Kennelly's theorem

The Kennelly theorem, named in honor of Arthur Edwin Kennelly, allows us to determine the equivalent charge in a star to one given in a triangle and vice versa. The theorem is also often called star-triangle transformation (written Y-Δ) or te-delta transformation (written T -Δ).

Transformation equations

The following table shows the transformation equations as a function of impedances and admittances.

Kennelly equations
Transformation Δ-Y
Depending on impedance Depending on admissions ${displaystyle {vec {Z}}_{at}={frac {{vec {Z}}_{AB}.{vec {Z}_{AC}}}{{vec {Z}}}}}_{AB+}{vec {Z}}}}_{BC}+{vec {Z}}}_{AC}}}}$ ${displaystyle {vec {Y}}_{AT}={vec {Y}}_{AB+}{vec {Y}}}_{AC}+{frac {{vec {Y}}}}{vec {Y}}}}{vec {Y}}}}{{vec {Y}}}}$ ${displaystyle {vec {Z}}_{BT}={frac {{vec {Z}}_{AB}.{vec {Z}}_{BC}}{{{vec {Z}}}}}}{{{vec {Z}}}}}}}$ ${displaystyle {vec {Y}}_{BT}={vec {Y}}_{AB}+}{vec {Y}}}_{BC}+{frac {{vec {Y}}}}}{{vec {Y}}}}}}}{{{vec {Y}}}}$ ${displaystyle {vec {Z}}_{CT}={frac {{vec {Z}}_{AC}.{vec {Z}_{BC}}}{{{vec {Z}}}}}}{{{AB+}{vec {Z}}}_{BC}$ ${displaystyle {vec {Y}}_{CT}={vec {Y}}_{AC+}{vec {Y}}_{BC}+{frac {{vec {Y}}}}{vec {Y}}}.{vec {Y}}}{{vec {Y}}}}}$
Transformation Y-Δ
Depending on impedance Depending on admissions ${displaystyle Z_{AB}={vec {Z}}_{AT}+{vec {Z}}}_{BT}+{frac {{vec {Z}}_{at}.{vec {Z}}}}{BT}}}{{vec {Z}}}}}}}}{{{vec}}}$ ${displaystyle {vec {Y}}_{AB}={frac {{vec {Y}}_{at}.{vec {Y}}_{BT}}{{vec {Y}}}}{{{vec {Y}}}}{vec {Y}}}}}}}{{vec}}}}}$ ${displaystyle Z_{BC}={vec {Z}}_{BT}+{vec {Z}}}}_{CT+{frac {{vec {Z}}_{BT}.{vec {Z}}}}{{{CT}}}{{{vec {Z}}}}}}}$ ${displaystyle {vec {Y}}_{BC}={frac {{vec {Y}}_{BT}.{vec {Y}}_{CT}}}{{{vec {Y}}}}{{vec {Y}}}}}}{BT}+{vec {Y}}}$ ${displaystyle Z_{AC}={vec {Z}}_{AT}+{vec {Z}}}_{CT+{frac {{vec {Z}}_{at}.{vec {Z}}}}}{{{CT}}{{{{{vec {Z}}}}}}}}$ ${displaystyle {vec {Y}}_{AC}={frac {{vec {Y}}_{at}.{vec {Y}}_{CT}}{{vec {Y}}}}{{vec {Y}}}}{vec {Y}}}}{vec {Y}}}}}}}}{{{{$

Demonstration

The Kennelly equations are analytically demonstrated below.

Triangle to star circuit

Let us assume the values Z_AB, Z_BC and Z_AC of the triangle load in figure 1 are known and we wish to obtain the values Z_AT, Z_BT and Z_CT of its star equivalent. To do this, we will obtain in both circuits the equivalent impedances with respect to points A-B, B-C and A-C and we will equalize them since they are equivalent charges (note that in the star there are always two impedances in series, while in the triangle there are two in series with the third in parallel):

${cHFFFF}{cH00FF}{cHFFFF}{cH00}{cH00FF}{cH00FF}{cH00FF}{cH00FF}{cH00}{cHFF}{cH}{cH00}{cHFF}{cH00}{cHFF}{cH00}{cH00}{cHFF}{cH00}{cH00}{cH00}{cH00}{cH00}{cH00}{cH00}{cH00}{cH00FFFFFFFF}{cH00}{cH00FF}{cH00}{cH00}{cH00FF}{cH00}{cH00}{cH00}{cH00}{cH00}{cH00}{cH00}{cH00}{cH$

${displaystyle}{vec {Z}}{B-C}={vec {Z}{B}{B}}{vec}{Z}{Z}{vec}{b}{c}{c}{b}{b}{b}}{vec}{?$

