Cube electrical resistance
Given a frame in the form of a cube, made of metal wire. The electrical resistance of each edge of the cube is equal to one ohm. What is the resistance of the cube during the passage of electric current from one vertex to another, if it is connected to a DC source as shown in the figure?
We consider the resistance of the circuit according to the formulas for parallel and series connection of resistances, we get the answer - the electrical resistance of the cube is 5/6 Ohm.
Interesting facts about the problem about the resistance of the cube of resistors
1. The solution to the problem about the resistance of a cube in general form can be found on the website of the Kvant magazine or see here: "At the end of the forties, the problem of the electrical resistance of a wire cube appeared in mathematical circles in Moscow. We don't know who invented it or found it in old textbooks. Problem was very popular, and everyone quickly learned about it. Very soon it began to be asked in exams and she became ...
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Consider a classical problem. A cube is given, the edges of which are conductors with some identical resistance. This cube is included in the electrical circuit between its various points. Question: what is the resistance of the cube in each of these cases? In this article, a tutor in physics and mathematics talks about how this classic problem is solved. There is also a video tutorial in which you will find not only a detailed explanation of the solution to the problem, but also a real physical demonstration that confirms all calculations.
So, the cube can be included in the circuit in three different ways.
Cube resistance between opposite vertices
In this case, the current, having reached point A, is distributed among the three edges of the cube. In this case, since all three edges are equivalent in terms of symmetry, none of the edges can be given more or less "significance". Therefore, the current between these ribs must be distributed equally. That is power...
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Weird..
You've answered your own question..
- Solder and "connect the ohmmeter probes to two points through which the main diagonal of the cube passes" "measure it"
Attached drawing: --
Enough simple reasoning. Enough school knowledge in physics. Geometry is not needed here, so let's move the cube to the plane and first mark the characteristic points.
Attached drawing: --
Still, it is better to give the logic of reasoning, and not just numbers at random. However, you didn't guess!
I propose to look for original solutions. You guessed it, but how did you decide? The answer is absolutely correct and you can close the topic. The only thing is that the problem can be solved in this way not only for the same R. It's simple if ...
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Let me comment on Master's statement
Let a voltage U be applied to the opposite edges of the cube A and C ", as a result of which a current I flows on the external section of the circuit with respect to the cube.
The figure shows the currents flowing along the faces of the cube. From symmetry considerations, it can be seen that the currents flowing along the faces AB, AA "and AD are equal - we denote this current as I1; in the same way, we obtain that the currents along the faces DC, DD", BC, BB", A"B", A"D "are equal to (I2)l; the currents in terms of CC", B"C" and D"C" are also equal to (I3).
We write Kirchhoff's laws (for example, for nodes A, B, C, C "):
( I = 3I1
( I1 = 2I2
( 2I2 = I3
( 3I3 = I
From here we get I1= I3 = I/3; I2 = I/6
Let the total resistance of the cube be r; then according to Ohm's law
(1) U = Ir.
On the other hand, when bypassing the contour ABCC" we obtain that
(2) U = (I1 + I2 + I3)R
From comparison (1) and (2) we have:
r = R*(I1 + I2 + I3)/I = R*(1/3 + 1/6 + 1/3) =...
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Students? These are school assignments. Ohm's law, series and parallel connections of resistances, the problem of three resistances and these at once.
Of course, I did not take into account the audience of the site, where most of the participants not only solve problems with pleasure, but also prepare tasks themselves. And, of course, he knows about classic puzzles that are at least 50 years old (I solved them from a collection older than the first edition of Irodov - 1979, as I understand it).
But still it is strange to hear that "problems are not Olympiad". IMHO, the "olympiad" of tasks is determined not so much and not even so much by complexity, but largely by the fact that when solving it is necessary (about something) to guess, after which the task becomes very simple from very complex.
The average student will write a system of Kirchoff equations and solve it. And no one can prove to him that the decision is wrong.
A smart student will guess the symmetry and solve problems faster than the average student.
P.S. However, "average students" are also different.
P.P.S....
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It is unreasonable to use universal mathematical packages in the presence of circuit analysis programs. The results can be obtained both in numerical form and in analytical form (for linear circuits).
I will try to give an algorithm for deriving the formula (R_eq = 3/4 R)
We cut the cube into 2 parts along the diagonals of the horizontal faces with a plane passing through the given points. We get 2 halves of the cube with a resistance equal to twice the desired resistance (the conductivity of half the cube is equal to half the desired conductivity). Where the cutting plane intersects the ribs, we divide their conductivities in half (we double the resistances). Expand half of the cube. We then obtain a scheme with two internal nodes. We replace one triangle with one star, since the numbers are integers. Well, then elementary arithmetic. It may be possible and even easier to decide, vague doubts gnaw ...
PS. In Mapple and/or Syrup, you can get a formula for any resistance, but looking at this formula, you will understand that only a computer wants with it...
