The Formula for the Equicalent Resistance of Complex Resistance
The Formula for the Equivalent Resistance of Complex Resistance Network Deli Zhang Email: [email protected] Date: 2015-04-06 Abstract: Propose the theoretical formula for calculating the equivalent resistance of complex resistance network, and prove. Key words: equivalent resistance; complex resistance network; matrix-tree theorem 1 Introduction In this paper the Matrix-Tree theorem is used for describing the formula simply. Actually in 2006 I have discovered Matrix-Tree independently and made some corresponding achievements. I found the term has already been accepted and used widely this year, so I won’t expound it in deep here, and just introduce some useful properties. For details, please refer to [1] if you still do not know it. A complex resistance network can be modeled by complete graph G with n vertices. 1 n 2 j 3 i In which the edge between vertex i and j stands for the resistance ri , j or conductance g i , j . Then the n n matrix of Matrix-Tree theorem for graph G as follows: g 1 ,* g 2 ,1 M n = g n ,1 g n ,n g n,2 of M n : g 1, n 2 ,* Where g i , j g j , i if i j , and g i ,* is the succinct form of Let T n 1 represent the determinant of minor M g 2,n g n ,* g 1, 2 n g t 1 ( t i ) i ,t . g T n 1 Property 1.1. T n 1 ( 1) n 1 g 2 ,1 = g 1, 2 1 .* g 2 .* g 1 , n 1 g 2 , n 1 g n 1 ,1 g n 1 , 2 ( g i, j g n 1 .* ) The expansion polynomial of T n 1 is corresponding to the spanning tree forest of G , says every term g i , j corresponds one spanning tree and g i , j is its branch. Take a typical example, if all edge g i , j is replaced by number 1 , then T n 1 = ( 1) n 1 n n 2 which is the well-known Cayley's formula for counting spanning trees [2]. And if T n 1 0 , it’s equivalent to all tree term is 0, then G is non-connected graph. Property 1.2. The determinant generated through replacing the ith row and the ith column with the nth row and the nth colum of M n , still equals T n 1 , due to the replacing operation is equivalent to processing of accumulating other rows to the ith row and then other columns to the ith column. Property 1.3. For i j , and assume i j , denote T n 1 ( g i , j x ) as the determinant in which all of the edge g i , j in T n 1 is replaced by x . For T n 1 apply property 1.2 to replace the jth row and the jth column, then g g 1, 2 1 ,* g 2 ,1 T n 1 g 2 ,* g 1, i g 1, n g 1, n 1 g 2 ,i g 2,n g 2 , n 1 g i,n g i , n 1 g n , n 1 g i ,1 g i,2 g n ,1 g n,2 g n ,i g n 1 ,1 g n 1, 2 g n 1, i g n 1, n g i ,* g n ,* You can see g i , j only exists in the element locates in the ith row and the ith column: Therefore g i ,* g i ,1 g i , 2 g i , j g i , n g n 1 ,* g g 1, 2 1 ,* g 2 ,1 T n 1 T n 1 ( g i , j 0 ) 2 ,* 0 g 1, n g 1, n 1 0 g 2,n g 2 , n 1 g i ,1 g i,2 g i, j g i,n g i , n 1 g n ,1 g n,2 0 g n , n 1 g n 1 ,1 g n 1, 2 0 g n 1, n g i , j 1 S i ,i g ii g n ,* g n 1 ,* S i ,i is the minor of the ith row and ith column element, specially which does not contain g i , j . 