# Star Delta (Y-Δ) Transformation with Example

In electrical networks, it’s often circuits are series combination and parallel combination. These complex circuits can be simplified to single resistor circuit theoretically using series and/or parallel formulas. But sometimes circuit’s element can’t be categorized as series or parallel. Take a look at the circuit below and you will observe that the R­3 is kind of confusing and can’t be categorized as series or parallel. Such circuits have multilateral shapes with each node at the corner. The simplest shape is three node shape, based on the shape they can be categorized as Star (Y) or Delta (Δ) connections. Each shape can be transformed into another shape using some transformational techniques. Arthur Edwin Kennelly first time proposed the Star-Delta transformation. The same transformation is also known as Star-Delta, T-π, and Star-Mesh transformation.

## Star to Delta Conversion:

If head and tails of three circuit elements are such connected that it makes a closed loop, such connection is called Delta Connection. Take a look at the diagram below and you will find two different geometries with a similar connection. To convert Star (Y) connection (on the left side) into delta (Δ) connection (on the right side), the following formulas can be used.
$$R_{ ab }=\frac { R_{ a }R_{ b }+R_{ b }R_{ c }+R_{ a }R_{ c } }{ R_{ c } } \\ R_{ bc }=\frac { R_{ a }R_{ b }+R_{ b }R_{ c }+R_{ a }R_{ c } }{ R_{ a } } \\ R_{ ac }=\frac { R_{ a }R_{ b }+R_{ c }R_{ c }+R_{ a }R_{ c } }{ R_{ b } }$$ R

Rb

Rc

Rab

Rbc

Rac

## Delta to Star Conversion:

If the heads or tails of three circuit elements are connected together which provide a common point, such connection is called Star (Y) Connection. Take a look at the diagram below, there are two different geometries for exactly the same connection. The conversion simplifies the circuit and converts delta connection to Star equivalent connection. We already know the resistances of Delta connection on left side and formula for right side Star equivalent connection resistances are given below.
$$R_{ ab }=\frac { R_{ a }R_{ b } }{ R_{ a }+R_{ b }+R_{ c } } \\ R_{ bc }=\frac { R_{ b }R_{ c } }{ R_{ a }+R_{ b }+R_{ c } } \\ R_{ ac }=\frac { R_{ a }R_{ c } }{ R_{ a }+R_{ b }+R_{ c } }$$ R

Rb

Rc

Rab

Rbc

Rac

## Example:

Suppose the above circuit that we want to compute equivalent resistance of the following circuit. It’s pretty difficult to find whether R3 is series or parallel. By observing, we can find that there two delta connections inside the circuit. I pick the upper delta connection and convert it to Star. 

Given values of R1, R2, R3, R4, and R5 are 5Ω, 10Ω, 15Ω, 20Ω and 25Ω respectively. The corresponding Star equivalent resistor R12, R23, R13, R4 and R5, where R4 and R5 are same as above.
$R_{ 12 }=\frac { 5×10 }{ 5+10+15 } =\frac { 5 }{ 3 } \\ R_{ 23 }=\frac { 10×15 }{ 5+10+15 } =5 \\ R_{ 13 }=\frac { 5×15 }{ 5+10+15 } =\frac { 5 }{ 2 }$ Now we can easily have identified all the resistances in the above diagram, such that R­­12 is in series with (R13 + R4) || (R23 + R5). So, the simple series-parallel calculation can be carried out as follow.

$R_{ Eq }=R_{ 12 }+\frac { (R_{ 1 }3+R_{ 4 })(R_{ 2 }3+R_{ 5 }) }{ (R_{ 2 }3+R_{ 5 })+(R_{ 1 }3+R_{ 4 }) } \\ R_{ Eq }=\frac { 5 }{ 3 } +\frac { \left( \frac { 5 }{ 2 } +20 \right) (5+25) }{ \left( \frac { 5 }{ 2 } +20 \right) +(5+25) } \\ R_{ Eq }=\frac { 5 }{ 3 } +\frac { \left( \frac { 45 }{ 2 } \right) 30 }{ \left( \frac { 45 }{ 2 } \right) +30 } =\frac { 5 }{ 3 } +\frac { 675 }{ \frac { 105 }{ 2 } } \\ R_{Eq}≅14.51\Omega$ ## Star Delta transformation in AC:

In AC circuits, impedance is playing the same role, resistance is playing in DC circuits. The complex quantity impedance is the combination of reactance and resistance and reactance is a combination of inductive and capacitive reactance. Fortunately, the above transformation is true for AC circuits having complex impedance instead of resistance.

For Δ-Y conversion:
$Z_{ ab }=\frac { Z_{ a }Z_{ b } }{ Z_{ a }+Z_{ b }+Z_{ c } } \\ Z_{ bc }=\frac { Z_{ b }Z_{ c } }{ Z_{ a }+Z_{ b }+Z_{ c } } \\ Z_{ ac }=\frac { Z_{ a }Z_{ c } }{ Z_{ a }+Z_{ b }+Z_{ c } }$

For Y-Δ conversion:
$\\ Z_{ ab }=\frac { Z_{ a }Z_{ b }+Z_{ b }Z_{ c }+Z_{ a }Z_{ c } }{ Z_{ c } } \\ Z_{ bc }=\frac { Z_{ a }Z_{ b }+Z_{ b }Z_{ c }+Z_{ a }Z_{ c } }{ Z_{ a } } \\ Z_{ a }c=\frac { Z_{ a }Z_{ b }+Z_{ b }Z_{ c }+Z_{ a }Z_{ c } }{ Z_{ b } }$


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