The meeting point of various components in a PS is called Bus.

The Bus or Bus bar is a conductor made of copper or aluminium having negligible resistances.

Hence the bus bar will have zero voltage drop when it conducts the rated current

Buses are considered as points of constant voltage in a PS

PS represented by impedance/reactance diagram is considered as a circuit or network.

Buses can be treated as nodes and the voltages of all buses (nodes) can be solved by conventional node analysis technique.

\[\mathrm{Y_{bus}}\cdot \mathrm{V} = \mathrm{I}\] where, \[\begin{aligned} \mathrm{Y_{bus}} & = \text{Bus admittance matrix of order}~(n \times n) \\ \mathrm{V} & = \text{Bus voltage matrix of order}~(n\times 1) \\ \mathrm{I} & = \text{Current sources matrix of order}~(n\times 1) \\ \mathrm{n} & = \text{Number of independent buses in the system} \end{aligned}\] In general, \[\begin{aligned} \mathrm{Y_{jj}} & = \text{Sum of all admittances connected to bus-j} \\ \mathrm{Y_{jk}} & = \text{-ve sum of all admittances connected between bus-j and bus-k} \\ \mathrm{Y_{jk}} & = \mathrm{Y_{kj}} \end{aligned}\] Note: If PS represented by reactance diagram, all elements are inductive susceptances (which are -ve). In this case, \(\mathrm{Y_{ij}}\) will be -ve and \(\mathrm{Y_{jk}}\) will be +ve

Voltage source with a source impedance and its Nortonâ€™s equivalent \[I_{s}=\dfrac{V_{s}}{Z_{s}} ~ \mbox{and} ~ Y_{s}=\dfrac{1}{Z_{s}}\]

\[\left[\begin{array}{c} I_{1}\\ I_{2}\\ 0\\ 0 \end{array}\right]=Y_{bus}\left[\begin{array}{c} V_{1}\\ V_{2}\\ V_{3}\\ V_{4} \end{array}\right]\Longrightarrow\left[\begin{array}{c} V_{1}\\ V_{2}\\ V_{3}\\ V_{4} \end{array}\right]=Z_{bus}\left[\begin{array}{c} I_{1}\\ I_{2}\\ 0\\ 0 \end{array}\right]\]

Applying KCL at node 1, \[\begin{aligned} I_{1} & =Y_{11}V_{1}+Y_{12}\left(V_{1}-V_{2}\right)+Y_{13}\left(V_{1}-V_{3}\right)\\ & =\left(Y_{11}+Y_{12}+Y_{13}\right)V_{1}-Y_{12}V_{2}-Y_{13}V_{3} \end{aligned}\]

In a similar way application of KCL at nodes 2, 3 and 4 results in the following equations \[\begin{aligned} I_{2} & =Y_{22}V_{2}+Y_{12}\left(V_{2}-V_{1}\right)+Y_{23}\left(V_{2}-V_{3}\right)+Y_{24}\left(V_{2}-V_{4}\right)\\ & =-Y_{12}V_{1}+\left(Y_{22}+Y_{12}+Y_{23}+Y_{24}\right)V_{2}-Y_{23}V_{3}-Y_{24}V_{4} \end{aligned}\]

\[\begin{aligned} 0 & =Y_{13}\left(V_{3}-V_{1}\right)+Y_{23}\left(V_{3}-V_{2}\right)+Y_{34}\left(V_{3}-V_{4}\right)\\ & =-Y_{13}V_{1}-Y_{23}V_{2}+\left(Y_{13}+Y_{23}+Y_{34}\right)V_{3}-Y_{34}V_{4} \end{aligned}\]

\[\begin{aligned} 0 & =Y_{24}\left(V_{4}-V_{2}\right)+Y_{34}\left(V_{4}-V_{3}\right)\\ & =-Y_{24}V_{2}-Y_{34}V_{3}+\left(Y_{24}+Y_{34}\right)V_{4} \end{aligned}\]

On combining \[ \left[\begin{array}{c} I_{1}\\ I_{2}\\ 0\\ 0 \end{array}\right]=\left[\begin{array}{cccc} Y_{11}+Y_{12}+Y_{13} & -Y_{12} & -Y_{13} & 0\\ -Y_{12} & Y_{22}+Y_{12}+Y_{23}+Y_{24} & -Y_{23} & -Y_{24}\\ -Y_{13} & -Y_{23} & Y_{13}+Y_{23}+Y_{34} & -Y_{34}\\ 0 & -Y_{24} & -Y_{34} & Y_{24}+Y_{34} \end{array}\right]\left[\begin{array}{c} V_{1}\\ V_{2}\\ V_{3}\\ V_{4} \end{array}\right] \]

\[ Y_{bus}=\left[\begin{array}{cccc} Y_{1}+Y_{12}+Y_{13} & -Y_{12} & -Y_{13} & 0\\ -Y_{12} & Y_{22}+Y_{12}+Y_{23}+Y_{24} & -Y_{23} & -Y_{24}\\ -Y_{13} & -Y_{23} & Y_{13}+Y_{23}+Y_{34} & -Y_{34}\\ 0 & -Y_{24} & -Y_{34} & Y_{24}+Y_{34} \end{array}\right]=\left[\begin{array}{ccccc} Y_{11} & -Y_{12} & -Y_{13} & \cdots & -Y_{1n}\\ -Y_{12} & Y_{2} & -Y_{23} & \cdots & -Y_{2n}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ -Y_{1n} & -Y_{2n} & -Y_{3n} & \cdots & Y_{n} \end{array}\right] \]

