# Star-Delta and Delta-Star Transformation

Many times we come across some circuits in which the resistances are neither in series nor in parallel. For example, a three terminal network such as delta network or star network.

In such situations, it is not possible to simplify the circuit by series and parallel circuit rules. Hence, we need to convert the circuit from one form to another to solve it for the unknown quantities. There are two such circuit configurations that are used to overcome these difficulties. They are :

1. Star or Wye (Y) or Tee (T) connection
2. Delta(Δ) or Pi (Π) connection

Star and delta connections are two types of electrical connections used in three-phase power systems.

• In a star connection, three phases are connected at a central point.
• In a delta connection, the three phases are connected in a loop.

The choice between these connections depends on the power requirements and the type of load being supplied.

#### Star or Wye (Y) or Tee (T) connection

A star circuit is one in which similar ends of three resistances are connected to a common point called a star point or neutral point. It is also called Wye or Tee (T) connection because of its shape, as shown in the figure below.

#### Delta(Δ) or Pi (Π)connection

A delta circuit is one in which non-similar ends of three elements are connected together as shown in the figure below. It is also called mesh connection or pi (π) connected because of its shape. Note that π is actually an inverted delta.

There is no neutral point in the case of the delta connection.

#### Delta- Star Transformation

Let’s consider three resistors RAB , RBC  and RCA connected in delta to three terminals A, B, C as shown in Fig.1.

Fig.1

We need to replace these three delta connected resistors by three resistors connected in series as shown in Fig.2 so that the two networks are electrically equivalent.

Fig.2

The two arrangements will be electrically equivalent if resistance between any two terminals of one network is equal to resistance between the corresponding terminals of other network.

Hence, Resistance between A and B for Star connection =  Resistance between A and B for Delta connection

RA+ RB = RAB II (RBC + RCA )

= [RAB × (RBC + RCA )] / [RAB +RBC + RCA ] …………….. (Eq.1)

Similarly,

RB+ RC = [RBC  × (RAB + RCA )] / [RAB +RBC + RCA ] …………….. (Eq.2)

And

RC+ RA= [RCA  × (RAB + RBC )] / [RAB +RBC + RCA ] …………….. (Eq.3)

Now, substracting Eq.(2) from Eq.(1)

RA – RC = [(RAB RCA ) – (RBC RCA)] / [RAB +RBC + RCA ] …………….(Eq.4)

Now, adding Eq.3 and Eq.4, we get

2RA = 2( RAB × RCA ) / (RAB  + RBC + RCA )

Hence, R = ( RAB × RCA )/(RAB  + RBC + RCA ) ……………(Eq.5)

Similarly the value of  RB and RC will be :

R = ( RAB × RBC )/(RAB  + RBC + RCA ) …………….. (Eq.6)

R = ( RBC × RCA )/(RAB  + RBC + RCA )……………….(Eq.7)

There is an easy way to remember this is :

Any arm of start connection = (Product of two adjacent arms of Δ)/ (Sum of all the arms of Δ )

#### Star-Delta Transformation

In order to replace  the star-connected network by an equivalent delta-connected network, we need to find out the values of  RAB, RBC and RCA .

Now, dividing the equations 5 and 6 , we get ;

RA / RB = RCA / RBC

Hence, RCA = RA RBC / RB

Now, further dividing Eq.5 and Eq.7, we get

RA / RC = RAB / RBC

Hence, RAB = RA RBC / RC

Substituting forRAB and RCA  in Eq.5 and simplifying  it gives,

RBC =  RB + RC + [( RB× RC )/ RA ] …………………… (Eq.8)

Similarly,

RAB =  RA + RB + [( RA× RB )/ R] ……………(Eq.9)

RCA =  RC + RA + [( RC× RA )/ RB]………………(Eq.10)

An easy way to remember this is :

Resistance between two terminals of delta = sum of star resistances connected to those terminals plus product of the same two resistances divided by the third resistance.