In addition to parallel and series, impedances also can make star and delta connections. Also are shown the transformations.
Star and delta connections
![star and delta connections](https://www.electricalelibrary.com/wp-content/uploads/2020/11/Delta-star-connection.png)
Some sources label delta connection as pi and star as T.
![Star connection T](https://www.electricalelibrary.com/wp-content/uploads/2020/11/T-star-1.jpg)
![pi-delta](https://www.electricalelibrary.com/wp-content/uploads/2020/11/pi-delta-1.jpg)
Star-delta transformation (Y-Δ)
![transformations](https://www.electricalelibrary.com/wp-content/uploads/2020/11/Rede-superpostaDESTAQUE-1.png)
Equations to transform a star configuration to delta.
Ra=\frac{R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}}{R_{1}}
Rb=\frac{R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}}{R_{2}}
Rc=\frac{R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}}{R_{3}}
Delta-star transformation (Δ-Y)
Equations for (Δ-Y) transformation.
R_{1}=\frac{RbRc}{Ra+Rb+Rc}
R_{2}=\frac{RaRc}{Ra+Rb+Rc}
R_{3}=\frac{RaRb}{Ra+Rb+Rc}
If all resistors are equal, the equations become much simpler. Considering R_{Y} as resistor value in star and R_{\Delta } of resistor in delta.
R_{Y}=\frac{R_{\Delta }}{3}
R_{\Delta }=3\cdot R_{Y}
Transformation with capacitors and coils
What if instead of resistors, we have capacitors and coil?
![Star delta configuration with capacitors](https://www.electricalelibrary.com/wp-content/uploads/2020/11/V.7-1.png)
Demonstrating the equation of (Δ-Y) transformation for capacitors.
\frac{1}{c_{1}}=\frac{\frac{1}{C_{2}C_{3}}}{\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}}
c_{1}=\frac{C_{2}C_{3}+C_{3}C_{1}+C_{1}C_{2}}{C_{1}}
c_{2}=\frac{C_{2}C_{3}+C_{3}C_{1}+C_{1}C_{2}}{C_{2}}
c_{3}=\frac{C_{2}C_{3}+C_{3}C_{1}+C_{1}C_{2}}{C_{3}}
Equations for inverse transformation (Y-Δ).
\frac{1}{C_{1}}=\frac{\frac{1}{c_{1}c_{2}}+\frac{1}{c_{1}c_{3}}+\frac{1}{c_{2}c_{3}}}{\frac{1}{c_{1}}}
C_{1}=\frac{c_{2}c_{3}}{c_{1}+c_{2}+c_{3}}
C_{2}=\frac{c_{1}c_{3}}{c_{1}+c_{2}+c_{3}}
C_{3}=\frac{c_{1}c_{2}}{c_{1}+c_{2}+c_{3}}
And for inductors, the calculations are similar to resistors.
![configurations with coils](https://www.electricalelibrary.com/wp-content/uploads/2020/11/https___haygot.s3.amazonaws.com_443_cheatsheet_21627-1.png)
Equivalent values from Y to Δ.
L_{RS}=\frac{L_{R}L_{S}+L_{S}L_{T}+L_{R}L_{T}}{L_{T}}
L_{RT}=\frac{L_{R}L_{S}+L_{S}L_{T}+L_{R}L_{T}}{L_{S}}
L_{ST}=\frac{L_{R}L_{S}+L_{S}L_{T}+L_{R}L_{T}}{L_{R}}
Equivalent values from Δ to Y.
L_{R}=\frac{L_{RS}L_{RT}}{L_{RS}+L_{RT}+L_{ST}}
L_{S}=\frac{L_{RS}L_{ST}}{L_{RS}+L_{RT}+L_{ST}}
L_{T}=\frac{L_{ST}L_{RT}}{L_{RS}+L_{RT}+L_{ST}}
And for impedances.
![impedances](https://www.electricalelibrary.com/wp-content/uploads/2020/11/Delta_Star_transformation-1.png)
Conversion from Y to Δ.
