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Circuit analysis via polyhedron
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[QUOTE="swampwiz, post: 6867262, member: 358750"] I've pondered my analysis here, and while the general idea is the same, there are some tweaks to the linear dependencies. The general idea is that due to the span restrictions in the EQs, there is a single linear dependence with the set of node/vertex EQs, no dependence of the loop/face EQs with either node/vertex EQs or with other loop/face EQs, and that the EQ for the remaining face would be the sum of all the other loop/faces, and hence a linear dependence. So there would be ( V + F ) EQs with 2 linear dependencies, or ( V + F - 2 ) independent EQs in E unknowns. @@@@ Every edge has a current (the direction would need to be arbitrarily chosen), and so there is a current continuity EQ for every node/vertex, which results in a simultaneous system of V (# of vertices) EQs in E (# of edges) unknowns. And because every branch/edge is between 2 node/vertices in which it will be considered positive in one and negative in the other, the net sum of all the EQs must be zero, and hence there is a linear dependence between the EQs, and so any one of these must be removed to get an independent set of EQ. Similarly, every face is an independent loop, and so that net voltage drop across that loop must be zero, which results in a simultaneous system of F (# of faces) EQ in E unknowns - and to give the circuit "life", there must be some EMF element in at least one of the branch/edges, and so via the principle of superposition, each EMF element could be examined one at a time, and such that the EMF element term would wind up as a non-zero term in the resultants. However, unlike the case for the EQs for the node/vertices, there must topographically be branch/edges on the outside of the resulting graph, and since these branch/edges will only be used one time, the EQs for the outer loop/faces must be linearly independent. If there are any loop/faces that are not outer, they will have branch/edges that are only used once aside from the outer loop/faces (i.e., they would be the next level of inner branch/edges) - and because these branch/edges had only been used once in the outer branches, these loop/faces EQs cannot be linearly dependent on the outer loop/face EQs, the net result is that they must linearly independent just as for the outer loop/faces. This can be done for any level of inner looping, and so the overall net result is that all these loop/face EQs must be linearly independent of each other - and since a branch/edge is only shared by a pair of loop/faces, and any branch/edge only has a span in a pair of node/vertex EQs that are linearly independent themselves, and such that EQs for node/vertices at most share a single branch/edge (i.e. the one between any pair of node/vertices), the rest of the span for the node/vertex EQs in terms of branch/vertices must be zero - and thus it must be that there is no way for there to be linear dependence of a loop/face EQ on any node/vertex EQ. Hence, the final net result must be that the # of EQs is ( V + F - 1 ) in E # of unknown edge/branch current values. Going back to the polyhedron model, topographically any polyhedron can be made into a something similar to a net in which all but one of the faces can be represented in a plane such that the remaining loop/face is on the other side of the plane, and thus a loop/face EQ for this remaining face would be a loop that contains all the other loops - and since the loop/faces have only 2 opposite direction conventions (i.e., CW or CCW), if all loop/faces use the same convention, each branch/edge that is not along the edge of the entire figure will be used in a pair of equal & opposite terms, and so the net result is that the sum of the EQs for the original loop/faces must be the EQ for the reamaining loop/face, and thus including this remaining loop/face would introduce another linear dependence. And so taken in total (i.e., including this final loop/face), there are ( V + F ) EQs with 2 linear dependencies, or a set of ( V + F - 2 ) linearly independent EQs in total in E unknowns - which wonderfully exactly matches Euler's polyhedron EQ (i.e., without holes), and so if all the non-EMF elements were of the same type, there would be a simple, well-conditioned matrix equation (with real values) to solve for the edge currents. [/QUOTE]
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