Assume all lines are represented by equivalent-\(\pi\) with the shunt admittance between the line \(i\) and \(j\) being denoted by \(Y_{chij}\)

Then equivalent admittance at the two end will be \(Y_{chij}/2\)

For e.g. shunt capacitance at two ends joining buses \(1\) and \(3\) will be \(Y_{ch13}/2\)

\(\Rightarrow\) We can then modify the admittance matrix as \[ Y_{bus}=\left[\begin{array}{cccc} Y_{1}+Y_{12}+Y_{13}+Y_{ch1} & -Y_{12} & -Y_{13} & 0\\ -Y_{12} & Y_{22}+Y_{12}+Y_{23}+Y_{24}+Y_{ch2} & -Y_{23} & -Y_{24}\\ -Y_{13} & -Y_{23} & Y_{13}+Y_{23}+Y_{34}+Y_{ch3} & -Y_{34}\\ 0 & -Y_{24} & -Y_{34} & Y_{24}+Y_{34}+Y_{ch4} \end{array}\right] \] where \[\begin{aligned} Y_{ch1} & =\dfrac{Y_{ch12}+Y_{ch13}}{2}\\ Y_{ch2} & =\dfrac{Y_{ch12}+Y_{ch23}+Y_{ch24}}{2}\\ Y_{ch3} & =\dfrac{Y_{ch13}+Y_{ch23}+Y_{ch34}}{2}\\ Y_{ch4} & =\dfrac{Y_{ch24}+Y_{ch34}}{2} \end{aligned}\]

\(Y_{bus}\) is a sparse matrix

Diagonal elements are dominating

Off diagonal elements are symmetric

The diagonal element of each node is the sum of the admittances connected to it

The off diagonal element is negated admittance

\[ \left[\begin{array}{c} I_{1}\\ I_{2}\\ I_{3}\\ I_{4} \end{array}\right]=\underset{\mbox{symmetric}}{\underbrace{\left[\begin{array}{cccc} Y_{11} & Y_{12} & Y_{13} & Y_{14}\\ Y_{21} & Y_{22} & Y_{23} & Y_{24}\\ Y_{31} & Y_{32} & Y_{33} & Y_{34}\\ Y_{41} & Y_{42} & Y_{43} & Y_{44} \end{array}\right]}}\left[\begin{array}{c} V_{1}\\ V_{2}\\ V_{3}\\ V_{4} \end{array}\right]\Longrightarrow\left[\begin{array}{c} V_{1}\\ V_{2}\\ V_{3}\\ V_{4} \end{array}\right]=\underset{\mbox{symmetric}}{\underbrace{\left[\begin{array}{cccc} Z_{11} & Z_{12} & Z_{13} & Z_{14}\\ Z_{21} & Z_{22} & Z_{23} & Z_{24}\\ Z_{31} & Z_{32} & Z_{33} & Z_{34}\\ Z_{41} & Z_{42} & Z_{43} & Z_{44} \end{array}\right]}}\left[\begin{array}{c} I_{1}\\ I_{2}\\ I_{3}\\ I_{4} \end{array}\right] \]

The self-inductance at bus-1 \[Y_{11}={\left.\dfrac{I_{1}}{V_{1}}\right|}_{V_{2}=V_{3}=V_{4}=0}\] \(Y_{11}\) is the admittance measured at bus-1 when buses 2, 3 and 4 are short circuited.

For example the mutual admittance between buses 1 and 2 is defined as \[Y_{12}={\left.\dfrac{I_{1}}{V_{2}}\right|}_{V_{1}=V_{3}=V_{4}=0}\] The mutual admittance \(Y_{12}\) is obtained as the ratio of the current injected in bus-1 to the voltage of bus-2 when buses 1, 3 and 4 are short circuited.

Driving point impedance at bus-1 \[Z_{11}={\left.\dfrac{V_{1}}{I_{1}}\right|}_{I_{2}=I_{3}=I_{4}=0}\] obtained by injecting a current at bus-1 while keeping buses 2, 3 and 4 open-circuited

Transfer impedance between buses 1 and 2 \[Z_{12}={\left.\dfrac{V_{1}}{I_{2}}\right|}_{I_{1}=I_{3}=I_{4}=0}\] by injecting a current at bus-2 while open-circuiting buses 1, 3 and 4.

NOTE: \(Z_{12}\) is not the reciprocal of \(Y_{12}\)

Sometimes it is desirable to reduce the network by eliminating the nodes in which the current do not enter or leave.

\[\left[\begin{array}{c} I_{A}\\ I_{x} \end{array}\right]=\left[\begin{array}{cc} K & L\\ L^{T} & M \end{array}\right]\left[\begin{array}{c} V_{A}\\ V_{x} \end{array}\right]\]

\(I_A\) is a vector containing the currents that are injected

\(I_x\) is a null vector

\(Y_{bus}\) is portioned with matrices \(K,~L,\text{and}~M\)

Note \(Y_{bus}\) contains both \(L\) and \(L^T\) due to its symmetric nature

\[\begin{aligned} I_{A} & =KV_{A}+LV_{x}\\ I_{x} & =0=L^{T}V_{A}+MV_{x}\Rightarrow V_{x}=-M^{-1}L^{T}V_{A}\\ \therefore ~I_{A} & =\left(K-LM^{-1}L^{T}\right)V_{A} \end{aligned}\] \[\boxed{Y_{bus}^{reduced}=K-LM^{-1}L^{T}}\]