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Electric Circuits Test 3

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Electric Circuits Test 3
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  • Question 1
    1 / -0

    The graph of a network has 8 nodes and 5 independent loops. The number of branches of the graph is

    Solution

    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

    ⇒ 5 = B – 8 + 1 ⇒ B = 12
  • Question 2
    1 / -0

    The reduced incidence matrix of a graph is given below

    A=[101110110001011110]

    The number of possible trees is

    Solution

    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] = [aij]n × b

    Where aij = 1, if jth branch is incident at ith node and oriented away.

    aij = -1, if jth branch is incident at ith node and oriented towards.

    aij = 0, if jth branch is not incident at ith 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 = nn-2, n > 2 for completely connected graph

    =|[Ar][Ar]T| for just connected graph

    Calculation:

    Given incidence matrix is

    A=[101110110001011110]

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

    It is reduced incidence matrix.

    Ar=[101110110001011110]

    [Ar]T=[110011101101101010]

    [Ar][Ar]T=[101110110001011110][110011101101101010]

    =[413131314]

    Number of trees =|[Ar][Ar]T|

    =|413131314|

    = 4 (12 – 1) + 1 (-4 – 3) – 3 (1 + 9)

    = 44 – 7 – 30 = 7

  • Question 3
    1 / -0
    The graph associated with an electrical network has 8 branches and 5 nodes. The rank of the cut-set matrix and tie-set matrix respectively can be no more than,
    Solution

    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

    Rank of tie-set matrix = b - (n – 1) = 8 – (5 – 1) = 4
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