${cHFFFFFF}{cHFFFF}{cHFFFFFF}{cHFFFFFF}{cH00}}{cHFFFF00}{cHFFFF00}{cH00}{cH00}}{cHFFFF}{cH00}{cHFFFFFF}{cH00}{cH00}{cH00}{cH00}}{cH00}{cH00}{cH00}{cH00}{cH00}}{cH00FFFFFFFFFFFFFFFF00}}}}{cHFFFFFFFFFFFFFFFFFF00}{cH00}{cH00}}}}{cH00}{cH00}{cH00}{cH00}{cH00}{cH00}{cH00}{cH00}{cH00FFFF00}{cH00}{$

The Kennelly equations are obtained from the previous ones in the following way:

1. Add the equations (1) and (3) and subtract the result of the (2)
2. Add the equations (1) and (2) and subtract the result of (3)
3. Add the equations (2) and (3) and subtract the result of (1)

Star to Triangle

Let us now assume the opposite case, that is, the values Z_AT, Z_BT and Z_CT of the star in the figure are known. 1, we want to obtain the values Z_AB, Z_BC and Z_AC of the equivalent delta load. To do this, the Δ-Y transformation equations will be taken, where for simplification of notation we will take

${displaystyle {vec {Z}}_{T}={vec {Z}}}_{AB+}{vec {Z}}}_{BC}+{vec {Z}}}}}$

leaving the following equations:

By performing the three possible binary multiplications between them, we obtain

${displaystyle {frac {({vec {Z}}_{AB})^{2}{vec {Z}_{AC}.{vec {Z}_{BC}}}}{{vec {Z}}}}}{vec}}{2}}}}{vec {Z}_{at}.{vec {Z}}{$

${displaystyle {frac {{vec {Z}_{AB}.$

${displaystyle {frac {{vec {Z}_{AB}.{vec {Z}}_{AC}.({vec {Z}}_{BC}}}{2}}}}{vec {Z}}}_{T}{vec}}}}{{vec {Z}}_{BT}{$

And adding them

${cHFFFFFF}{cH00FF00}{cH00FF00}{cH00FF00}{cH00FF00}{cH00FF00}{cH00FFFF00}{cH00FFFF00}{cH00FF00}{cH00FF00}{cH00FFFFFF00}{cH00}{cH00FFFFFFFF00}{cH00}{cH00}{cH00}{cH00FFFFFFFFFF00}{cH00}{cH00FFFFFFFFFF00}{cH00}{cH00FFFFFFFF00FF00FFFFFFFFFFFFFFFFFFFFFF00}{cH00}{cH00}{cH00}{cH00}{cH00}{cH00FFFF00}{cH00}{cH00FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF$

${displaystyle ={vec {Z}}_{at}.{vec {Z}}_{BT}+{vec {Z}}_{at}.{vec {Z}}}}_{CT}}$

We split the first member for the value of ${displaystyle {vec {Z}}_{AT}}$:

${cHFFFFFF}{cHFFFFFF}{cHFFFFFF00}{cHFFFF00}{cH00FF00}{cH00FF00}{cH00FFFF00}{cHFFFFFFFFFF00}{cHFFFF00}{cH00}{cHFFFFFFFFFFFFFFFFFFFFFFFF00}}{cH00}{cH00}}{cH00}{cH00}{cH00}{cH00}{cH00}{cHFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00}}{cH00}{cH00}{c}{cH00}{cH00}{cH00}{cH00}{cH00}{cHFFFFFFFFFFFFFFFFFFFFFFFFFFFF00}{$

${cHFFFFFF}{cHFFFF}{cHFFFF}{cHFFFF}{cHFFFF}{cHFFFF}{cHFFFFFF}{cHFF}{cHFF}{cHFF}{cHFF}{cHFF}{cHFF}{cHFF}{cHFF}{cHFFFF}{cH}{cH00}{cH00}{cH00}{cH00}{cH00}{cH00}}{cH00}{cH00}{cHFFFF}{cH00}{cH00}{cH00}{cH00}}{cH00}{cH00}{cHFFFF}{cHFF}{cH00}{cH00}{cHFF}{cH00}{cH00}{cH00}{c$

And dividing the second member by ${displaystyle {vec {Z}}_{AT}}$:

${cHFFFFFF}{cHFFFF}{cHFFFFFF}{cHFFFFFF}{cHFFFF}{cHFFFF}{cHFFFFFF}{cHFFFF}{cHFFFF}}{cHFFFF}{cHFF}{cHFFFF}{cHFFFF}{cHFF}{cH00}{cH}{cH}{cH00}{cH00}{cH00}{cH00}{cHFFFFFFFFFFFFFFFFFFFFFFFF}}}}{cHFFFFFFFFFFFF}{cH00}{cHFFFFFF}{cH}}{cH00}{cH00}{cH00}{cH00}{cH00}{cHFFFFFFFFFF}{cH00}{cH}{cH00}{cH00}{$

By matching both results we get one of the transformation equations. The other two can be obtained in the same way by dividing by ${displaystyle {vec {Z}}_{BT}}$ and ${displaystyle {vec {Z}_{CT}}}$