Question

Figure 2.33 gives the general $\mathbf{∆}\mathbf{-}\mathbf{Y}$transformation. (a) Show that the general transformation reduces to that given in Figure 2.16 for a balanced three-phase load. (b) Determine the impedances of the equivalent Y for the following impedances: localid="1655112851334" ${\mathbf{Z}}_{\mathbf{AB}}\mathbf{=}\mathbf{j}\mathbf{}\mathbf{10}\mathbf{,}{\mathbf{Z}}_{\mathbf{BC}}\mathbf{=}\mathbf{j}\mathbf{20}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{Z}}_{\mathbf{CA}}\mathbf{=}\mathbf{-}\mathbf{j}\mathbf{25}\mathbf{\Omega }$.

Short answer

(a) From the equation (1) and (2), results are same. Therefore, all the equation are verified.

(b) The impedances of the equivalent Y for the $∆$impedances are $-\mathrm{j}50\mathrm{\Omega },\mathrm{j}40\mathrm{\Omega }\mathrm{and}-\mathrm{j}100\mathrm{\Omega }$.

Step-by-step solution

Step 1: Show that the general transformation reduces to that given in Figure 2.16 for a balanced three-phase load.

(a)

For the balance three phase Y- connected load.

$$\begin{gathered} {\mathbf{Z}}_{\mathbf{A}}\mathbf{=}{\mathbf{Z}}_{\mathbf{B}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}{\mathbf{Z}}_{\mathbf{c}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}{\mathbf{Z}}_{\mathbf{Y}} \end{gathered}$$

For the balance three phase $∆-$connected load.

$$\begin{gathered} {\mathrm{Z}}_{\mathrm{AB}}={\mathrm{Z}}_{\mathrm{Bc}} \\ ={\mathrm{Z}}_{\mathrm{CA}} \\ ={\mathrm{Z}}_{∆} \end{gathered}$$

Substitute the value in any of the $\mathrm{Y}-∆$expression given

$$\begin{gathered} {\mathrm{Z}}_{∆}=\frac{{\mathrm{Z}}_{\mathrm{Y}}{\mathrm{Z}}_{\mathrm{Y}}+{\mathrm{Z}}_{\mathrm{Y}}{\mathrm{Z}}_{\mathrm{Y}}+{\mathrm{Z}}_{\mathrm{Y}}{\mathrm{Z}}_{\mathrm{Y}}}{{\mathrm{Z}}_{\mathrm{Y}}+{\mathrm{Z}}_{\mathrm{Y}}+{\mathrm{Z}}_{\mathrm{Y}}} \\ {\mathrm{Z}}_{∆}=\frac{{\mathrm{Z}}_{\mathrm{Y}}^2+{\mathrm{Z}}_{\mathrm{Y}}^2+{\mathrm{Z}}_{\mathrm{Y}}^2}{+3{\mathrm{Z}}_{\mathrm{Y}}} \\ {\mathrm{Z}}_{∆}=\frac{3{\mathrm{Z}}_{\mathrm{Y}}^2}{3{\mathrm{Z}}_{\mathrm{Y}}} \\ {\mathrm{Z}}_{∆}=3{\mathrm{Z}}_{\mathrm{Y}} \\ \left(1\right) \end{gathered}$$

Now substitute any of the expression of $∆-\mathrm{Y}$

$$\begin{gathered} \mathrm{ZY}=\frac{{\mathrm{Z}}_{∆}{\mathrm{Z}}_{∆}}{{\mathrm{Z}}_{∆}+{\mathrm{Z}}_{∆}+{\mathrm{Z}}_{∆}} \\ {\mathrm{Z}}_{\mathrm{Y}}=\frac{{\mathrm{Z}}_{∆}^2}{3{\mathrm{Z}}_{∆}} \\ {\mathrm{Z}}_{\mathrm{Y}}=\frac{{\mathrm{Z}}_{∆}}{3} \\ {\mathrm{Z}}_{∆}=3{\mathrm{Z}}_{\mathrm{Y}} \\ \left(1\right) \end{gathered}$$

From the equation (1) and (2), results are same. Therefore, all the equation are verified.

Calculates the impedances of the equivalent for the impedances.

(b)

Calculate the impedance$Z_A$as,

$$\begin{gathered} Z_A=\frac{Z_{AB}{Z}_{CA}}{Z_{AB}+Z_{BC}+Z_{CA}} \\ {Z}_A=\frac{\left(j10\right)(-j25)}{j10+j20-j25} \\ {Z}_A=\frac{-250}{j5} \\ {Z}_A=-50\Omega \end{gathered}$$

Calculate the impedance$Z_{\mathit{B}}$as

localid="1655112965966" $$\begin{gathered} {\mathrm{Z}}_B=\frac{{\mathrm{Z}}_{\mathrm{AB}}{\mathrm{Z}}_{BA}}{{\mathrm{Z}}_{\mathrm{AB}}+{\mathrm{Z}}_{\mathrm{BC}}+{\mathrm{Z}}_{\mathrm{CA}}} \\ {\mathrm{Z}}_B=\frac{\left(\mathrm{j}10\right)(-\mathrm{j}25)}{\mathrm{j}10+\mathrm{j}20-\mathrm{j}25} \\ {\mathrm{Z}}_B=\frac{-200}{\mathrm{j}5} \\ {\mathrm{Z}}_B=j40\mathrm{\Omega } \end{gathered}$$

Calculate the impedance$Z_C$as,

localid="1655112971637" $$\begin{gathered} {\mathrm{Z}}_{\mathrm{C}}=\frac{{\mathrm{Z}}_{BC}{\mathrm{Z}}_{CA}}{{\mathrm{Z}}_{\mathrm{AB}}+{\mathrm{Z}}_{\mathrm{BC}}+{\mathrm{Z}}_{\mathrm{CA}}} \\ {\mathrm{Z}}_C=\frac{\left(\mathrm{j}20\right)(-\mathrm{j}25)}{\mathrm{j}10+\mathrm{j}20-\mathrm{j}25} \\ {\mathrm{Z}}_C=\frac{500}{\mathrm{j}5} \\ {\mathrm{Z}}_C=-j100\mathrm{\Omega } \end{gathered}$$

Hence the impedances of the equivalent Y for the $∆$impedances are $-j50\Omega ,j40\Omega \mathrm{and}-j100\Omega .$.