Solutions for Resistive Networks

F. Poyo (yoshi.tamori@gmail.com)


To think about a symmetry is likely to provide us a drastically easy answer. The problems are also the two of such problems. The first one uses concentric symmetry. The second one uses translational symmetry.


Solution of the second question.

Let the just after the next nodes of the first terminals (C) and (D) as follows

                (C)
     (A)o---R---+-+-R---+---R--- ...        
                |       |
                R       R      ..... (repeat infinitely)
                |       |
     (B)o-------+-+-----+------- ...
                (D)
Then, cut the ladder at the points as follows.

     (A)o---R---+-o           (C) o--R---+---R--- ...        
                |                        |
                R                        R      ..... (repeat infinitely)
                |                        |
     (B)o-------+-o           (D) o------+------- ...

Since the obtained ladder is also infinite ladder, the equivalent resistance between (C) and (D) must equal to the one between (A) and (B) (thus

                     (C)
     (A)o---R---+-o--+        (A) o
                |    |            |
                R    Rcd  ==      Rab
                |    |            |
     (B)o-------+-o--+        (B) o
                     (D)
where
            Re=Rab=Rcd.

) Therefore,

                   R x Re
         Re = R + --------.
                   R + Re

This leads to

          2             2
         Re - R x Re - R  = 0

It follows that

               1 + SQRT(5)
         Re = -------------R
                    2
The ladder surely has irrational resistance.


Solution of the first question.

Put a current source between (A) and infinity point which provides current I.


            ... 
                  _______________
             |   /              |
        (I/4)R  /              (I)
             | /                |
  ... ---R--(A)--R---...      -----(infinity point)
       (I/4) |  (I/4)          ---
             R                  -
             |(I/4)

            ... 
In the next, put a current source between (B) and infinity point which provides negative current -I.

            ... 
                  _______________
             |   /              |
       (-I/4)R  /              (-I)
             | /                |
  ... ---R--(B)--R---...      -----(infinity point)
      (-I/4) |  (-I/4)         ---
             R                  -
             |(-I/4)

            ... 
Since in a linear electric circuit, we can always superimpose several passive circuits of single active node, we obtain the amount of current between (A) and (B) by I/2.

            ... 
                  _______________
             |   /              |
             R  /              (I)
             | /                |
  ... ---R--(A)--R---...        +  (infinity point)
             |                  | 
             |                  |
             R(I/4)-(-I/4)=I/2  |
             |  ----------------+
             | /                
  ... ---R--(B)--R---...     
             |                 
             R                ## Note the direction of
             |                   the currents. Negative
                                 current has opposite direction.
            ... 

Therefore, the voltage between (A) and (B) is

               I
     V =  R x ---.
               2

Even if we substitute the infinite circuit by the equivalent
resistance such that

                ________________  
               /                |
            (A)                (I) (infinity point)
             |                  | 
             |                  |
             Re (I)             |
             |                  |
             |  ----------------+
             | /                
            (B)              
, the voltage between (A) and (B) should be the same
as V. 

        V = Re x I

Therefore, from
                         I
       V = Re x I = R x ---,
                         2

we get
                R
          Re = ---
                2

The network circuit has rational resistance.


Overviewing the above results, irrationality (whether or not the value is irrational) seems to be dependent on the symmetry of the system. Thus non-concentric circuit gives irrational number. While concentric circuit gives rational. Though this is not a strict rule, it is likely to seen in physics systems. It is easy to show the resistive network on a D-dimensional hyper-cube has the equivalent resistance

                   R
                  ---
                   D

Mathemusement

yoshi.tamori@gmail.com