Electric Circuits Test 3

Given that,

Number of nodes (N) = 8

Number of independent loops (L) = 5

Let number of branches are B.

We know that,

L = B – N + 1

Incidence matrix:

• It is the matrix which gives relation between branches and nodes.
• The rows of matrix represent the number of nodes and the columns of matrix represents the numbers of branches.
• We can construct the incidence matrix for the directed graph. We can draw a graph with the help of incidence matrix.
• The algebraic sum of elements of all the columns is zero.
• The rank of incidence matrix is (n–1).

The elements of incidence matrix are given by [A] = [a_ij]_{n × b}

Where a_ij = 1, if j^th branch is incident at i^th node and oriented away.

a_ij = -1, if j^th branch is incident at i^th node and oriented towards.

a_ij = 0, if j^th branch is not incident at i^th node

Reduced incidence matrix:

• If one of the nodes in the given graph is considered as reference node, then that row can be neglected by writing incidence matrix is called as reduced incidence matrix.
• The order of reduced incidence matrix is (n–1) × b.
• The Algebraic sum of some of the columns is not zero.

Number of trees of a graph can be calculate by using the following formula.

Number of trees = n^n-2, n > 2 for completely connected graph

\(= \left| {\left[ {{A_r}} \right]{{\left[ {{A_r}} \right]}^T}} \right|\) for just connected graph

Calculation:

Given incidence matrix is

\(A = \left[ {\begin{array}{*{20}{c}} { - 1}&0&1&1&{ - 1}&0\\ 1&{ - 1}&0&0&0&{ - 1}\\ 0&1&{ - 1}&{ - 1}&1&0 \end{array}} \right]\)

For the above matrix, the algebraic sum of elements of all the columns is not zero.

It is reduced incidence matrix.

\({A_r} = \left[ {\begin{array}{*{20}{c}} { - 1}&0&1&1&{ - 1}&0\\ 1&{ - 1}&0&0&0&{ - 1}\\ 0&1&{ - 1}&{ - 1}&1&0 \end{array}} \right]\)

\({\left[ {{A_r}} \right]^T} = \left[ {\begin{array}{*{20}{c}} { - 1}&1&0\\ 0&{ - 1}&1\\ 1&0&{ - 1}\\ 1&0&{ - 1}\\ { - 1}&0&1\\ 0&{ - 1}&0 \end{array}} \right]\)

\(\left[ {{A_r}} \right]{\left[ {{A_r}} \right]^T} = \left[ {\begin{array}{*{20}{c}} { - 1}&0&1&1&{ - 1}&0\\ 1&{ - 1}&0&0&0&{ - 1}\\ 0&1&{ - 1}&{ - 1}&1&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 1}&1&0\\ 0&{ - 1}&1\\ 1&0&{ - 1}\\ 1&0&{ - 1}\\ { - 1}&0&1\\ 0&{ - 1}&0 \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} 4&{ - 1}&{ - 3}\\ { - 1}&3&{ - 1}\\ { - 3}&{ - 1}&4 \end{array}} \right]\)

Number of trees \(= \left| {\left[ {{A_r}} \right]{{\left[ {{A_r}} \right]}^T}} \right|\)

\(= \left| {\begin{array}{*{20}{c}} 4&{ - 1}&{ - 3}\\ { - 1}&3&{ - 1}\\ { - 3}&{ - 1}&4 \end{array}} \right|\)

= 4 (12 – 1) + 1 (-4 – 3) – 3 (1 + 9)

= 44 – 7 – 30 = 7

Concept:

Tie set matrix:

• It gives the relation between tie-set currents and branch currents.
• The rows of a matrix represent the tie-set currents.
• The columns of a matrix represent branches of the graph.
• The order of the tie set matrix is (b – n + 1) × b
• The rank of a tie-set matrix is (b – n + 1)

Cut-set matrix:

• It gives the relation between cut-set voltages and branch voltages.
• The rows of a matrix represent the cut-set voltages.
• The columns of a matrix represent the branches of the graph.
• The order of the cut set matrix is (n – 1) × b.
• The rank of a cut-set matrix is (n – 1)

Calculation:

For the given graph,

Number of branches (b) = 8

Number of nodes (n) = 5

Rank of cut-set matrix = (n – 1) = 5 – 1 = 4