electric power (section)

== Balanced circuits ==
In the perfectly balanced case all three lines share equivalent loads. Examining the circuits, we can derive relationships between line voltage and current, and load voltage and current for wye- and delta-connected loads.

In a balanced system each line will produce equal voltage magnitudes at phase angles equally spaced from each other. With V<sub>1</sub> as our reference and V<sub>3</sub> lagging V<sub>2</sub> lagging V<sub>1</sub>, using [[angle notation]], and V<sub>LN</sub> the voltage between the line and the neutral we have:<ref name="GloverSarma2011">{{cite book |author1=J. Duncan Glover |author2=Mulukutla S. Sarma |author3=Thomas J. Overbye |title=Power System Analysis & Design |url=https://books.google.com/books?id=HrtXToLEbVoC |date=April 2011 |publisher=Cengage Learning |isbn=978-1-111-42579-1 |pages=60–68}}</ref>

:<math>\begin{align}
  V_1 &= V_\text{LN}\angle 0^\circ, \\
  V_2 &= V_\text{LN}\angle{-120}^\circ, \\
  V_3 &= V_\text{LN}\angle{+120}^\circ.
\end{align}</math>

These voltages feed into either a wye- or delta-connected load.

===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>

=== Delta (Δ) ===
[[File:3 Phase Power Connected to Delta Load.svg|thumb|right|Three-phase AC generator connected as a wye source to a delta-connected load]]

In the delta circuit, loads are connected across the lines, and so loads see line-to-line voltages:<ref name="GloverSarma2011" />
: <math>\begin{align}
  V_{12} &= V_1 - V_2 = (V_\text{LN}\angle 0^\circ) - (V_\text{LN}\angle {-120}^\circ) \\
         &= \sqrt{3}V_\text{LN}\angle 30^\circ = \sqrt{3}V_{1}\angle (\phi_{V_1} + 30^\circ), \\

  V_{23} &= V_2 - V_3 = (V_\text{LN}\angle {-120}^\circ) - (V_\text{LN}\angle 120^\circ) \\
         &= \sqrt{3}V_\text{LN}\angle {-90}^\circ = \sqrt{3}V_{2}\angle (\phi_{V_2} + 30^\circ), \\

  V_{31} &= V_3 - V_1 = (V_\text{LN}\angle 120^\circ) - (V_\text{LN}\angle 0^\circ) \\
         &= \sqrt{3}V_\text{LN}\angle 150^\circ = \sqrt{3}V_{3}\angle (\phi_{V_3} + 30^\circ).
\end{align}</math>

(Φ<sub>v1</sub> is the phase shift for the first voltage, commonly taken to be 0°; in this case, Φ<sub>v2</sub> = −120° and Φ<sub>v3</sub> = −240° or 120°.)

Further:
:<math>\begin{align}
  I_{12} &= \frac{V_{12}}{|Z_\Delta|} \angle ( 30^\circ - \theta), \\
  I_{23} &= \frac{V_{23}}{|Z_\Delta|} \angle (-90^\circ - \theta), \\
  I_{31} &= \frac{V_{31}}{|Z_\Delta|} \angle ( 150^\circ - \theta),
\end{align}</math>

where ''θ'' is the phase of delta impedance (''Z''<sub>Δ</sub>).

Relative angles are preserved, so ''I''<sub>31</sub> lags ''I''<sub>23</sub> lags ''I''<sub>12</sub> by 120°. Calculating line currents by using KCL at each delta node gives

:<math>\begin{align}
  I_1 &= I_{12} - I_{31} = I_{12} - I_{12}\angle 120^\circ \\
      &= \sqrt{3}I_{12} \angle (\phi_{I_{12}} - 30^\circ) = \sqrt{3}I_{12} \angle (-\theta)
\end{align}</math>

and similarly for each other line:

: <math>\begin{align}
  I_2 &= \sqrt{3}I_{23} \angle (\phi_{I_{23}} - 30^\circ) = \sqrt{3}I_{23} \angle (-120^\circ - \theta), \\
  I_3 &= \sqrt{3}I_{31} \angle (\phi_{I_{31}} - 30^\circ) = \sqrt{3}I_{31} \angle (120^\circ - \theta),
\end{align}</math>

where, again, ''θ'' is the phase of delta impedance (''Z''<sub>Δ</sub>).

[[File:Delta connection currents.png|thumb|A delta configuration and a corresponding phasor diagram of its currents. Phase voltages are equal to line voltages, and currents are calculated as{{ubl
| ''I''<sub>a</sub> {{=}} ''I''<sub>ab</sub> − ''I''<sub>ca</sub> {{=}} {{sqrt|3}}{{nnbsp}}''I''<sub>ab</sub>∠−30°,
| ''I''<sub>b</sub> {{=}} ''I''<sub>bc</sub> − ''I''<sub>ab</sub>,
| ''I''<sub>c</sub> {{=}} ''I''<sub>ca</sub> − ''I''<sub>bc</sub>.
}}{{paragraph}}
The overall power transferred is
{{ubl|''S''<sub>3Φ</sub> {{=}} 3''V''<sub>phase</sub>''I''*<sub>phase</sub>.}}
]]
Inspection of a phasor diagram, or conversion from phasor notation to complex notation, illuminates how the difference between two line-to-neutral voltages yields a line-to-line voltage that is greater by a factor of {{sqrt|3}}. As a delta configuration connects a load across phases of a transformer, it delivers the line-to-line voltage difference, which is {{sqrt|3}} times greater than the line-to-neutral voltage delivered to a load in the wye configuration. As the power transferred is ''V''<sup>2</sup>/''Z'', the impedance in the delta configuration must be 3 times what it would be in a wye configuration for the same power to be transferred.