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.
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,
$$Z_T = Z_3 + Z_{T3}\\
=-1.92Ω - j 2.56Ω + 3.15 Ω j 0.56 Ω\\
= 1.23 Ω - j 2.0 Ω = 2.35 \angle -58.41^\circ$$
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