Electrons fly into a flat capacitor of length L at an angle a to the plane of the plates, and fly out at an angle β. Determine the initial kinetic energy of the electrons if the field strength of the capacitor is E.
The resistance of any edge of the wire frame of the cube is equal to R. Find the resistance between the vertices of the cube that are furthest apart from each other.
When a current of 1.4 A was passed for a long time through the wire, the latter heated up to 55°C, and with a current of 2.8 A - up to 160°C. To what temperature does the wire heat up at a current of 5.6A? Wire resistance does not depend on temperature. The ambient temperature is constant. Heat transfer is directly proportional to the temperature difference between the wire and air.
A lead wire with diameter d melts when current I1 is passed for a long time. At what current will a wire with diameter 2d melt? The heat loss by the wire in both cases is considered proportional to the surface of the wire.
How much heat will be released in the circuit after switch K is opened? The circuit parameters are shown in the figure.
An electron flies into a uniform magnetic field, the direction of which is perpendicular to the direction of its movement. Electron speed v = 4·107 m/s. Magnetic field induction B = 1 mT. Find the tangential aτ and normal an acceleration of the electron in a magnetic field.
In the circuit shown in the figure thermal power, released in the external circuit, is the same with the key K closed and open. Determine the internal resistance of the battery r if R1 = 12 Ohm, R2 = 4 Ohm. 
Two particles with a charge ratio q1/q2 = 2 and a mass ratio m1/m2 = 4 fly into a uniform magnetic field perpendicular to its induction lines and move in circles with a radius ratio R1/R2 = 2. Determine the ratio of the kinetic energies W1/W2 of these particles.
The oscillatory circuit consists of a capacitor with a capacity C = 400 pF and a coil with an inductance L = 10 mH. Find the amplitude of current oscillations Im if the amplitude of voltage oscillations Um = 500 V.
After what time (in fractions of the period t/T) will the capacitor of the oscillating circuit first have a charge equal to half the amplitude value? (the time dependence of the charge on the capacitor is given by the equation q = qm cos ω0t)
How many electrons are emitted from the cathode surface in 1 s at a saturation current of 12 mA? q = 1.6·10-19 Cl.
Current strength in the circuit electric stove is equal to 1.4 A. What electric charge passes through the cross section of its spiral in 10 minutes?
Determine the cross-sectional area and length of a copper conductor if its resistance is 0.2 Ohm and its mass is 0.2 kg. The density of copper is 8900 kg/m3, the resistivity is 1.7 * 10-8 Ohm * m.
In the figure of the AB circuit section, the voltage is 12 V, the resistances R1 and R2 are equal to 2 Ohms and 23 Ohms, respectively, the resistance of the voltmeter is 125 Ohms. Determine the voltmeter readings. 
Determine the resistance value of the ammeter shunt to expand the current measurement limits from 10 milliamps (I1) to 10 Amps (I). The internal resistance of the ammeter is 100 Ohms (R1).
What thermal power is released in resistor R1 in the circuit, the circuit of which is shown in the figure, if the ammeter shows power direct current I = 0.4 A? Resistor resistance values: R1 = 5 Ohm, R2 = 30 Ohm, R3 = 10 Ohm, R4 = 20 Ohm. The ammeter is considered ideal. 
Two identical small metal balls are charged so that the charge of one of them is 5 times greater than the charge of the other. The balls were brought into contact and moved apart to the same distance. How many times has the force of their interaction changed in magnitude if: a) the balls are charged in the same way; b) are the balls oppositely charged?
The length of a cylindrical copper wire is 10 times greater than the length of an aluminum wire, and their masses are the same. Find the resistance ratio of these conductors.
The wire ring is included in a circuit through which a current of 9 A passes. The contacts divide the length of the ring in a ratio of 1:2. At the same time, a power of 108 W is released in the ring. What power will be released in the ring at the same current strength in the external circuit if the contacts are placed along the diameter of the ring?
Two balls of the same volume, each having a mass of 0.6 ∙ 10 -3 g, are suspended on silk threads 0.4 m long so that their surfaces touch. The angle at which the threads diverged when imparting equal charges to the balls is 60°. Find the magnitude of the charges and the force of electric repulsion.
Two identical balls, one charged with a negative charge of 1.5 μC, the other with a positive charge of 25 μC, are brought into contact and again moved apart to a distance of 5 cm. Determine the charge of each ball after contact and the force of their interaction.
To develop students' creative abilities, problems involving solving DC resistor circuits using the equipotential node method are of interest. The solution to these problems is accompanied by a consistent transformation original circuit. Moreover, it undergoes the greatest change after the first step when this method is used. Further transformations involve equivalent replacement of series or parallel resistors.
