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

ii

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
ii
'
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
'
'
'
ii
'
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
ii
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