# Star and delta configurations

In addition to parallel and series, impedances also can make star and delta connections. Also are shown the transformations.

### Star and delta connections

Some sources label delta connection as pi and star as T.

### Star-delta transformation (Y-Δ)

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?

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.

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.

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.

Converting internal star in delta.

$R_{\Delta }=3\cdot R_{Y}$

$R_{\Delta }=3\cdot 6=18 \Omega$

3 resistor pairs are in parallel. Just simplify the circuit to find I current.

$Rt=\frac{9\cdot 18}{9+18}=6\Omega$

$I=\frac{42}{6}=8A$

How to calculate I current in this circuit?

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.

$Rt=\frac{9,33k\cdot 7,39k}{16,72k}=4,12k\Omega$

$I=\frac{8}{0,41k+4,12k}=1,76mA$

How to find total resistance of this resistor association?

Putting this cube in a shape that allows to visualize a group of resistors for conversion.

$R_{Y}=\frac{R_{\Delta }}{3}=\frac{9}{3}=3\Omega$

Converting delta on the left in star.

$R_{a}=R_{b}=\frac{R1\cdot R2}{R1+R2+R7}=\frac{9\cdot 12}{33}=3,27\Omega$

$R_{c}=\frac{12\cdot 12}{9+12+12}=4,36\Omega$

$R7+R8=12,27\Omega$

$R2+R6=7,36\Omega$

$\frac{12,27\cdot 7,36}{12,27+7,36}=4,6\Omega$

$3,27+4,6=7,87\Omega$

Finally, the total resistance is:

$Rt=\frac{9\cdot 7,87}{9+7,87}=4,19\Omega$ 