# 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

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