Question

The sequence impedance matrix ${\mathit{Z}}_{\mathbf{s}}$for a balanced-Y load is a diagonal matrix and the sequence networks are uncoupled.

(a) True

(b) False

Short answer

Therefore, the correct option is (a): True.

Step-by-step solution

Determine the formula to calculate the diagonal matrix for balanced Y-connected load.

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

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

Here ${\mathit{Z}}_{\mathbf{P}}$is the phase impedance matrix and Ais the transformation matrix.

The transformation matrix is given as follows.

$\mathit{A}\mathbf{=}\mathbf{\left[}\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& {\mathbf{a}}^{\mathbf{2}}& \mathbf{a}\\ \mathbf{1}& \mathbf{a}& {\mathbf{a}}^{\mathbf{2}}\end{array}\mathbf{\right]}$

The inverse of transformation matrix is given as follows.

${\mathit{A}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\left[}\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& \mathbf{a}& {\mathbf{a}}^{\mathbf{2}}\\ \mathbf{1}& {\mathbf{a}}^{\mathbf{2}}& \mathbf{a}\end{array}\mathbf{\right]}$

The phase impedance matrix is given as follows.

${\mathit{Z}}^{\mathbf{p}}\mathbf{=}\mathbf{\left[}\begin{array}{ccc}{\mathbf{Z}}_{\mathbf{Y}}\mathbf{+}{\mathbf{Z}}_{\mathbf{n}}& {\mathbf{Z}}_{\mathbf{n}}& {\mathbf{Z}}_{\mathbf{n}}\\ {\mathbf{Z}}_{\mathbf{n}}& {\mathbf{Z}}_{\mathbf{Y}}\mathbf{+}{\mathbf{Z}}_{\mathbf{n}}& {\mathbf{Z}}_{\mathbf{n}}\\ {\mathbf{Z}}_{\mathbf{n}}& {\mathbf{Z}}_{\mathbf{n}}& {\mathbf{Z}}_{\mathbf{y}}\mathbf{+}{\mathbf{Z}}_{\mathbf{n}}\end{array}\mathbf{\right]}$

Calculate the diagonal matrix for balanced Y-connected load.

Calculate the sequence impedance matrix.

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

${\mathit{Z}}_{\mathbf{s}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\left[}\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& \mathbf{a}& {\mathbf{a}}^{\mathbf{2}}\\ \mathbf{1}& {\mathbf{a}}^{\mathbf{2}}& \mathbf{a}\end{array}\mathbf{\right]}\mathbf{\left[}\begin{array}{ccc}{\mathbf{Z}}_{\mathbf{y}}\mathbf{+}{\mathbf{Z}}_{\mathbf{n}}& {\mathbf{Z}}_{\mathbf{n}}& {\mathbf{Z}}_{\mathbf{n}}\\ {\mathbf{Z}}_{\mathbf{n}}& {\mathbf{Z}}_{\mathbf{y}}\mathbf{+}{\mathbf{Z}}_{\mathbf{n}}& {\mathbf{Z}}_{\mathbf{n}}\\ {\mathbf{Z}}_{\mathbf{n}}& {\mathbf{Z}}_{\mathbf{n}}& {\mathbf{Z}}_{\mathbf{y}}\mathbf{+}{\mathbf{Z}}_{\mathbf{n}}\end{array}\mathbf{\right]}\mathbf{\left[}\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& {\mathbf{a}}^{\mathbf{2}}& \mathbf{a}\\ \mathbf{1}& \mathbf{a}& {\mathbf{a}}^{\mathbf{2}}\end{array}\mathbf{\right]}$

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

${\mathit{Z}}_{\mathbf{s}}\mathbf{=}\mathbf{\left[}\begin{array}{ccc}{\mathbf{Z}}_{\mathbf{y}}\mathbf{+}\mathbf{3}{\mathbf{Z}}_{\mathbf{n}}& \mathbf{0}& \mathbf{0}\\ \mathbf{0}& {\mathbf{Z}}_{\mathbf{y}}& \mathbf{0}\\ \mathbf{0}& \mathbf{0}& {\mathbf{Z}}_{\mathbf{y}}\end{array}\mathbf{\right]}$

Hence, the sequence matrix $\left[\begin{array}{ccc}{Z}_y+3Z_n& 0& 0\\ 0& Z_y& 0\\ 0& 0& Z_y\end{array}\right]$is the diagonal matrix.