Editing Three-phase electric power (section)

=== 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.