Ther e ar e many applications of linear algebra for example chemists might use ow eduction to get a lear er pictur e of what elements go into a complicated eaction In this lectur e we explor e the linear algebra associated with electrical networks G ID: 25596 Download Pdf

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Ther e ar e many applications of linear algebra for example chemists might use ow eduction to get a lear er pictur e of what elements go into a complicated eaction In this lectur e we explor e the linear algebra associated with electrical networks G

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Graphs, networks, incidence matrices When we use inear algebra to understand physical systems, we often ﬁnd mor e str uctur e in the matrices and vectors than appears in the exampl es we make up in class. Ther e ar e many applications of linear algebra; for example, chemists might use ow eduction to get a lear er pictur e of what elements go into a complicated eaction. In this lectur e we explor e the linear algebra associated with electrical networks. Graphs and networks A graph is a collection of nodes joined by edges; Figur e 1 shows one small graph. 23 Figur e 1: A

graph with 4 nodes and 5 edges. e put an arr ow on each edge to indicate the positive dir ection for curr ents unning thr ough the graph. 23 Figur e 2: The graph of Figur e 1 with a dir ection on each edge. Incidence matrices The incidence matrix of this dir ected graph has one column for each node of the graph and one r ow for each edge of the graph: 1 1 0 0 1 0 1 0 0 . 1 0 1 If an edge uns fr om node to node , the ow corr esponding to that edge has 1 in column and 1 in column ; all other entries in that ow ar e 0. If we wer e 1 0 1

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studying a lar ger graph we would get a lar

ger matrix but it would be sparse ; most of the entries in that matrix would be 0. This is one of the ways matrices arising fr om applications might have extra str uctur e. Note that nodes 1, 2 and 3 and edges , and form a loop. The matrix describing just those nodes and edges looks like: 1 0 0 0 . 1 0 Note that the thir d ow is the sum of the ﬁrst two ows; loops in the graph corr espond to linearly dependent ows of the matrix. o ﬁnd the nullspace of , we solve : 2 1 3 2 3 1 4 1 4 3 0 0 0 0 . 0 If the components of the vector describe the electrical potential at the nodes of the

graph, then is a ve ctor describing the differ ence in potential acr oss each edge of th e graph. e see when 1 2 3 , so the nullspace has dimension 1. In terms of an electrical network, the potential dif fer ence is zer o on ach edge if each node has the same potential. e can’t tell what that potential is by observing the ﬂow of electricity thr ough the network, but if one node of the network is gr ounded then its potential is zer o. Fr om that we can determine the potential of all other nodes of the graph. The matrix has 4 columns and a 1 dimensional nullspace, so its rank is 3. The

ﬁrst, second and fourth columns ar e its pivot columns; these edges connect all the nodes of the graph without forming a loop a graph with no loops is called a tr ee . The left nullspace of consists of the solutions to the equation: . Since has 5 columns and rank 3 we know that the dimension of is 2. Note that 2 is the number of loops in the graph and is the number of edges. The rank is 1, one less than the number of nodes. This gives us # loops # edges # nodes , or: number of nodes number of edges number of loops 1. This is Euler ’s formula for connected graphs. Kirchhof law In our

example of an electri cal network, we started with the potentials of the nodes. The matrix then told us something about potential dif fer ences. An engineer could cr eate a matri x using Ohm’s law and information about 2

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23 Figur e 3: The curr ents in our graph. the conductance of the edges and use that matrix to determine the curr ent on each edge. Kir chhof ’s Curr ent Law then says that , wher e is the vector with components , , , , . ectors in the nullsp ace of corr espond to collections of curr ents that satisfy Kir chhof ’s law . , , , 4 0 potentials at nodes Kir chhof

’s Curr ent Law 2 , etc. potential dif fer ences Ohm’s Law , , , , 5 curr ents on edges ritten out, looks like: 0 1 0 1 0 2 3 4 1 1 0 0 0 0 0 1 1 0 5 Multiplying the ﬁrst ow by the column vector we get 1 3 4 0. This tells us that the total curr ent ﬂowing out of node 1 is zer o it’s a balance equation, or a conservation law . Multiplying the second ow by tells us 1 2 0; the curr ent coming into node 2 is balanced with the curr ent going out. Multiplying the bottom ows, we get 2 3 5 0 and 4 5 0. e could use the method of elimination on to ﬁnd its column space, bu t we alr

eady know the rank. o get a basis for we just need to ﬁnd two independent vectors in this space. Looking t the equations 1 2 0 we might guess 1 2 1. Then we could use the conservation laws for node 3 to guess 3 1 and 0. e satisfy the conservation conditions on node 4 1 with 4 0, giving us a basis vector . his vector epr esents one unit 1 0 0 3

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of curr ent ﬂowing ar ound the loop joining nodes 1, 2 and 3; a multiple of this vector epr esents a dif fer ent amount of curr ent ar ound the same loop. e ﬁnd a second basis vector for by looking at the loop formed

by 1 nodes 1, 3 and 4: 0 1 1 1 . The vector 1 0 1 1 that epr esents a curr ent ar ound the outer loop is also in the nullspace, but it is the sum of the ﬁrst two vectors we found. e’ve almost completely cover ed the mathematics of simple cir cuits. Mor e complex cir cuits might have batteries in the edges, or curr ent sour ces between nodes. Adding curr ent sour ces changes the in Kir chhof ’s curr ent law to . Combining the equations , and gives us: . 4

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