Question

For a three-phase symmetrical impedance load, the sequence impedance matrix is ___________ and hence the sequence networks are (a) coupled or (b) uncoupled.

Short answer

Therefore, the correct option is (a).

Step-by-step solution

Determine the equation to calculate the sequence impedance matrix.

The phase impedance matrix for the three-phase star connected load is given as follows.

${\mathit{Z}}_{\mathbf{p}}\mathbf{=}\left[\begin{array}{ccc}{Z}_y+Z_n& Z_n& Z_n\\ Z_n& Z_y+Z_n& Z_n\\ Z_n& Z_n& Z_y+Z_n\end{array}\right]$ ……. (1)

Here ${\mathit{Z}}_{\mathbf{y}}$is the per phase impedance and ${\mathit{Z}}_{\mathbf{n}}$is the neutral impedance.

The equation to calculate the sequence impedance matrix is given as follows.

${\mathit{Z}}_{\mathbf{s}}\mathbf{=}{\mathit{A}}^{\mathbf{-}\mathbf{1}}{\mathit{Z}}_{\mathbf{p}}\mathit{A}$ ……. (2)

Here, $\mathit{A}$is the transformation matrix.

Consider the identities.

role="math" localid="1654168104322" $$\begin{gathered} \mathbf{1}\mathbf{+}\mathit{a}\mathbf{+}{\mathit{a}}^{\mathbf{2}}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}{\mathit{a}}^{\mathbf{4}}\mathbf{=}\mathit{a} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{a}}^{\mathbf{3}}\mathbf{=}\mathbf{1} \end{gathered}$$

Determine the sequence impedance matrix.

Calculate the sequence impedance matrix.

Substitute $\frac{\mathbf{1}}{\mathbf{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]$for ${\mathit{A}}^{\mathbf{-}\mathbf{1}}$, $\left[\begin{array}{ccc}{Z}_y+Z_n& Z_n& Z_n\\ Z_n& Z_y+Z_n& Z_n\\ Z_n& Z_n& Z_y+Z_n\end{array}\right]$for ${\mathit{Z}}_{\mathbf{p}}$and $\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]$for $\mathit{A}$into equation (2).

$$\begin{gathered} {\mathit{Z}}_{\mathbf{s}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{ccc}{Z}_y+Z_n& Z_n& Z_n\\ Z_n& Z_y+Z_n& Z_n\\ Z_n& Z_n& Z_y+Z_n\end{array}\right]\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right] \\ {\mathit{Z}}_{\mathbf{s}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{ccc}{Z}_y+Z_n& Z_n& Z_n\\ Z_n& Z_y+Z_n& Z_n\\ Z_n& Z_n& Z_y+Z_n\end{array}\right] \\ {\mathit{Z}}_{\mathbf{s}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left[\begin{array}{ccc}3\left(Z_y+3Z_n\right)& Z_y\left(1+a+a^2\right)& Z_y\left(1+a+a^2\right)\\ \left(Z_y+3Z_n\right)\left(1+a+a^2\right)& Z_y\left(1+a^3+a^3\right)& Z_y\left(1+a^4+a^2\right)\\ \left(Z_y+3Z_n\right)\left(1+a+a^2\right)& Z_y\left(1+a^4+a^2\right)& Z_y\left(1+a^3+a^3\right)\end{array}\right] \end{gathered}$$

Substitute $\mathit{a}$for ${\mathit{a}}^{\mathbf{4}}$, $\mathbf{1}$for ${\mathit{a}}^{\mathbf{3}}$and $\mathbf{0}$for $\mathbf{1}\mathbf{+}\mathit{a}\mathbf{+}{\mathit{a}}^{\mathbf{2}}$into above matrix.

$$\begin{gathered} {\mathit{Z}}_{\mathbf{s}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left[\begin{array}{ccc}3\left(Z_y+3Z_n\right)& 0& Z_y\left(1+a+a^2\right)\\ 0& Z_y\left(1+1+1\right)& Z_y\left(1+a+a^2\right)\\ 0& 0& Z_y\left(1+1+1\right)\end{array}\right] \\ {\mathit{Z}}_{\mathbf{s}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left[\begin{array}{ccc}3\left(Z_y+3Z_n\right)& 0& 0\\ 0& 3Z_y& 0\\ 0& 0& 3Z_y\end{array}\right] \\ {\mathit{Z}}_{\mathbf{s}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left[\begin{array}{ccc}3\left(Z_y+3Z_n\right)& 0& 0\\ 0& Z_y& 0\\ 0& 0& Z_y\end{array}\right] \end{gathered}$$

The sequence impedance matrix is the diagonal matrix and independent on the other phase’s impedance, therefore sequence networks are uncoupled.