In the perfectly balanced case all three lines share equivalent loads. Examining the circuits, we can derive relationships between line-to-line voltage and current, and load voltage and current for star- and delta-connected loads. In a balanced system each line will produce equal voltage magnitudes at phase angles equally spaced from each other. With V1 as our reference and V3 lagging V2 lagging V1, using
angle notation, and VLN the line-to-neutral voltage we have: :\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} These voltages feed into either a star- or delta-connected load.
Star . The voltage seen by the load will depend on the load connection; for the star case, connecting each load to a line-to-neutral voltage gives :\begin{align} I_1 &= \frac{V_1}\angle (-\theta), \\ I_2 &= \frac{V_2}\angle (-120^\circ - \theta), \\ I_3 &= \frac{V_3}\angle ( 120^\circ - \theta), \end{align} where
Ztotal is the sum of line and load impedances (
Ztotal =
ZLN +
ZY), and
θ is the phase of the total impedance (
Ztotal). The phase angle difference between voltage and current of each phase is not necessarily 0 and depends on the type of load impedance,
Zy. 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°. 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: : I_1 + I_2 + I_3 = I_\text{N} = 0.
Delta In the delta circuit, loads are connected across the lines, and so loads see line-to-line voltages: : \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} (Φv1 is the phase shift for the first voltage, commonly taken to be 0°; in this case, Φv2 = −120° and Φv3 = −240° or 120°.) Further: :\begin{align} I_{12} &= \frac{V_{12}} \angle ( 30^\circ - \theta), \\ I_{23} &= \frac{V_{23}} \angle (-90^\circ - \theta), \\ I_{31} &= \frac{V_{31}} \angle ( 150^\circ - \theta), \end{align} where
θ is the phase of delta impedance (
ZΔ). Relative angles are preserved, so
I31 lags
I23 lags
I12 by 120°. Calculating line currents by using KCL at each delta node gives :\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} and similarly for each other line: : \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} where, again,
θ is the phase of delta impedance (
ZΔ). 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 . As a delta configuration connects a load across phases of a transformer, it delivers the line-to-line voltage difference, which is times greater than the line-to-neutral voltage delivered to a load in the star configuration. As the power transferred is
V2/
Z, the impedance in the delta configuration must be 3 times what it would be in a star configuration for the same power to be transferred. == Single-phase loads ==