Question

(a) Consider three equal impedances $\mathit{j}\mathbf{27}\mathit{\Omega }$of connected in$\mathit{\Delta }$ . Obtain the sequence networks. (b) Now, with a mutual impedance of$\mathit{\Delta }$ between each pair of adjacent branches in the $\mathit{\Delta }$-connected load of part (a), how would the sequence networks change?

Short answer

(a) Therefore, the sequence network.

(b) Therefore, the sequence network.

Step-by-step solution

Write the given data from the question:

The equation impedance,$Z_{\Delta }=j27\Omega$

The mutual impedance,$=j6\Omega$

Determine the network sequence.

(a)

The line to line voltages related to current as,

$\left[\begin{array}{c}{V}_{ab}\\ V_{bc}\\ V_{ca}\end{array}\right]=\left[\begin{array}{ccc}j27& 0& 0\\ 0& j27& 0\\ 0& 0& j27\end{array}\right]\left[\begin{array}{c}{I}_{ab}\\ I_{bc}\\ I_{ca}\end{array}\right]$

The relation between line to line voltages and sequence voltages can be define by following.

$\begin{array}{l}\left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}{V}_{ab}\\ V_{bc}\\ V_{ca}\end{array}\right]\\ \left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=A\left[\begin{array}{c}{V}_{ab}\\ V_{bc}\\ V_{ca}\end{array}\right]\end{array}$

Similarly, the relation between sequence voltages and current can be define by following.

$\begin{array}{l}A\left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=\left[\begin{array}{ccc}j27& 0& 0\\ 0& j27& 0\\ 0& 0& j27\end{array}\right]A\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\\ A\left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=j27\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\\ A\left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=j27IA\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\end{array}$

Here $I$$I$is the identity matrix.

Multiply both the sides with localid="1656767349235" $A^{-1}$.

localid="1656767353767" $\begin{array}{l}{A}^{-1}A\left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=j27A^{-1}A\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\\ \left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=j27I\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\\ \left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=j27\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\\ \left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=j27\left[\begin{array}{ccc}j27& 0& 0\\ 0& j27& 0\\ 0& 0& j27\end{array}\right]\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\end{array}$

Calculate thelocalid="1656767360713" $\text{Y}$-connected load fromlocalid="1656767420521" $\Delta$-connected load.

localid="1656767426327" $\begin{array}{l}{Z}_Y=\frac{Z_{\Delta }}{3}\\ Z_Y=\frac{j27}{3}\\ Z_Y=j9\Omega \end{array}$

Draw the sequence network.

Determine the sequence network with mutual impedance.

(b)

The mutual impedance is present between each pair of adjacent.

$\left[\begin{array}{c}{V}_{ab}\\ V_{bc}\\ V_{ca}\end{array}\right]=\left[\begin{array}{ccc}j27& j6& j6\\ j6& j27& j6\\ j6& j6& j27\end{array}\right]\left[\begin{array}{c}{I}_{ab}\\ I_{bc}\\ I_{ca}\end{array}\right]$

The relation between line to line voltages and sequence voltages can be define by following.

$\begin{array}{l}\left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}{V}_{ab}\\ V_{bc}\\ V_{ca}\end{array}\right]\\ \left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=A\left[\begin{array}{c}{V}_{ab}\\ V_{bc}\\ V_{ca}\end{array}\right]\end{array}$

Similarly, the relation between sequence voltages and current can be define by following.

$\begin{array}{l}A\left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=\left[\begin{array}{ccc}j27& j6& j6\\ j6& j27& j6\\ j6& j6& j27\end{array}\right]A\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\\ A\left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=\left\{j21\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]+j6\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]\right\}A\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\\ A\left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=j21IA\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]+j6\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]A\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\end{array}$

Multiply both the sides with$A^{-1}$.

$A^{-1}A\left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=j21A^{-1}A\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]+j6A^{-1}\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]A\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]$

Substitute $\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle -120°& 1\angle 120°\\ 1& 1\angle 120°& 1\angle -120°\end{array}\right]$for $A^{-1}$into above equation

$\begin{array}{l}\left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=j21\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]+j6\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle -120°& 1\angle 120°\\ 1& 1\angle 120°& 1\angle -120°\end{array}\right]\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]A\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\\ \left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=\left[\begin{array}{ccc}j21& 0& 0\\ 0& j21& 0\\ 0& 0& j21\end{array}\right]\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]+j6\left[\begin{array}{ccc}3& 3& 3\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]A\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\end{array}$

Substitute $\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]$for into above equation.

$\begin{array}{l}\left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=\left[\begin{array}{ccc}j21& 0& 0\\ 0& j21& 0\\ 0& 0& j21\end{array}\right]\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]+j6\left[\begin{array}{ccc}3& 3& 3\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\\ \left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=\left[\begin{array}{ccc}j21& 0& 0\\ 0& j21& 0\\ 0& 0& j21\end{array}\right]\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]+j6\times 3\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\\ \left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=\left[\begin{array}{ccc}j21& 0& 0\\ 0& j21& 0\\ 0& 0& j21\end{array}\right]\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]+j6\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\\ \left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=\left[\begin{array}{ccc}j21& 0& 0\\ 0& j21& 0\\ 0& 0& j21\end{array}\right]\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]+j6\left[\begin{array}{ccc}1+1+1& 1+a+a^2& 1+a+a^2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\end{array}$

Substitute$0$for$1+a+a^2$into above equation.

.$\begin{array}{l}\left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=\left[\begin{array}{ccc}j21& 0& 0\\ 0& j21& 0\\ 0& 0& j21\end{array}\right]\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]+j6\left[\begin{array}{ccc}3& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\\ \left[\begin{array}{c}{V}_{ab0}\\ V_{ab1}\\ V_{ab2}\end{array}\right]=\left[\begin{array}{ccc}j39& 0& 0\\ 0& j21& 0\\ 0& 0& j21\end{array}\right]\left[\begin{array}{c}{I}_{ab0}\\ I_{ab1}\\ I_{ab2}\end{array}\right]\end{array}$

Calculate the $\text{Y}$-connected load from $\Delta$-connected load.

$Z_Y=\frac{Z_{\Delta }}{3}$

Calculate the zero sequence.

$\begin{array}{l}{Z}_0=\frac{j39}{3}\\ Z_0=j13\end{array}$

Calculate the positive and negative sequence.

$\begin{array}{l}{Z}_1=Z_2\\ Z_1=\frac{j21}{3}\\ Z_1=j7\end{array}$

Draw the sequence network.