Z_{ac}=\frac{Z_{a}Z_{c}+Z_{a}Z_{b}+Z_{b}Z_{c}}{Z_{b}}
Z_{ab}=\frac{Z_{a}Z_{c}+Z_{a}Z_{b}+Z_{b}Z_{c}}{Z_{c}}
Z_{bc}=\frac{Z_{a}Z_{c}+Z_{a}Z_{b}+Z_{b}Z_{c}}{Z_{a}}
Conversion from Δ to Y.
Z_{a}=\frac{Z_{ab}Z_{ac}}{Z_{ab}+Z_{ac}+Z_{bc}}
Z_{b}=\frac{Z_{ab}Z_{bc}}{Z_{ab}+Z_{ac}+Z_{bc}}
Z_{c}=\frac{Z_{ac}Z_{bc}}{Z_{ab}+Z_{ac}+Z_{bc}}
3 problem examples
Let’s find the value of I current in this circuit.
![Problem 1 with star and delta](https://www.electricalelibrary.com/wp-content/uploads/2020/11/Problem-1-star-delta-1.png)
Converting internal star in delta.
R_{\Delta }=3\cdot R_{Y}
R_{\Delta }=3\cdot 6=18 \Omega
![Problem 1 with 2 deltas](https://www.electricalelibrary.com/wp-content/uploads/2020/11/Problem-1-star-delta-2-1.png)
3 resistor pairs are in parallel. Just simplify the circuit to find I current.
![Simplified problem 1](https://www.electricalelibrary.com/wp-content/uploads/2020/11/Problem-1-star-delta-3-1.png)
Rt=\frac{9\cdot 18}{9+18}=6\Omega
I=\frac{42}{6}=8A
How to calculate I current in this circuit?
![](https://www.electricalelibrary.com/wp-content/uploads/2020/11/Problema-2-1.png)
Replacing R1, R2 and R3 with (Δ-Y) tranformation.
Ra=\frac{4,7k\cdot 1,1k}{4,7k+1,1k+6,8k}=\frac{5,17k}{12,6}=0,41k\Omega
Rb=\frac{4,7k\cdot 6,8k}{4,7k+1,1k+6,8k}=\frac{31,96k}{12,6}=2,53k\Omega
Rc=\frac{1,1k\cdot 6,8k}{4,7k+1,1k+6,8k}=\frac{7,48k}{12,6}=0,59k\Omega
With simplified circuit, becomes easier to calculate current.
![Problem 2 2](https://www.electricalelibrary.com/wp-content/uploads/2020/11/Problem-2-2-1.png)
![Problem-2-3-1](https://www.electricalelibrary.com/wp-content/uploads/2020/11/Problem-2-3-1.png)
Rt=\frac{9,33k\cdot 7,39k}{16,72k}=4,12k\Omega
I=\frac{8}{0,41k+4,12k}=1,76mA
How to find total resistance of this resistor association?
![Problem 3 of star and delta](https://www.electricalelibrary.com/wp-content/uploads/2020/11/Problem-3-1.png)
Putting this cube in a shape that allows to visualize a group of resistors for conversion.
![Problem-3-2 about star and delta](https://www.electricalelibrary.com/wp-content/uploads/2020/11/Problem-3-2.png)
R_{Y}=\frac{R_{\Delta }}{3}=\frac{9}{3}=3\Omega
![](https://www.electricalelibrary.com/wp-content/uploads/2020/11/Problem-3-3-1.png)
![](https://www.electricalelibrary.com/wp-content/uploads/2020/11/Problem-3-4-1.png)
Converting delta on the left in star.
R_{a}=R_{b}=\frac{R1\cdot R2}{R1+R2+R7}=\frac{9\cdot 12}{33}=3,27\Omega
R_{c}=\frac{12\cdot 12}{9+12+12}=4,36\Omega
![](https://www.electricalelibrary.com/wp-content/uploads/2020/11/Problem-3-5-1.png)
R7+R8=12,27\Omega
R2+R6=7,36\Omega
\frac{12,27\cdot 7,36}{12,27+7,36}=4,6\Omega
3,27+4,6=7,87\Omega
Finally, the total resistance is:
Rt=\frac{9\cdot 7,87}{9+7,87}=4,19\Omega