Star Delta Transformation

There are two types of system available in electric circuit, single phase and three phase system. In single phase circuit, there will be only one phase, i.e the current will flow through only one wire and there will be one return path called neutral line to complete the circuit. So in single phase minimum amount of power can be transported.
In 1882, new invention has been done on polyphase system, that more than one phase can be used for generating, transmitting and for load system. Three phase circuit is the polyphase system where three phases are send together from the generator to the load.
In three phase circuit, connections can be given in two types:
  • Star or Wye(Y) connection
  • Delta(Δ) connection
Wye Delta Configuration.
Fig. 1: Star Delta Configuration.
Circuit configurations are often encountered in which the resistors do not appear to be in series or parallel. Under these conditions, it may be necessary to convert the circuit from one form to another to solve for any unknown quantities if mesh or nodal analysis is not applied. Two circuit configurations that often account for these difficulties are the Star (Y) and Delta (Δ) configurations depicted in [Fig. 1].
The purpose of this section is to develop the equations for converting from Star to Δ, or vice versa. This type of conversion normally leads to a network that can be solved using techniques such as those described in previous chapters.
Introducing the concept of Δ-Star or Star-Δ conversions.
Fig. 2: Introducing the concept of Δ-Star or Star-Δ conversions.
It is our purpose (referring to [Fig. 2]) to find some expression for $R_1$, $R_2$, and $R_3$ in terms of $R_A$, $R_B$, and $R_C$, and vice versa, that will ensure that the resistance between any two terminals of the Y configuration will be the same with the Δ configuration inserted in place of the Y configuration (and vice versa). If the two circuits are to be equivalent, the total resistance between any two terminals must be the same. Consider terminals a-c in the $Δ-Y$ configurations in [Fig. 3].
Introducing the concept of Δ-Y or Y-Δ conversions.
Fig. 3: Finding the resistance Rac for the Y and Δ configurations.
Let us first assume that we want to convert the Δ ($R_A$, $R_B$, $R_C$) to the Y ($R_1$, $R_2$, $R_3$). This requires that we have a relationship for $R_1$, $R_2$, and $R_3$ in terms of $R_A$, $R_B$, and $R_C$. If the resistance is to be the same between terminals a-c for both the Δ and the Y, the following must be true:
$$R_{a-c}(Y) = R_{a-c}(Δ)$$
so that
$$R_{a-c}= R_1 + R_3 = {R_B(R_A + R_C) \over R_B + (R_A + R_C)} \tag{1}$$
Using the same approach for a-b and b-c, we obtain the following relationships:
$$R_{a-b} = R_1 + R_2 = {R_C (R_A + R_B) \over R_C + (R_A + R_B)} \tag{2}$$
$$R_{b-c} = R_2 + R_3 = {R_A(R_B + R_C) \over R_A + (R_B + R_C)} \tag{3}$$
Subtracting Eq.(1) from Eq.(2), we have
$$ \begin{split} (R_1 + R_2) - (R_1 + R_3) &= ({R_C R_B + R_C R_A \over R_A + R_B + R_C}) \\ & \qquad - ({R_BR_A + R_BR_A \over R_A + R_B + R_C}) \end{split} $$
so that
$$ R_2 - R_3 ={R_AR_C - R_BR_A \over RA + RB + RC} \tag{4}$$
Subtracting Eq. (4) from Eq. (3) yields
$$ \begin{split} (R_2 + R_3) - (R_2 - R_3) &= ({ R_AR_B + R_AR_C \over R_A + R_B + R_C})\\ \qquad & - ({R_AR_C - R_BR_A \over R_A + R_B + R_C}) \end{split} $$
so that
$$\bbox[5px,border:1px solid grey] {R_3 = {R_AR_B \over R_A + R_B + R_C}} \tag{5}$$
Following the same procedure for $R_1$ and $R_2$, we have
$$\bbox[5px,border:1px solid grey] {R_1 = {R_BR_C \over R_A + R_B + R_C}} \tag{6}$$
$$\bbox[5px,border:1px solid grey]{R_2 = {R_AR_C \over R_A + R_B + R_C}} \tag{7}$$
Note that each resistor of the Y is equal to the product of the resistors in the two closest branches of the Δ divided by the sum of the resistors in the Δ.

Wye to Delta Conversion

To obtain the relationships necessary to convert from a Y to a Δ, first divide Eq. (5) by Eq. (6):
$$\begin{split} {R_3 \over R_1} &= {(R_AR_B)/(R_A + R_B + R_C) \over (R_BR_C)/(R_A + R_B + R_C) } \\ &= {R_A \over R_C}\\ \end{split}$$
$$R_A = {R_CR_3 \over R_1}$$
Then divide Eq. (5) by Eq. (7):
$$\begin{split} {R_3 \over R_2} &= {(R_AR_B)/(R_A + R_B + R_C) \over (R_AR_C)/(R_A + R_B + R_C)}\\ &={R_B \over R_C}\\ \end{split}$$
$$R_B = {R_3R_C \over R_2}$$
Substituting for $R_A$ and $R_B$ in Eq. (7) yields
$$\begin{split} R_2 &= {(R_CR_3/R_1)R_C \over (R_3R_C/R_1) + (R_CR_3/R_2) + R_C}\\ & = {(R_3/R1)R_C \over (R_3/R_2) + (R_3/R_1) + 1} \end{split}$$
Placing these over a common denominator, we obtain
$$ \begin{split} R_2 &= {(R_3R_C/R_1) \over (R_1R_2 + R_1R_3 + R_2R_3)/(R_1R_2)}\\ &= {R_2R_3R_C \over R_1R_2 + R_1R_3 + R_2R_3} \end{split}$$
$$\bbox[5px,border:1px solid grey] {R_C = {R_1R_2 + R_1R_3 + R_2R_3 \over R_3}} \tag{8}$$
We follow the same procedure for RB and RA:
$$\bbox[5px,border:1px solid grey] {R_A = {R_1R_2 + R_1R_3 + R_2R_3 \over R_1}} \tag{9}$$
$$\bbox[5px,border:1px solid grey] {R_B = {R_1R_2 + R_1R_3 + R_2R_3 \over R_2}} \tag{10}$$
Let us consider what would occur if all the values of a Δ or Y were the same. If $R_A = R_B = R_C$, Eq. (8) would become (using RA only) the following:
$$ \begin{split} R_3 &= {R_AR_B \over R_A + R_B + R_C} \\ &= {R_AR_A \over R_A + R_A + R_A} \\ &= {(R_A)^2 \over 3R_A} \\ R_3&= {R_A \over 3}\\ \end{split} $$
In general, therefore,
$$\bbox[5px,border:1px solid grey]{R_Y = {R_{Δ} \over 3}} \tag{11}$$
$$\bbox[5px,border:1px solid grey] {R_{Δ} = 3R_Y} \tag{12}$$
which indicates that for a Star(Wye) of three equal resistors, the value of each resistor of the Δ is equal to three times the value of any resistor of the Star (Y).

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