STAR-DELTA CONVERSION

The Y-Δ transform, also written wye-delta and also 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.^[1] It is widely used in analysis of three-phase electric power circuits.

The Y-Δ transform can be considered a special case of the star-mesh transform for three resistors. In mathematics, the Y-Δ transform plays an important role in theory of circular planar graphs.^[2

Basic symbolic representation 

Basic Y-Δ transformation[edit]

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 to Y[edit]

The general idea is to compute the impedance ${\displaystyle R_{\text{Y}}}$ at a terminal node of the Y circuit with impedances ${\displaystyle R'}$, ${\displaystyle R''}$ to adjacent nodes in the Δ circuit by

${\displaystyle R_{\text{Y}}={\frac {R'R''}{\sum R_{\Delta }}}}$

where ${\displaystyle R_{\Delta }}$ are all impedances in the Δ circuit. This yields the specific formulae,

${\displaystyle {\begin{aligned}R_{1}&={\frac {R_{\text{b}}R_{\text{c}}}{R_{\text{a}}+R_{\text{b}}+R_{\text{c}}}}\\[3pt]R_{2}&={\frac {R_{\text{a}}R_{\text{c}}}{R_{\text{a}}+R_{\text{b}}+R_{\text{c}}}}\\[3pt]R_{3}&={\frac {R_{\text{a}}R_{\text{b}}}{R_{\text{a}}+R_{\text{b}}+R_{\text{c}}}}\end{aligned}}}$

Equations for the transformation from Y to Δ[edit]

The general idea is to compute an impedance ${\displaystyle R_{\Delta }}$ in the Δ circuit by

${\displaystyle R_{\Delta }={\frac {R_{P}}{R_{\text{opposite}}}}}$

where ${\displaystyle R_{P}=R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}$ is the sum of the products of all pairs of impedances in the Y circuit and ${\displaystyle R_{\text{opposite}}}$ is the impedance of the node in the Y circuit which is opposite the edge with ${\displaystyle R_{\Delta }}$. The formulae for the individual edges are thus

${\displaystyle {\begin{aligned}R_{\text{a}}&={\frac {R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}{R_{1}}}\\[3pt]R_{\text{b}}&={\frac {R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}{R_{2}}}\\[3pt]R_{\text{c}}&={\frac {R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}{R_{3}}}\end{aligned}}}$

Or, if using admittance instead of resistance:

${\displaystyle {\begin{aligned}Y_{\text{a}}&={\frac {Y_{3}Y_{2}}{\sum Y_{\text{Y}}}}\\[3pt]Y_{\text{b}}&={\frac {Y_{3}Y_{1}}{\sum Y_{\text{Y}}}}\\[3pt]Y_{\text{c}}&={\frac {Y_{1}Y_{2}}{\sum Y_{\text{Y}}}}\end{aligned}}}$

Note that the general formula in Y to Δ using admittance is similar to Δ to Y using resistance.

Δ-load to Y-load transformation equations[edit]

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

To relate ${\displaystyle \left\{R_{\text{a}},R_{\text{b}},R_{\text{c}}\right\}}$ from Δ to ${\displaystyle \left\{R_{1},R_{2},R_{3}\right\}}$ 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:

${\displaystyle {\begin{aligned}R_{\Delta }\left(N_{1},N_{2}\right)&=R_{\text{c}}\parallel (R_{\text{a}}+R_{\text{b}})\\[3pt]&={\frac {1}{{\frac {1}{R_{\text{c}}}}+{\frac {1}{R_{\text{a}}+R_{\text{b}}}}}}\\[3pt]&={\frac {R_{\text{c}}\left(R_{\text{a}}+R_{\text{b}}\right)}{R_{\text{a}}+R_{\text{b}}+R_{\text{c}}}}\end{aligned}}}$

To simplify, let ${\displaystyle R_{\text{T}}}$ be the sum of ${\displaystyle \left\{R_{\text{a}},R_{\text{b}},R_{\text{c}}\right\}}$.

${\displaystyle R_{\text{T}}=R_{\text{a}}+R_{\text{b}}+R_{\text{c}}}$

Thus,

${\displaystyle R_{\Delta }\left(N_{1},N_{2}\right)={\frac {R_{\text{c}}(R_{\text{a}}+R_{\text{b}})}{R_{\text{T}}}}}$

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

${\displaystyle R_{\text{Y}}\left(N_{1},N_{2}\right)=R_{1}+R_{2}}$

hence:

${\displaystyle R_{1}+R_{2}={\frac {R_{\text{c}}(R_{\text{a}}+R_{\text{b}})}{R_{\text{T}}}}}$   (1)

Repeating for ${\displaystyle R(N_{2},N_{3})}$:

${\displaystyle R_{2}+R_{3}={\frac {R_{\text{a}}(R_{\text{b}}+R_{\text{c}})}{R_{\text{T}}}}}$   (2)

and for ${\displaystyle R\left(N_{1},N_{3}\right)}$:

${\displaystyle R_{1}+R_{3}={\frac {R_{\text{b}}\left(R_{\text{a}}+R_{\text{c}}\right)}{R_{\text{T}}}}.}$   (3)

From here, the values of ${\displaystyle \left\{R_{1},R_{2},R_{3}\right\}}$ can be determined by linear combination (addition and/or subtraction).

