**Igor Feliksovich Spivak-Lavrov*, Maksat
Serikbayuly Kurmanbay, Аidana N. Mazhit**

Department of
Physics, Aktobe Regional State University named after K. Zhubanov, Аktobe,
Kazakhstan

***Corresponding author:** Igor Feliksovich Spivak-Lavrov, Department of
Physics, Aktobe Regional State University named after K. Zhubanov, Аktobe,
Kazakhstan. Tel: +787012503562; Email: spivakif@rambler.ru

**Received Date:** 07 August, 2018; **Accepted Date:** 24 August,
2018; **Published Date:** 03 September, 2018

Precise
solutions of problems on finding the electrical resistance between adjacent
nodes of infinite two-dimensional grids with square and triangular cells are
obtained in the work. The calculation technique used is based on the use of
equivalent circuits of infinite resistance chains. It is shown that the method
of calculating the resistance of such systems, described in the collection of
tasks, Irodov I.E. and based on the use of the principles of superposition and
symmetry is only approximate. The discrepancy with the resistance values given
in the collection of Irodov I.E. is about 10%.

**Keywords:**
Infinite two-dimensional network of resistances; Resistance calculation

1. Introduction

Calculation
of the resistance of complex resistor connections has always attracted the
attention of physicists. Thus, they developed original methods for calculating
infinite chains of resistances. At present, in connection with the development
of nanotechnology, the problems of calculating the resistances of infinite
resistor networks have become especially topical. Let us dwell in more detail
on one problem placed in the collection [1].
«There is a boundless wire mesh with square cells (рис. 1). The resistance of each
conductor between neighboring grid nodes is *r*. Find the
resistance of this grid between two adjacent grid nodes *A* and *B*». In the 2003 collection, this task is numbered
3.154. To this problem is given the Instruction: "To use the principles of
symmetry and superposition".

In
the collection [1] the solution of this problem
is given. We will quote it below. "Connect mentally to points A and B the
voltage source U. Then,
where current in the supply wires, is the current in the conductor *АВ*. The current can be represented as a superposition of two
currents. If the current "flowed into" point *А* and spread out along the grid to
infinity, then the conductor *АВ* – from symmetry – would go current.
Similarly, if the current entered the grid from infinity and
"flowed out" of point *В*, then
the conductor *АВ*
would also have a current.
By imposing these decisions on each other, we get therefore " (Figure 1).

**2.
Research Hypothesis**

But
this solution is only approximate, because it is obtained from the principles
of superposition and symmetry, which in this case are performed only
approximately. Indeed, from the same superposition principle it follows that
when the current "flows into"
point *А*,
then a voltage must be applied to this point. Similarly, a voltage is applied to point *В*. If we apply both these voltages
simultaneously, then the symmetry in the distribution of the currents at points
*A* and *B* is
violated because of the mutual influence of the potentials and the response of
the problem will change. Moreover, it can be said that the mutual influence of
potentials will lead to the fact that the current in the conductor *АВ* will be greater than and, consequently, the resistance.

2.1. Calculation of
the Resistance of Grid with Square Cells

We
give below an alternative solution to this problem. First, we divide the entire
plane into two identical half-planes, cutting the grid along a straight line
passing through points *A* and *B*, to the
left of point *A*
and to the right of *B*. For this, each of the resistances *r*, lying
to the left of point *А*
and to the right of *В*, is replaced by two parallel-connected resistance of each. We obtain the following picture, shown
in Figure 2. In Figure 3
shows an equivalent scheme of a cut mesh, where the resistance of the
half-planes obtained as a result of cutting is denoted by *R*:

Then
the resistance between adjacent nodes *A* and *B* of an infinite two-dimensional grid with square
cells is equal to:

To do
this, we similarly cut the grid along a straight line passing through points *C* and *D*. The resulting picture is shown in Figure
4, where you can see two infinite half-planes with resistance *R*, located below points *A*, *B* and
above points *C*, *D*, and
also two identical infinite chains going to the left of points *A* and *C* and to the right of points *B* and *D*. We denote the resistance of such an
infinite chain by *r**. Then, comparing the schemes
in Figures 4, we can draw up the equivalent circuit depicted in Figure 5.

