Ac Wye Delta Transformation

Example 1: Find the total impedance $Z_T$ of the network of Fig. 3. 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): $$Z_1 = 3.84 Ω + j 1.11 Ω$$ $$Z_2 = 3.84 Ω + j 1.11 Ω$$ $$Z_3 = -1.92 Ω - j 2.56 Ω$$ $$Z_4 =2Ω \,\, Z_5 = 3 Ω$$ 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, $$Z_T = Z_3 + Z_{T3}\\ =-1.92Ω - j 2.56Ω + 3.15 Ω j 0.56 Ω\\ = 1.23 Ω - j 2.0 Ω = 2.35 \angle -58.41^\circ$$