Y-Δ transform

For the device which transforms three-phase electric power without a neutral wire into three-phase power with a neutral wire, see delta-wye transformer.

For application in statistical mechanics see Yang-Baxter equation.

The Y-Δ transform, also written wye-delta and known by many other names, is a mathematical technique to simplify the analysis of an electrical network. The name derives from the shapes of the circuit diagrams, which look respectively like the letter Y and the Greek capital letter Δ. This circuit transformation theory was published by Arthur Edwin Kennelly in 1899.

The Y-Δ transform can be considered a special case of the star-mesh transform for three resistors.

Names

Illustration of the transform in its T-Π representation,

The Y-Δ transform is known by a variety of other names, mostly based upon the two shapes involved, listed in either order. The Y, spelled out as wye, can also be called T or star; the Δ, spelled out as delta, can also be called triangle, Π (spelled out as pi), or mesh. Thus, common names for the transformation include wye-delta or delta-wye, star-delta, star-mesh, or T-Π.

Basic Y-Δ transformation

Δ and Y circuits with the labels which are used in this article.

The transformation is used to establish equivalence for networks with three terminals. Where three elements terminate at a common node and none are sources, the node is eliminated by transforming the impedances. For equivalence, the impedance between any pair of terminals must be the same for both networks. The equations given here are valid for complex as well as real impedances.

Equations for the transformation from Δ-load to Y-load 3-phase circuit

The general idea is to compute the impedance R_y at a terminal node of the Y circuit with impedances R', R'' to adjacent nodes in the Δ circuit by

$R_y = \frac{R'R''}{\sum R_\Delta}$

where R_Δ are all impedances in the Δ circuit. This yields the specific formulae

$R_1 = \frac{R_bR_c}{R_a + R_b + R_c},$

$R_2 = \frac{R_aR_c}{R_a + R_b + R_c},$

$R_3 = \frac{R_aR_b}{R_a + R_b + R_c}.$

Equations for the transformation from Y-load to Δ-load 3-phase circuit

The general idea is to compute an impedance R_Δ in the Δ circuit by

$R_\Delta = \frac{R_P}{R_\mathrm{opposite}}$

where R_P = R₁R₂ + R₂R₃ + R₃R₁ is the sum of the products of all pairs of impedances in the Y circuit and R_opposite is the impedance of the node in the Y circuit which is opposite the edge with R_Δ. The formula for the individual edges are thus

$R_a = \frac{R_1R_2 + R_2R_3 + R_3R_1}{R_1},$

$R_b = \frac{R_1R_2 + R_2R_3 + R_3R_1}{R_2},$

$R_c = \frac{R_1R_2 + R_2R_3 + R_3R_1}{R_3}.$

Usefulness

Resistive networks between two terminals can theoretically be simplified to a single equivalent resistor (more generally, the same is true of impedance). Series and parallel transforms are basic tools for doing so, but for complex networks such as the bridge illustrated here, they do not suffice.

The Y-Δ transform can be used to eliminate one node at a time and produce a network that can be further simplified, as shown.

Transformation of a bridge resistor network, using the Y-Δ transform to eliminate node D, yields an equivalent network that may readily be simplified further.

The reverse transformation, Δ-Y, which adds a node, is often handy to pave the way for further simplification as well.

Transformation of a bridge resistor network, using the Δ-Y transform, also yields an equivalent network that may readily be simplified further.

Graph theory

In graph theory, the Y-Δ transform means replacing a Y subgraph of a graph with the equivalent Δ subgraph. The transform preserves the number of edges in a graph, but not the number of vertices or the number of cycles. Two graphs are said to be Y-Δ equivalent if one can be obtained from the other by a series of Y-Δ transforms in either direction. For example, the Petersen family is a Y-Δ equivalence class.

Demonstration

Δ-load to Y-load transformation equations

Δ and Y circuits with the labels that are used in this article.

To relate {R_a,R_b,R_c} from Δ to {R₁,R₂,R₃} from Y, the impedance between two corresponding nodes is compared. The impedance in either configuration is determined as if one of the nodes is disconnected from the circuit.

