Question

(a) Given the circuit diagram in Figure 2.32 showing admittances and current sources at nodes 3 and 4, set up the nodal equations in matrix format. (b) If the parameters are given by :${\mathbf{Y}}_{\mathbf{a}}\mathbf{=}\mathbf{-}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{S}\mathbf{,}\mathbf{}{\mathbf{Y}}_{\mathbf{b}}\mathbf{=}\mathbf{-}\mathbf{j}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{S}\mathbf{,}\mathbf{}{\mathbf{Y}}_{\mathbf{c}\mathbf{=}}\mathbf{-}\mathbf{j}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{S}\mathbf{,}\mathbf{}{\mathbf{Y}}_{\mathbf{d}}\mathbf{=}\mathbf{-}\mathbf{j}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{S}\mathbf{,}$${\mathbf{Y}}_{\mathbf{e}}\mathbf{=}\mathbf{-}\mathbf{j}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{S}\mathbf{,}\mathbf{}{\mathbf{Y}}_{\mathbf{f}}\mathbf{=}\mathbf{-}\mathbf{j}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{S}\mathbf{,}\mathbf{}{\mathbf{Y}}_{\mathbf{g}}\mathbf{=}\mathbf{-}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{S}\mathbf{,}{\mathbf{l}}_{\mathbf{3}\mathbf{=}}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\angle }\mathbf{-}\mathbf{90}\mathbf{°}\mathbf{A}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{l}}_{\mathbf{4}\mathbf{=}}\mathbf{0}\mathbf{.}\mathbf{62}\mathbf{\angle }\mathbf{-}\mathbf{135}\mathbf{°}\mathbf{A}\mathbf{,}$set up the nodal equations and suggest how you would go about solving for the voltages at the nodes.

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Short answer

(a) Therefore, the format for the system showing current and voltages sources at node 3 and 4, nodal equation matrix format is

$J\left[\begin{array}{cccc}-1.52& 8& 4& 2.5\\ 8& -17& 4& 5\\ 4& 4& -8.8& 0\\ 2.5& 5& 5& -8.3\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\\ V_3\\ V_4\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 1\angle -90°\\ 0.62\angle -135°\end{array}\right]$

(b) The modal voltages can be calculated by using the expression V =${Y^{\mathit{-}\mathit{1}}}_{\mathit{b}\mathit{u}\mathit{s}}l\mathit{.}$

Step-by-step solution

Find the 4 X 4 matrix for nodal equation of the given system

(a)

The 4 X 4 matrix for the nodal equation is given by

$$\begin{gathered} {\mathbf{Y}}_{\mathbf{bus}}\mathbf{V}\mathbf{=}\mathbf{I} \\ \left[\begin{array}{cccc}{Y}_{11}& Y_{12}& Y_{13}& Y_{14}\\ Y_{21}& Y_{22}& Y_{23}& Y_{24}\\ Y_{31}& Y_{32}& Y_{33}& Y_{34}\\ Y_{41}& Y_{42}& Y_{43}& Y_{44}\end{array}\right]\left[\begin{array}{c}{Y}_1\\ Y_2\\ Y_3\\ Y_4\end{array}\right]\mathbf{=}\left[\begin{array}{c}{I}_1\\ I_2\\ I_3\\ I_4\end{array}\right] \end{gathered}$$

Here the diagonal admittance of the matrix can be calculated as

$Y_{\mathit{K}\mathit{K}}$=sum of the admittance connected to the node K

The non-diagonal admittance can be calculated as

$Y_{\mathit{k}\mathit{n}}$₌-(The sum of the admittance connected between the k and n.

$\left[\begin{array}{cccc}{Y}_c+Y_d+Y_f& -Y_d& -Y_c& -Y_f\\ -Y_d& Y_b+Y_d+Y_e& -Y_e& -Y_a\\ -Y_c& -Y_b& Y_a+Y_b+Y_c& 0\\ -Y_f& -Y_c& 0& Y_e+Y_f+Y_g\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\\ V_3\\ V_4\end{array}\right]=\left[\begin{array}{c}{I}_1\\ I_2\\ I_3\\ I_4\end{array}\right]$

Substitute the values and solve as

$$\begin{gathered} j\left[\begin{array}{cccc}-4-8-2.5& 8& 4& 2.5\\ 8& -4-8-5& 4& 5\\ 4& 4& -0.8-4-4& 0\\ 2.5& 5& 5& -5-2.5-0.8\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\\ V_3\\ V_4\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 1\angle -90°\\ 0.62\angle -135°\end{array}\right] \\ j\left[\begin{array}{cccc}-1.54& 8& 4& 2.5\\ 8& -17& 4& 5\\ 4& 4& -8.8& 0\\ 2.5& 5& 5& -8.3\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\\ V_3\\ V_4\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 1\angle -90°\\ 0.62\angle -135°\end{array}\right] \end{gathered}$$

Hence the format for the system showing current and voltages sources at node 3 and 4, nodal equation matrix format is

$\left[\begin{array}{cccc}-1.54& 8& 4& 2.5\\ 8& -17& 4& 5\\ 4& 4& -8.8& 0\\ 2.5& 5& 5& -8.3\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\\ V_3\\ V_4\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 1\angle -90°\\ 0.62\angle -135°\end{array}\right]$

 suggestion for solving for the voltages at the nodes.

(b)

Nodal equation in matrix format is given by

${\mathrm{Y}}_{\mathrm{bus}}\mathrm{V}=\mathrm{I}$

Multiply both the sides with ${\mathrm{Y}}_{\mathrm{bus}}^{-1}.$

$$\begin{gathered} {\mathrm{Y}}_{\mathrm{bus}}^{-1}{\mathrm{Y}}_{\mathrm{bus}}\mathrm{V}={\mathrm{Y}}_{\mathrm{bus}}^{-1}\mathrm{l} \\ \mathrm{V}={\mathrm{Y}}_{\mathrm{bus}}^{-1}\mathrm{l} \end{gathered}$$

The ${\mathrm{Y}}_{\mathrm{bus}}^{-1}$can be calculated as

localid="1655116007840" $Y_{\mathit{b}\mathit{u}\mathit{s}}^{\mathit{-}\mathit{1}}\mathit{=}\frac{\mathit{a}\mathit{d}\mathit{j}\mathit{\left(}{\mathit{Y}}_{\mathit{b}\mathit{u}\mathit{s}}\mathit{\right)}}{\left|Y_{bus}\right|}$

Here adj(localid="1655116022021" $Y_{bus}$) is the of the adjoint of thelocalid="1655116060149" ${\mathrm{Y}}_{\mathrm{bus}}$and localid="1655116129659" ${\mathrm{IY}}_{\mathrm{bus}}\mathrm{I}$is the determinant of the matrix Y_bus.

Now the nodal voltages can be calculated by using the expression $V\mathit{=}{Y^{\mathit{-}\mathit{1}}}_{\mathit{b}\mathit{u}\mathit{s}}l\mathit{}\mathit{.}$