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funny quotes
xxx: Yes! YES! Faster, even faster! I want two at once, no, three! And this one too! Oh yeah!
yyy: ... man, what are you doing there?
xxx: Finally unlimited download torrents :D
type_2: interesting, what if he put a cast iron cube painted in a Rubik's cube into it? :)
A discussion of a Lego robot that solves a Rubik's Cube in 6 seconds.
type_2: I wonder if he puts a cast-iron cube painted into a Rubik's cube there? :)
punky: Guess the country from the comments...
xxx: did you try on the new shorts?
yyy: no)
YY: Tomorrow...
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Solving problems for the calculation of electrical resistance using models
Sections: Physics
Objectives: educational: to systematize the knowledge and ability of students to solve problems and calculate equivalent resistances using models, frameworks, etc.
Developing: development of logical thinking skills of abstract thinking, the ability to replace equivalence schemes, simplify the calculation of schemes.
Educational: fostering a sense of responsibility, independence, the need for skills acquired in the lesson in the future
Equipment: a wire frame of a cube, a tetrahedron, an infinite chain of resistance grids.
DURING THE CLASSES
Update:
1. Teacher: "Remember the series connection of resistances."
Students draw a diagram on the board.
and write down
Teacher: remember the parallel connection of resistances.
A student on the blackboard draws an elementary ...
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For the development of students' creative abilities, the tasks of solving DC resistor circuits by the method of equipotential nodes are of interest. The solution of these problems is accompanied by a sequential transformation of the original scheme. Moreover, it undergoes the greatest change after the first step, when this method is used. Further conversions are associated with the equivalent replacement of series or parallel resistors.
To transform a chain, they use the property that in any chain, points with the same potentials can be connected into nodes. And vice versa: the nodes of the chain can be divided if after that the potentials of the points included in the node do not change.
In the methodological literature, they often write like this: if the circuit contains conductors with the same resistances, located symmetrically about any axis or plane of symmetry, then the points of these conductors, symmetrical about this axis or plane, have the same potential. But the whole difficulty is that no one designates such an axis or plane in the diagram and it is not easy to find it.
I propose another, simplified way of solving such problems.
Task 1. A wire cube (Fig. 1) is included in the chain between the points A to V.
Find its total resistance if the resistance of each edge is R.
Let's put the cube on the edge AB(Fig. 2) and "cut" it into twoparallel halves plane AA 1 B 1 Bpassing through the lower and upper edges.
Consider the right half of the cube. We take into account that the lower and upper ribs split in half and became 2 times thinner, and their resistances increased 2 times and became 2 R(Fig. 3).
1) Find resistanceR1the top three conductors connected in series:
4) Find the total resistance of this half of the cube (Fig. 6):
Find the total resistance of the cube:
It turned out to be relatively simple, understandable and accessible to everyone.
Task 2. The wire cube is connected to the circuit not by an edge, but by a diagonal AC any edge. Find its total resistance if the resistance of each edge is R (Fig. 7).
Place the cube on edge AB again. "Saw" the cube into twoparallel halvesthe same vertical plane (see Fig. 2).
Again, consider the right half of the wire cube. We take into account that the upper and lower ribs split in half and their resistances became 2 R.
Taking into account the conditions of the problem, we have the following connection (Fig. 8).
Consider a classical problem. A cube is given, the edges of which are conductors with some identical resistance. This cube is included in the electrical circuit between its various points. Question: what is cube resistance in each of these cases? In this article, a tutor in physics and mathematics talks about how this classic problem is solved. There is also a video tutorial in which you will find not only a detailed explanation of the solution to the problem, but also a real physical demonstration that confirms all calculations.
So, the cube can be included in the circuit in three different ways.
Cube resistance between opposite vertices
In this case, the current, reaching the point A, is distributed among the three edges of the cube. In this case, since all three edges are equivalent in terms of symmetry, none of the edges can be given more or less "significance". Therefore, the current between these ribs must be distributed equally. That is, the current strength in each rib is equal to:
As a result, it turns out that the voltage drop on each of these three ribs is the same and equal to , where is the resistance of each rib. But the voltage drop between two points is equal to the potential difference between these points. That is, the potentials of the points C, D and E the same and equal. For reasons of symmetry, the potentials of the points F, G and K are also the same.