Using
the equivalent circuit in Figure 5, we write the
equation for determining the resistance of the half-plane *R*:

Solving
this quadratic equation with respect to *R*, we find

The
resistance of an infinite periodic chain *r* *is found by the standard method, solving
equation

Location

Substituting
(5) into (3), we obtain

Substitution
of this value of *R* into (1) leads to
the final result:

This
result, although not very strong, is still different from the result of 0.5 *r *obtained
in the approximation of the principles of symmetry and superposition. In this
connection it is interesting to look at the distribution of currents at point *A*. If
current *I*
approaches the point *A*, then it will be distributed as follows:
according to the resistance going up and down from point *A*, the same currents will go,
according to the resistance between nodes *A*
and *B*, there will be a current,
and in the opposite direction there will be a current. It's a pity, the beauty
is gone, the symmetry has disappeared, and everything has become very prosaic.
Well, in life it often happens that beauty deceives us and then it is difficult
to get rid of beautiful illusions.

2.1. Calculation of
the Resistance of Grid with Triangular Cells

Here
we also give a solution to the problem of an infinite net by the same
resistances *r*
forming regular triangles, as shown in Figure 6.
Suppose it is necessary to find the resistance *R*_{AB}
of the lattice between the two nearest nodes *A* and *B*. Using for the solution of the
problem the principles of symmetry and superposition, by analogy with a square
grid, it is easy to obtain the following simple result of,
which is also only approximate.

In
Figure 6 shows the part of the grid and the bold lines show the directions
along which it is necessary to make cuts in order to break it into the same
half-planes and endless chains. First, we cut along the rays issuing from the
nodes *A* and *B*, thus breaking the entire grid into two
half-planes with the same resistances *R*. As a result, just as in the case of a
rectangular grid, we obtain the equivalent circuit shown in Figure 3, and
formula (1) for the resistance of an infinite net.

Qn
figure 6 each side of the triangle has a resistance r. The thick lines indicate
the directions of the sections. To determine the resistance of the half-plane *R*, we cut
once more the lattice along the rays emanating from the nodes *C* and *D*. We obtain the equivalent circuit shown in Figure 7 and, respectively, equation (8):

After
simple transformations, we obtain the following quadratic equation for the
determination of R:

Location

Here,
the resistance of the infinite chain *r** is found from the equivalent circuit shown in Figure 8, which leads to the equation (11).

Solving
this equation, we find

*r** into formula (10), we obtain for R the expression:

3. Conclusion

Thus, in this paper we develop a method for
calculating infinite networks of resistances, which makes it possible to find
the exact resistance between the nearest nodes of such networks. It is shown
that the calculation method used in [1], based
on the principles of superposition and symmetry, gives only an approximate
underestimated result for this resistance. And for a grid with square cells,
the difference of results does not exceed 5%, and for a network with triangular
elements it approaches 20%.

Figure
1: Infinite wire mesh with square cells.

Figure
2: Infinite wire mesh, divided into two half-planes.

Figure
3: Equivalent scheme of unlimited wire mesh.

Figure
4: Infinite wire mesh, cut along two straight lines.

Figure
5: An equivalent circuit for the resistance of the half-plane *R *in the case of a square grid.

Figure
6: Infinite grid of regular triangles.

Figure
7: An equivalent circuit for the resistance of the half-plane *R *in the case of a triangular grid.

**Figure 8:** Equivalent circuit
for calculation of resistance *r* *infinite chain in the case of a triangular
grid.

1. Irodov IE (2003) Problems in
general physics. Tutorial. 6 ^{th} ed.,
Sr. – SPb . Izd. "Lan", 2003.Pg. No. 416.

2.
Bessonov LA (2002) Theoretical bases of electrical
engineering. Electric circuits. – Moscow: Gardariki, 2002. Pg. No. 638.

**Citation: **Spivak-Lavrov IF,
Kurmanbai MS, Mazhit АN (2018) About One Method of Calculation of Resistance of
Two-dimensional Infinite Grid Systems. Educ Res Appl: ERCA-157. DOI: 10.29011/2575-7032/100057

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