The impedance between N₁ and N₂ with N₃ disconnected in Δ:

$\begin{align} R_\Delta(N_1, N_2) &= R_c \parallel (R_a+R_b) \\[8pt] &= \frac{1}{\frac{1}{R_c}+\frac{1}{R_a+R_b}} \\[8pt] &= \frac{R_c(R_a+R_b)}{R_a+R_b+R_c}. \end{align}$

To simplify, let R_T be the sum of {R_a,R_b,R_c}.

R_T = R_a + R_b + R_c

Thus,

$R_\Delta(N_1, N_2) = \frac{R_c(R_a+R_b)}{R_T}$

The corresponding impedance between N₁ and N₂ in Y is simple:

R_Y(N₁,N₂) = R₁ + R₂

hence:

$R_1+R_2 = \frac{R_c(R_a+R_b)}{R_T}$   (1)

Repeating for R(N₂,N₃):

$R_2+R_3 = \frac{R_a(R_b+R_c)}{R_T}$   (2)

and for R(N₁,N₃):

$R_1+R_3 = \frac{R_b(R_a+R_c)}{R_T}.$   (3)

From here, the values of {R₁,R₂,R₃} can be determined by linear combination (addition and/or subtraction).

For example, adding (1) and (3), then subtracting (2) yields

$R_1+R_2+R_1+R_3-R_2-R_3 = \frac{R_c(R_a+R_b)}{R_T} + \frac{R_b(R_a+R_c)}{R_T} - \frac{R_a(R_b+R_c)}{R_T}$

$2R_1 = \frac{2R_bR_c}{R_T}$

thus,

$R_1 = \frac{R_bR_c}{R_T}.$

where R_T = R_a + R_b + R_c

For completeness:

$R_1 = \frac{R_bR_c}{R_T}$ (4)

$R_2 = \frac{R_aR_c}{R_T}$ (5)

$R_3 = \frac{R_aR_b}{R_T}$ (6)

Y-load to Δ-load transformation equations

Let

R_T = R_a + R_b + R_c.

We can write the Δ to Y equations as

$R_1 = \frac{R_bR_c}{R_T}$   (1)

$R_2 = \frac{R_aR_c}{R_T}$   (2)

$R_3 = \frac{R_aR_b}{R_T}.$   (3)

Multiplying the pairs of equations yields

$R_1R_2 = \frac{R_aR_bR_c^2}{R_T^2}$   (4)

$R_1R_3 = \frac{R_aR_b^2R_c}{R_T^2}$   (5)

$R_2R_3 = \frac{R_a^2R_bR_c}{R_T^2}$   (6)

and the sum of these equations is

$R_1R_2 + R_1R_3 + R_2R_3 = \frac{R_aR_bR_c^2 + R_aR_b^2R_c + R_a^2R_bR_c}{R_T^2}$   (7)

Factor R_aR_bR_c from the right side, leaving R_T in the numerator, canceling with an R_T in the denominator.

$R_1R_2 + R_1R_3 + R_2R_3 = \frac{(R_aR_bR_c)(R_a+R_b+R_c)}{R_T^2}$

$R_1R_2 + R_1R_3 + R_2R_3 = \frac{R_aR_bR_c}{R_T}$ (8)

-Note the similarity between (8) and {(1),(2),(3)}

Divide (8) by (1)

$\frac{R_1R_2 + R_1R_3 + R_2R_3}{R_1} = \frac{R_aR_bR_c}{R_T}\frac{R_T}{R_bR_c},$

$\frac{R_1R_2 + R_1R_3 + R_2R_3}{R_1} = R_a,$

which is the equation for R_a. Dividing (8) by (2) or (3) (expressions for R₂ or R₃) gives the remaining equations.

See also

• Star-mesh transform
• Analysis of resistive circuits
• Electrical network, single-phase electric power, alternating-current electric power, three-phase power, polyphase systems for examples of Y and Δ connections
• AC motor for a discussion of the Y-Δ starting technique
• Nikola Tesla
• John Hopkinson
• Mikhail Dolivo-Dobrovolsky
• Charles Proteus Steinmetz

References

• William Stevenson, Elements of Power System Analysis 3rd ed., McGraw Hill, New York, 1975, ISBN 0070612854

External links

• Star-Triangle Conversion: Knowledge on resistive networks and resistors
• Calculator of Star-Triangle transform