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:
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.
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].
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:
so that
Using the same approach for a-b and b-c, we obtain the following
relationships:
and
Subtracting Eq.(1) from Eq.(2), we have
so that
Subtracting Eq. (4) from Eq. (3) yields
so that
Following the same procedure for $R_1$ and $R_2$, we have
and
or
Then divide Eq. (5) by Eq. (7):
or
Substituting for $R_A$ and $R_B$ in Eq. (7) yields
Placing these over a common denominator, we obtain
and
We follow the same procedure for RB and RA:
and
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:
In general, therefore,
or
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).
- Star or Wye(Y) connection
- Delta(Δ) connection
Fig. 1: Star Delta Configuration.
Fig. 2: Introducing the concept of Δ-Star or Star-Δ conversions.
Fig. 3: Finding the resistance Rac for the Y and Δ configurations.
$$R_{a-c}(Y) = R_{a-c}(Δ)$$
$$R_{a-c}= R_1 + R_3 = {R_B(R_A + R_C) \over R_B + (R_A + R_C)} \tag{1}$$
$$R_{a-b} = R_1 + R_2 = {R_C (R_A + R_B) \over R_C + (R_A + R_B)} \tag{2}$$
$$R_{b-c} = R_2 + R_3 = {R_A(R_B + R_C) \over R_A + (R_B + R_C)} \tag{3}$$
$$
\begin{split}
(R_1 + R_2) - (R_1 + R_3) &= ({R_C R_B + R_C R_A \over R_A + R_B + R_C}) \\
& \qquad - ({R_BR_A + R_BR_A \over R_A + R_B + R_C})
\end{split}
$$
$$ R_2 - R_3 ={R_AR_C - R_BR_A \over RA + RB + RC} \tag{4}$$
$$ \begin{split}
(R_2 + R_3) - (R_2 - R_3) &= ({ R_AR_B + R_AR_C \over R_A + R_B + R_C})\\
\qquad & - ({R_AR_C - R_BR_A \over R_A + R_B + R_C})
\end{split} $$
$$\bbox[5px,border:1px solid grey] {R_3 = {R_AR_B \over R_A + R_B + R_C}} \tag{5}$$
$$\bbox[5px,border:1px solid grey] {R_1 = {R_BR_C \over R_A + R_B + R_C}} \tag{6}$$
$$\bbox[5px,border:1px solid grey]{R_2 = {R_AR_C \over R_A + R_B + R_C}} \tag{7}$$
Wye to Delta Conversion
To obtain the relationships necessary to convert from a Y to a Δ, first divide Eq. (5) by Eq. (6):$$\begin{split}
{R_3 \over R_1} &= {(R_AR_B)/(R_A + R_B + R_C) \over
(R_BR_C)/(R_A + R_B + R_C) } \\
&= {R_A \over R_C}\\
\end{split}$$
$$R_A = {R_CR_3 \over R_1}$$
$$\begin{split}
{R_3 \over R_2} &= {(R_AR_B)/(R_A + R_B + R_C) \over (R_AR_C)/(R_A + R_B + R_C)}\\
&={R_B \over R_C}\\
\end{split}$$
$$R_B = {R_3R_C \over R_2}$$
$$\begin{split}
R_2 &= {(R_CR_3/R_1)R_C \over (R_3R_C/R_1) + (R_CR_3/R_2) + R_C}\\
& = {(R_3/R1)R_C \over (R_3/R_2) + (R_3/R_1) + 1}
\end{split}$$
$$ \begin{split}
R_2 &= {(R_3R_C/R_1) \over (R_1R_2 + R_1R_3 + R_2R_3)/(R_1R_2)}\\
&= {R_2R_3R_C \over R_1R_2 + R_1R_3 + R_2R_3}
\end{split}$$
$$\bbox[5px,border:1px solid grey] {R_C = {R_1R_2 + R_1R_3 + R_2R_3 \over R_3}} \tag{8}$$
$$\bbox[5px,border:1px solid grey] {R_A = {R_1R_2 + R_1R_3 + R_2R_3 \over R_1}} \tag{9}$$
$$\bbox[5px,border:1px solid grey] {R_B = {R_1R_2 + R_1R_3 + R_2R_3 \over R_2}} \tag{10}$$
$$ \begin{split}
R_3 &= {R_AR_B \over R_A + R_B + R_C} \\
&= {R_AR_A \over R_A + R_A + R_A} \\
&= {(R_A)^2 \over 3R_A} \\
R_3&= {R_A \over 3}\\
\end{split}
$$
$$\bbox[5px,border:1px solid grey]{R_Y = {R_{Δ} \over 3}} \tag{11}$$
$$\bbox[5px,border:1px solid grey] {R_{Δ} = 3R_Y} \tag{12}$$
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