Node Elimination by Matrix Partitioning
Sometimes it is desirable to reduce the network by eliminating the nodes in which the current do not enter or leave. Let (3.3) be written as
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(3.13) |
In the above equation IA is a vector containing the currents that are injected, Ix is a null vector and the Ybus matrix is portioned with the matrices K , L and M . Note that the Ybus matrix contains both L and LT due to its symmetric nature.
We get the following two sets of equations from (3.13)
|
(3.14) |
 |
(3.15) |
Substituting (3.15) in (3.14) we get
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(3.16) |
Therefore we obtain the following reduced bus admittance matrix
 |
(3.17) |
Do an example |
