# تحويل واي-دلتا

(تم التحويل من Y-Δ transform)

تحويل واي دلتا, Y-Δ transform وتكتب Y-delta، Wye-delta، Kennelly’s delta-star transformation, star-mesh transformation, T-Π or T-pi transform، هي تقنية رياضية لتبسيط تحليل الشبكة الإلكترونية.

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## 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 3 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 real as well as complex impedances.

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

$R_{1}={\frac {R_{a}R_{b}}{R_{a}+R_{b}+R_{c}}},$
$R_{2}={\frac {R_{b}R_{c}}{R_{a}+R_{b}+R_{c}}},$
$R_{3}={\frac {R_{a}R_{c}}{R_{a}+R_{b}+R_{c}}}.$

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

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

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

$R_{a}={\frac {R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}{R_{2}}},$
$R_{b}={\frac {R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}{R_{3}}},$
$R_{c}={\frac {R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}{R_{1}}}.$

## 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 graph family is a Y-Δ equivalence class.

## Demonstration

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

To relate {$R_{a},R_{b},R_{c}$ } from Δ to {$R_{1},R_{2},R_{3}$ } 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{aligned}R_{\Delta }(N_{1},N_{2})&=R_{b}\parallel (R_{a}+R_{c})\\&={\frac {1}{{\frac {1}{R_{b}}}+{\frac {1}{R_{a}+R_{c}}}}}\\&={\frac {R_{b}(R_{a}+R_{c})}{R_{a}+R_{b}+R_{c}}}.\end{aligned}}

To simplify, let's call $R_{T}$  the sum of {$R_{a},R_{b},R_{c}$ }.

$R_{T}=R_{a}+R_{b}+R_{c}$

Thus,

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

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

$R_{Y}(N_{1},N_{2})=R_{1}+R_{2}$

hence:

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

Repeating for $R(N_{2},N_{3})$ :

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

and for $R(N_{1},N_{3})$ :

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

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

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

$R_{1}+R_{2}+R_{1}+R_{3}-R_{2}-R_{3}={\frac {R_{b}(R_{a}+R_{c})}{R_{T}}}+{\frac {R_{a}(R_{b}+R_{c})}{R_{T}}}-{\frac {R_{c}(R_{a}+R_{b})}{R_{T}}}$
$2R_{1}={\frac {2R_{b}R_{a}}{R_{T}}}$

thus,

$R_{1}={\frac {R_{b}R_{a}}{R_{T}}}.$

where $R_{T}=R_{a}+R_{b}+R_{c}$

For completeness:

$R_{1}={\frac {R_{b}R_{a}}{R_{T}}}$  (4)
$R_{2}={\frac {R_{b}R_{c}}{R_{T}}}$  (5)
$R_{3}={\frac {R_{a}R_{c}}{R_{T}}}$  (6)

Let

$R_{T}=R_{a}+R_{b}+R_{c}$ .

We can write the Δ to Y equations as

$R_{1}={\frac {R_{a}R_{b}}{R_{T}}}$    (1)
$R_{2}={\frac {R_{b}R_{c}}{R_{T}}}$    (2)
$R_{3}={\frac {R_{a}R_{c}}{R_{T}}}.$    (3)

Multiplying the pairs of equations yields

$R_{1}R_{2}={\frac {R_{a}R_{b}^{2}R_{c}}{R_{T}^{2}}}$    (4)
$R_{1}R_{3}={\frac {R_{a}^{2}R_{b}R_{c}}{R_{T}^{2}}}$    (5)
$R_{2}R_{3}={\frac {R_{a}R_{b}R_{c}^{2}}{R_{T}^{2}}}$    (6)

and the sum of these equations is

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

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

$R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}={\frac {(R_{a}R_{b}R_{c})(R_{a}+R_{b}+R_{c})}{R_{T}^{2}}}$
$R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}={\frac {R_{a}R_{b}R_{c}}{R_{T}}}$  (8)

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

Divide (8) by (1)

${\frac {R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}}{R_{1}}}={\frac {R_{a}R_{b}R_{c}}{R_{T}}}{\frac {R_{T}}{R_{a}R_{b}}},$
${\frac {R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}}{R_{1}}}=R_{c},$

which is the equation for $R_{c}$ . Dividing (8) by $R_{2}$  or $R_{3}$  gives the other equations.

## المصادر

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