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

${\displaystyle {\begin{aligned}R_{1}+R_{2}+R_{1}+R_{3}-R_{2}-R_{3}&={\frac {R_{\text{c}}(R_{\text{a}}+R_{\text{b}})}{R_{\text{T}}}}+{\frac {R_{\text{b}}(R_{\text{a}}+R_{\text{c}})}{R_{\text{T}}}}-{\frac {R_{\text{a}}(R_{\text{b}}+R_{\text{c}})}{R_{\text{T}}}}\\[3pt]{}\Rightarrow 2R_{1}&={\frac {2R_{\text{b}}R_{\text{c}}}{R_{\text{T}}}}\\[3pt]{}\Rightarrow R_{1}&={\frac {R_{\text{b}}R_{\text{c}}}{R_{\text{T}}}}.\end{aligned}}}$

For completeness:

${\displaystyle R_{1}={\frac {R_{\text{b}}R_{\text{c}}}{R_{\text{T}}}}}$ (4)

${\displaystyle R_{2}={\frac {R_{\text{a}}R_{\text{c}}}{R_{\text{T}}}}}$ (5)

${\displaystyle R_{3}={\frac {R_{\text{a}}R_{\text{b}}}{R_{\text{T}}}}}$ (6)

Y-load to Δ-load transformation equations[edit]

Let

${\displaystyle R_{\text{T}}=R_{\text{a}}+R_{\text{b}}+R_{\text{c}}}$.

We can write the Δ to Y equations as

${\displaystyle R_{1}={\frac {R_{\text{b}}R_{\text{c}}}{R_{\text{T}}}}}$   (1)

${\displaystyle R_{2}={\frac {R_{\text{a}}R_{\text{c}}}{R_{\text{T}}}}}$   (2)

${\displaystyle R_{3}={\frac {R_{\text{a}}R_{\text{b}}}{R_{\text{T}}}}.}$   (3)

Multiplying the pairs of equations yields

${\displaystyle R_{1}R_{2}={\frac {R_{\text{a}}R_{\text{b}}R_{\text{c}}^{2}}{R_{\text{T}}^{2}}}}$   (4)

${\displaystyle R_{1}R_{3}={\frac {R_{\text{a}}R_{\text{b}}^{2}R_{\text{c}}}{R_{\text{T}}^{2}}}}$   (5)

${\displaystyle R_{2}R_{3}={\frac {R_{\text{a}}^{2}R_{\text{b}}R_{\text{c}}}{R_{\text{T}}^{2}}}}$   (6)

and the sum of these equations is

${\displaystyle R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}={\frac {R_{\text{a}}R_{\text{b}}R_{\text{c}}^{2}+R_{\text{a}}R_{\text{b}}^{2}R_{\text{c}}+R_{\text{a}}^{2}R_{\text{b}}R_{\text{c}}}{R_{\text{T}}^{2}}}}$   (7)

Factor ${\displaystyle R_{\text{a}}R_{\text{b}}R_{\text{c}}}$ from the right side, leaving ${\displaystyle R_{\text{T}}}$ in the numerator, canceling with an ${\displaystyle R_{\text{T}}}$ in the denominator.

${\displaystyle {\begin{aligned}R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}&={}{\frac {\left(R_{\text{a}}R_{\text{b}}R_{\text{c}}\right)\left(R_{\text{a}}+R_{\text{b}}+R_{\text{c}}\right)}{R_{\text{T}}^{2}}}\\&={}{\frac {R_{\text{a}}R_{\text{b}}R_{\text{c}}}{R_{\text{T}}}}\end{aligned}}}$ (8)

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

Divide (8) by (1)

${\displaystyle {\begin{aligned}{\frac {R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}}{R_{1}}}&={}{\frac {R_{\text{a}}R_{\text{b}}R_{\text{c}}}{R_{\text{T}}}}{\frac {R_{\text{T}}}{R_{\text{b}}R_{\text{c}}}}\\&={}R_{\text{a}},\end{aligned}}}$

which is the equation for ${\displaystyle R_{\text{a}}}$. Dividing (8) by (2) or (3) (expressions for ${\displaystyle R_{2}}$ or ${\displaystyle R_{3}}$) gives the remaining equations.