# Strange Delta-Wye Transform

In power theory (notably three-phase AC power theory), the delta-wye transform can be used to convert a delta impedance network (left in diagram) into an equivalent wye network (right in diagram). The word "equivalent" here means that when supplied with identical node-to-node voltages ($$\vec{V_{AB}},\vec{V_{BC}},\vec{V_{CA}}$$), the network draws identical line currents ($$\vec{I_{A}},\vec{I_{B}},\vec{I_{C}}$$).

When doing these transforms, we are often converting a purely resistive delta network to a purely resistive wye network. Such wye networks are equivalent to their respective delta networks for all possible sets of boundary voltages.

However, we can derive a wye network consisting of inductive, capacitive, and resistive components which is equivalent to a purely resistive delta network for one particular set of boundary voltages.

Consider the purely resistive delta network on the left (with accompanying boundary voltages). Its properties are below:

$\large{\vec{Z_{AB}} = \vec{Z_{BC}} = \vec{Z_{CA}} = (10 + j0) \Omega \\ \vec{V_{AB}} = 10 \angle 0^\circ \hspace{1cm} \vec{V_{BC}} = 10 \angle {-120}^\circ \hspace{1cm} \vec{V_{CA}} = 10 \angle 120^\circ}$

Suppose we transform the delta network into a (provisionally) equivalent wye network with the following properties:

$\large{\vec{Z_{AN}} = (5 + j0)\Omega \hspace{1cm} \vec{Z_{BN}} = \vec{Z} \hspace{1cm} \vec{Z_{CN}} = \vec{Z}^* \\ \vec{V_{AB}} = 10 \angle 0^\circ \hspace{1cm} \vec{V_{BC}} = 10 \angle {-120}^\circ \hspace{1cm} \vec{V_{CA}} = 10 \angle 120^\circ}$

To 3 decimal places, what is the magnitude of $$\vec{Z}$$, in ohms?

Details and Assumptions:
- $$j = \sqrt{-1}$$ (the imaginary unit)
- The $$\angle$$ symbol denotes complex phase angle
- The $$*$$ superscript denotes complex conjugation

×