2 Formula for the Equivalent Resistance Lemma 2.1. Star-Mesh transform [3]: The equivalent conductance for the star graph between vertex i and j, given: D i, j g i ,k g j ,k n g , i j k t ,k t 1 So after the transforming, the vertex k will be removed in network and other conductance g i , j obtains an increment D i , j , thus the current conductance for graph G : g i , j g i , j D i , j . ' Theorem 2.1. The equivalent resistance between vertex i and j for the whole network given: T n 1 g i , j 0 r R i, j 1 i, j T n 1 n 2 , i j n , ri , j 1 g i, j (1) , T n 1 0 . 1 n 2 j 3 Ri,j i Proof: If n 2 , assume i 1 , j 2 , then T n 1 g i , j 0 T1 g 1, 2 0 0 0 . Obviously R1, 2 r1, 2 , so the formula holds. If n k , assume R i , j 1 T k 1 g i , j 0 r holds, now let’s prove the formula holds as well for i, j T k 1 n k 1. 1 k 2 j 3 k+1 i Ri,j ' According to Lemma 2.1, remove vertex k 1 from network to get the sub-network G with k vertices, then the conductance between vertex i and j becomes: g ' i, j g i, j g i , k 1 g j , k 1 k g , 1 i, j k . t , k 1 t 1 Due to the assumption for n k , so R i , j ' ' T k 1 g 0 i, j ' 1 r i, j ' Tk 1 Therefore, to prove the formula holds, just need to prove below equation: ' ' T k 1 g 0 T g 0 i , j ' 1 k i, j r 1 r i, j ' i, j Tk Tk 1 (2) g g 1, 2 1 ,* g 1, 2 For T k g 2 ,* g 1, k 1 g 1, k g 2 , k 1 g 2,k g 1, k 1 g 2 , k 1 g 1, k g 2,k g g k 1, k k 1 ,* g k 1, k g k ,* Apply property 1.2, to say can replace the kth row and the kth column with the k 1 th row and the k 1 th column of M g , then g 1, 2 1 ,* g g 1, 2 k 1 2 ,* g 1, k 1 g 1, k 1 g 2 , k 1 g 2 , k 1 g 1, k 1 g 2 , k 1 g 1, k 1 g 2 , k 1 g k 1, k 1 g 1, 2 g 1, k 1 g 1, k 1 k 1 ,* g 2 , k 1 g 2 , k 1 k 1 ,* g 1 ,* g g 1, 2 2 ,* g 1, k 1 g 2 , k 1 g 1, k 1 g 2 , k 1 g 1 ,* g 1, k 1 g 1, 2 g 1, k 1 g g k 1 ,* k 1 ,* g k 1, k 1 g 1, 2 g 2 , k 1 g k 1 ,* g g 1 ,* g g 2 , k 1 2 ,* g 1, k 1 k 1 ,* g 1, k 1 k 1 ,* g k 1 ,* g 2 , k 1 g k 1 ,* g k 1 ,* 0 g k 1 ,* g k 1 ,* g 2 , k 1 0 k 1 ,* g 2 , k 1 g 2 , k 1 g k 1, k 1 g 1, k 1 g 1, k 1 g k 1, k 1 g k 1, k 1 g 2 , k 1 g 2 , k 1 g k 1 ,* g k 1 ,* g k 1, k 1 g 1, k 1 g 1, k 1 g g × 1 × g g k 1 g 2 , k 1 g g k 1, k 1 g 1, k 1 g g k 1, k 1 k 1 ,* g k 1, k 1 g k 1, k 1 0 g k 1 ,* g k 1, k 1 1 g 1, k 1 g 1, 2 g 1, k 1 g 1, k 1 g k 1 g 2 , k 1 g g 2 ,* g 1, k 1 g g 2 , k 1 k 1 ,* g 1, k 1 g 1, k 1 g 1, 2 g 2 , k 1 k 1 ,* g 2 , k 1 g g k 1 ,* g 2 , k 1 g 2 , k 1 g 2 , k 1 g k 1, k 1 k 1 ,* g k 1, k 1 g 1, k 1 g k 1, k 1 g k 1, k 1 g k 1 ,* g 1, k 1 g k 1 ,* g 2 , k 1 g k 1 ,* g k 1 ,* g k 1, k 1 g k 1, k 1 g k 1 ,* × g k 1 ,* Let a i , j be the element of above determinant, and 1 i , j k 1， If i j , then a i , j g i , j g If i j , then a i , j a j ,i g i , k 1 g j , k 1 g ' j ,i g g i , k 1 g k i ,t ' j , k 1 t 1 ,t i g t , k 1 k g i ,t t 1 , t i j Therefore T k ' g i ,* ' g g g ) k 1 ,* ' g k 1 ,* g i , k 1 ( g i , t g t , k 1 t 1 ,t i j k 1 ,* g t 1 ,t j k 1 ,* g i , k 1 k g g i , k 1 g k 1 ,* k g i ,t k 1 ,* g i , k 1 j , k 1 g k g j , k 1 t 1, t i k 1 ,* ' i, j i, j k g i ,t g i , k 1 g t 1 , t i g g g i , k 1 If i j , then a i , j g i ,* g g 1 ,* ' 1, 2 ' ' ' 1, k 1 ' = T k 1 g g g 1, 2 g g ' 2 ,* 2 , k 1 k 1 ,* ' 1, k 1 ' 2 , k 1 g ' k 1 ,* ' So T k T k ( g i , j 0 ) ( T k 1 T k 1 ( g i , j 0 ) ) ( g k 1,* ) ' k 1 ,* And rightly T k ( g i , j 0 ) T k 1 ( g i , j 0 ) ( g k 1,* ) ' g (3) According to property 1.3, got T k' 1 T k' 1 ( g ' 0) i, j g ' i, j 1 ii ' S i ,i Because g i , j only exists in g i , j , similarly T k 1 T k 1 ( g i , j 0 ) g i , j 1 ' ' ' ii ' S i ,i . So T k T k ( g i , j 0 ) ( T k 1 T k 1 ( g i , j 0 ) ) ( g k 1,* ) ' ' Tk 1 Tk 1 ( g i , j 0 ) ' ( g i, j ' ' g i, j ' ) ( g k 1,* ) (4) The expression (3) and (4) are substituted into the equation (2), then proved the formula holds for n k 1 as well. Therefore the formula (1) holds. Done！ 3 Corollaries Corollary 3.1. According to property 1.1, ( T k T k ( g i , j 0 ) ) is corresponding to all trees which contains the edge g i , j , thus ( T k T k ( g i , j 0 ) ) 1 g i, j T k ( g i , j 1) T k ( g i , j 0 ) . Then the more elegant formula than (1) given G i, j T n 1 (5) T n 1 ( g i , j 1) T n 1 ( g i , j 0 ) n 2 , i j n , T n 1 0 . Gi, j 1 Ri , j is the equivalent conductance between vertex i and j.. Corollary 3.2. Subsituting the expression T n 1 T n 1 ( g i , j 0 ) g i , j 1 into formula (1), and due to both T n 1 and S i ,i natural way to express the formula with only M G i, j n C i, j are minors of M n ii S i ,i get from property 1.3 (the determinant of the matrix M n ), the ’s cofactors [4] is given (6) C i , j ,i , j n 2 , i j n , C i, j 0 . C i , j is the (i,j) cofactor of M C i , j ( 1) i j T n 1 T n 1 . n by removing the ith row and the jth column. Particularly if i j then The cofactor of M n by removing the two rows of ith and jth and the two columns of ith and jth is denoted as C i , j , i , j here, which is the replacement for S i , i just like the relationship between C i , j and T n 1 . Corollary 3.3. Obviously, The formulas (1) (5) and (6) hold in the field of complex number and it’s applicable to impedance [5] as well. References [1] https://math.berkeley.edu/~mhaiman/math172-spring10/matrixtree.pdf [2] http://en.wikipedia.org/wiki/Cayley%27s_formula [3] http://en.wikipedia.org/wiki/Star-mesh_transform [4] http://en.wikipedia.org/wiki/Minor_%28linear_algebra%29 [5] http://en.wikipedia.org/wiki/Electrical_impedance
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