Encyclopedia of Electrical Engineering

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:
**Fig.1: **Wye 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 wye (Y) and Delta (Δ) configurations depicted in Fig. 1.
The purpose of this section is to develop the equations for converting
from Δ to Y, 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.
**Fig. 2: **Introducing the concept of Δ-Y or Y-Δ conversions.
It is our purpose (referring to Fig. 2) to find some expression for
R1, R2, and R3 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.
**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:
$$R_{a-b} = R_1 + R_2 = {R_C (R_A + R_B) \over R_C + (R_A + R_B)} \tag{2}$$
and
$$R_{b-c} = R_2 + R_3 = {R_A(R_B + R_C) \over R_A + (R_B + R_C)} \tag{3}$$
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
$$\bbox[5px,border:1px solid blue] {\color{blue}{R_3 = {R_AR_B \over R_A + R_B + R_C}}} \tag{5}$$
Following the same procedure for R1 and R2, we have
$$\bbox[5px,border:1px solid blue] {\color{blue}{R_1 = {R_BR_C \over R_A + R_B + R_C}}} \tag{6}$$
and
$$\bbox[5px,border:1px solid blue] {\color{blue}{R_2 = {R_AR_C \over R_A + R_B + R_C}}} \tag{7}$$
#### Wye to Delta Conversion

To obtain the relationships necessary to convert from a Y to a Δ, first
divide Eq. (5) by Eq. (6):
$$\begin{split}
{R_3 \over R_1} &= {(R_AR_B)/(R_A + R_B + R_C) \over
(R_BR_C)/(R_A + R_B + R_C) } \\
&= {R_A \over R_C}\\
\end{split}$$
or
$$R_A = {R_CR_3 \over R_1}$$
Then divide Eq. (5) by Eq. (7):
$$\begin{split}
{R_3 \over R_2} &= {(R_AR_B)/(R_A + R_B + R_C) \over (R_AR_C)/(R_A + R_B + R_C)}\\
&={R_B \over R_C}\\
\end{split}$$
or
$$R_B = {R_3R_C \over R_2}$$
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
$$ \begin{split}
R_2 &= {(R_3R_C/R_1) \over (R_1R_2 + R_1R_3 + R_2R_3)/(R_1R_2)}\\
&= {R_2R_3R_C \over R_1R_2 + R_1R_3 + R_2R_3}
\end{split}$$
and
$$\bbox[5px,border:1px solid blue] {\color{blue}{R_C = {R_1R_2 + R_1R_3 + R_2R_3 \over R_3}}} \tag{8}$$
We follow the same procedure for RB and RA:
$$\bbox[5px,border:1px solid blue] {\color{blue}{R_A = {R_1R_2 + R_1R_3 + R_2R_3 \over R_1}}} \tag{9}$$
and
$$\bbox[5px,border:1px solid blue] {\color{blue}{R_B = {R_1R_2 + R_1R_3 + R_2R_3 \over R_2}}} \tag{10}$$
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,
$$\bbox[5px,border:1px solid blue] {\color{blue}{R_Y = {R_{Δ} \over 3}}} \tag{11}$$
or
$$\bbox[5px,border:1px solid blue] {\color{blue}{R_{Δ} = 3R_Y}} \tag{12}$$
which indicates that for a Y of three equal resistors, the value of each
resistor of the Δ is equal to three times the value of any resistor of the Y.
##### Wye Delta Transformation Related Questions

- Wye(Y) or Star connection
- Delta(Δ) connection

Bridge Networks
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Wye Delta Transformation