# Ac Wye Delta Transformation

The $\Delta -Y$, $Y- \Delta$ (or $\pi -T$, $T- \pi$ as defined in Section 7.12) conversions for ac circuits will not be derived here since the development corresponds exactly with that for dc circuits. Taking the $\Delta -Y$ configuration shown in Fig. 1,
Fig. 1: $\Delta -Y$ configuration
we find the general equations for the impedances of the Y in terms of those for the $\Delta$: $$Z_1 = {Z_BZ_C \over Z_A + Z_B + Z_C}$$ $$Z_2 = {Z_CZ_A \over Z_A + Z_B + Z_C}$$ $$Z_3 = {Z_AZ_B \over Z_A + Z_B + Z_C}$$ For the impedances of the $\Delta$ in terms of those for the Y, the equations are $$Z_A = {Z_1Z_2 + Z_1Z_3 +Z_2Z_3 \over Z_1}$$ $$Z_B = {Z_1Z_2 + Z_1Z_3 +Z_2Z_3 \over Z_2}$$ $$Z_C = {Z_1Z_2 + Z_1Z_3 +Z_2Z_3 \over Z_3}$$
Note that each impedance of the Y is equal to the product of the impedances in the two closest branches of the $\Delta$, divided by the sum of the impedances in the $\Delta$.
Further, the value of each impedance of the $\Delta$ is equal to the sum of the possible product combinations of the impedances of the Y, divided by the impedances of the Y farthest from the impedance to be determined.
Drawn in different forms (Fig. 2), they are also referred to as the T and $\pi$ configurations.
Fig. 2: The T and $\pi$ configurations.
In the study of dc networks, we found that if all of the resistors of the $\Delta$ or Y were the same, the conversion from one to the other could be accomplished using the equation $$R_{\Delta} = 3 R_Y \,\, or \,\, R_Y = {R_{\Delta} \over 3}$$ For ac networks, $$Z_{\Delta} = 3 Z_Y \,\, or \,\, Z_Y = {Z_{\Delta} \over 3}$$ Be careful when using this simplified form. It is not sufficient for all the impedances of the $\Delta$ or Y to be of the same magnitude: The angle associated with each must also be the same.
Example 1: Find the total impedance $Z_T$ of the network of Fig. 3.
Fig. 3: Converting the upper $\Delta$ of a bridge configuration to a Y
Solution:
$Z_B = -j4$, $Z_A = -j4$, $Z_C = 3 + j4$ $$\begin{split} Z_1 &= {Z_BZ_C \over Z_A +Z_B + Z_C}\\ &= {(-j 4 Ω)(3 Ω + j 4 Ω) \over (-j 4 Ω) + (-j 4 Ω)+ (3 Ω + j 4 Ω) + Z_C}\\ &={(4 \angle -90^\circ)(5 \angle 53.13^\circ) \over (5 \angle -53.13^\circ) }\\ &=4Ω \angle 16.13^\circ = 3.84 Ω + j 1.11 Ω\\ Z_2 &= {Z_AZ_C \over Z_A +Z_B + Z_C}\\ &= {(-j 4 Ω)(3 Ω + j 4 Ω) \over (-j 4 Ω) + (-j 4 Ω)+ (3 Ω + j 4 Ω) + Z_C}\\ &=4Ω \angle 16.13^\circ = 3.84 Ω + j 1.11 Ω\\ \end{split}$$ In this example, $Z_A = Z_B$. Therefore, $Z_1 = Z_2$, and $$\begin{split} Z_3 &= {Z_AZ_B \over Z_A +Z_B + Z_C}\\ &= {(-j 4 Ω)(-j 4 Ω) \over (-j 4 Ω) + (-j 4 Ω)+ (3 Ω + j 4 Ω) + Z_C}\\ &=3.2 Ω \angle -126.87^\circ = -1.92 Ω - j 2.56 Ω\\ \end{split}$$ Replace the $\Delta$ by the Y (Fig. 4):
Fig. 4: The network of Fig. 3 following the substitution of the Y configuration.
$$Z_1 = 3.84 Ω + j 1.11 Ω$$ $$Z_2 = 3.84 Ω + j 1.11 Ω$$ $$Z_3 = -1.92 Ω - j 2.56 Ω$$ $$Z_4 =2Ω \,\, Z_5 = 3 Ω$$ Impedances $Z_1$ and $Z_4$ are in series: $$Z_{T1} = Z_1+ Z_4 = 3.84 Ωj + j 1.11 + 2Ω \\ =5.94Ω \angle 10.75^\circ$$ Impedances $Z_2$ and $Z_5$ are in series: $$Z_{T2} = Z_2+ Z_5 = 3.84 Ωj + j 1.11 + 3Ω \\ =6.93Ω \angle 9.22^\circ$$ Impedances $Z_{T1}$ and $Z_{T2}$ are in parallel: $$\begin{split} Z_{T3} &= {Z_{T1} Z_{T2} \over Z_{T1} +Z_{T2} }\\ &={(5.94Ω \angle 10.75^\circ)(6.93Ω \angle 9.22^\circ) \over (5.94Ω \angle 10.75^\circ) +(6.93Ω \angle 9.22^\circ) }\\ &= { 41.16Ω \angle 19.98^\circ \over 12.87\angle 9.93^\circ }\\ &= 3.198 Ω \angle 10.05^\circ= 3.15Ω + j 0.56 Ω\\ \end{split}$$ Impedances $Z_3$ and $Z_{T3}$ are in series. Therefore, $$Z_T = Z_3 + Z_{T3}\\ =-1.92Ω - j 2.56Ω + 3.15 Ω j 0.56 Ω\\ = 1.23 Ω - j 2.0 Ω = 2.35 \angle -58.41^\circ$$