Points with the same potential can be connected by conductors. This will not change anything, because no current will flow through these conductors anyway:
As a result, we get that the edges AC, AD and AE T. Similarly, ribs Facebook, GB and KB connect at one point. Let's call it a point. M. As for the remaining 6 edges, all their "beginnings" will be connected at the point T, and all ends are at the point M. As a result, we get the following equivalent circuit:
Resistance of a cube between opposite corners of one face
In this case, the edges are equivalent AD and AC. They will carry the same current. In addition, the equivalent are also KE and KF. They will carry the same current. We repeat once again that the current between the equivalent edges must be distributed equally, otherwise the symmetry will be broken:
Thus, in this case, the points have the same potential C and D, as well as points E and F. So these points can be combined. Let the points C and D unite at a point M, and the points E and F- at the point T. Then we get the following equivalent circuit:
On the vertical section (directly between the points T and M) current does not flow. Indeed, the situation is analogous to a balanced measuring bridge. This means that this link can be excluded from the chain. After that, it will not be difficult to calculate the total resistance:
The resistance of the upper link is , the lower one is . Then the total resistance is:
Cube resistance between adjacent vertices of the same face
This is the last possible option for connecting the cube to an electrical circuit. In this case, the equivalent edges through which the same current will flow are the edges AC and AD. And, accordingly, the same potentials will have points C and D, as well as points symmetrical to them E and F:
Again we connect in pairs the points with the same potentials. We can do this because no current will flow between these points, even if we connect them with a conductor. Let the points C and D merge into a dot T, and the points E and F- exactly M. Then we can draw the following equivalent circuit:
The total resistance of the resulting circuit is calculated by standard methods. Each segment of two resistors connected in parallel is replaced by a resistor with resistance . Then the resistance of the "upper" segment, consisting of series-connected resistors , and , is equal to .
This segment is connected to the "middle" segment, consisting of a single resistor with resistance , in parallel. The resistance of a circuit consisting of two resistors connected in parallel with resistance and is equal to:
That is, the scheme is simplified to an even simpler form:
As you can see, the resistance of the "upper" U-shaped segment is:
Well, the total resistance of two resistors connected in parallel with resistance and is equal to:
Experiment to measure the resistance of a cube
To show that all this is not a mathematical trick and that there is real physics behind all these calculations, I decided to conduct a direct physical experiment to measure the resistance of a cube. You can watch this experiment in the video at the beginning of the article. Here I will post photos of the experimental setup.
Especially for this experiment, I soldered a cube, the edges of which are the same resistors. I also have a multimeter, which I turned on in resistance measurement mode. The resistance of a single resistor is 38.3 kOhm:
Sections: Physics
Goals: educational: to systematize the knowledge and skills of students to solve problems and calculate equivalent resistances using models, frames, etc.
Developing: development of logical thinking skills of abstract thinking, the ability to replace equivalence schemes, simplify the calculation of schemes.
Educational: fostering a sense of responsibility, independence, the need for skills acquired in the lesson in the future
Equipment: a wire frame of a cube, a tetrahedron, an infinite chain of resistance grids.
DURING THE CLASSES
Update:
1. Teacher: "Remember the series connection of resistances."
Students draw a diagram on the board.
and write down
U about \u003d U 1 + U 2
Y about \u003d Y 1 \u003d Y 2
Teacher: remember the parallel connection of resistances.
The student draws an elementary diagram on the board:
Y about \u003d Y 1 \u003d Y 2
; for for n equal
Teacher: And now we will solve problems for calculating the equivalent resistance, a section of the circuit is presented in the form of a geometric figure, or a metal mesh.
Task #1
Wire frame in the form of a cube, the edges of which represent equal resistance R. Calculate the equivalent resistance between points A and B. To calculate the equivalent resistance of this frame, it is necessary to replace it with an equivalent circuit. Points 1, 2, 3 have the same potential, they can be connected into one node. And the points (vertices) of the cube 4, 5, 6 can be connected to another node for the same reason. Students have a model on each desk. After performing the described steps, an equivalent circuit is drawn.
On the AC section, the equivalent resistance is ; on CD; on DB ; and finally for the series connection of resistances we have: 
By the same principle, the potentials of points A and 6 are equal, B and 3 are equal. Students combine these points on their model and get the equivalent circuit:
The calculation of the equivalent resistance of such a circuit is simple. 
Task #3
The same cube model, with inclusion in the circuit between points 2 and B. Students connect points with equal potentials 1 and 3; 6 and 4. Then the circuit will look like this:
Points 1.3 and 6.4 have equal potentials, and the current through the resistances between these points will not flow, and the circuit is simplified to the form; the equivalent resistance of which is calculated as follows:
Task #4
An equilateral triangular pyramid whose edge has resistance R. Calculate the equivalent resistance when included in the circuit.
Points 3 and 4 have an equal potential, so no current will flow along edge 3.4. Students remove it.
Then the diagram will look like this:
The equivalent resistance is calculated as follows:
Task number 5
Metal mesh with link resistance R. Calculate the equivalent resistance between points 1 and 2.
At point 0, you can separate the links, then the circuit will look like:
- resistance of one half symmetrical in 1-2 points. Parallel to it is the same branch, therefore 
Task number 6
The star consists of 5 equilateral triangles, the resistance of each 
.
Between points 1 and 2 one triangle is parallel to four connected in series
Having experience in calculating the equivalent resistance of wire frames, you can begin to calculate the resistance of a circuit containing an infinite number of resistances. For example:
If you separate the link
from the general scheme, then the scheme will not change, then it can be represented as
or 
,
we solve this equation with respect to R equiv.
The result of the lesson: we have learned how to abstractly represent circuit sections of the circuit, replace them with equivalent circuits that make it easy to calculate the equivalent resistance.
Note: This model should be represented as:
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