To transform a circuit, they use the property that in any circuit points with the same potentials can be connected into nodes. And vice versa: the nodes of the circuit can be divided if after this the potentials of the points included in the node do not change.
IN methodological literature often written like this: if a circuit contains conductors with equal resistances located symmetrically relative to any axis or plane of symmetry, then the points of these conductors, symmetrical relative to this axis or plane, have the same potential. But the whole difficulty is that no one indicates such an axis or plane on the diagram and it is not easy to find it.
I propose another, simplified way to solve such problems.
Problem 1. A wire cube (Fig. 1) is included in the circuit between the points A to B.
Find its total resistance if the resistance of each edge is equal R.
Place the cube on its edge AB(Fig. 2) and “cut” it into twoparallel halves plane AA 1 B 1 B, passing through the lower and upper edge.
Let's look at the right half of the cube. Let's take into account that the lower and upper ribs split in half and became 2 times thinner, and their resistance increased 2 times and became 2 times R(Fig. 3).
1) Find resistanceR 1three upper 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.
Problem 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 equal R (Fig. 7).
Place the cube on edge AB again. “Saw” the cube into twoparallel halvesthe same vertical plane (see Fig. 2).
Again we look at 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 each R.
Taking into account the conditions of the problem, we have the following connection (Fig. 8).
Sections: Physics
Goals: educational: systematize students’ knowledge and skills in solving problems and calculating equivalent resistances using models, frames, etc.
Developmental: development of logical thinking skills, abstract thinking, skills to replace equivalence schemes, simplify the calculation of schemes.
Educational: fostering a sense of responsibility, independence, and the need for skills acquired in class in the future
Equipment: wire frame of a cube, tetrahedron, mesh of an endless chain of resistance.
DURING THE CLASSES
Update:
1. Teacher: “Let’s remember the series connection of resistances.”
Students draw a diagram on the board.
and write down
U rev =U 1 +U 2
Y rev =Y 1 =Y 2
Teacher: let's remember parallel connection resistance.
A student sketches a basic diagram on the board:
Y rev =Y 1 =Y 2
; for for n equal
Teacher: Now we will solve problems on calculating the equivalent resistance of a section of the circuit presented in the form geometric figure, or metal mesh.
Task No. 1
A wire frame in the form of a cube, the edges of which represent equal resistances R. Calculate the equivalent resistance between points A and B. To calculate the equivalent resistance of a given 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 points (vertices) of the cube 4, 5, 6 can be connected into another node for the same reason. Students have such a model on each desk. After completing the described steps, draw an equivalent circuit.
In 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 an equivalent diagram:
Calculating the equivalent resistance of such a circuit is simple 
Problem No. 3
The same model of a cube, with inclusion in the circuit between points 2 and B. Students connect points with equal potentials 1 and 3; 6 and 4. Then the diagram will look like this:
Points 1,3 and 6,4 have equal potentials, and no current will flow through the resistances between these points and the circuit is simplified to the form; the equivalent resistance of which is calculated as follows:
Problem No. 4
An equilateral triangular pyramid, the edge of which has a resistance R. Calculate the equivalent resistance when connected to the circuit.
Points 3 and 4 have equal potential, so no current will flow along edge 3.4. The students clean it up.
Then the diagram will look like this:
The equivalent resistance is calculated as follows:
Problem No. 5
Metal mesh with link resistance equal to R. Calculate the equivalent resistance between points 1 and 2.
At point 0 you can separate the links, then the diagram will look like:
- the resistance of one half is symmetrical at 1-2 points. There is a similar branch parallel to it, so 
Problem No. 6
The star consists of 5 equilateral triangles, the resistance of each 
.
Between points 1 and 2, one triangle is parallel to four triangles 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 circuit, then the circuit will not change, then it can be represented in the form
or 
,
solve this equation for R eq.
Lesson summary: we learned to abstractly represent circuit sections of a circuit and replace them with equivalent circuits, which make it easy to calculate the equivalent resistance.
Instructions: This model can be represented as:
Let's consider a classic problem. Given a cube, the edges of which represent conductors with some identical resistance. This cube is included in an electrical circuit between all its possible points. Question: what is equal cube resistance in each of these cases? In this article, a physics and mathematics tutor 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 confirming all the calculations.
So, a cube can be connected to a circuit in three different ways.
Resistance of a cube between opposite vertices
In this case, the current, having reached the point A, is distributed between three edges of the cube. Moreover, since all three edges are equivalent in terms of symmetry, no edge can be given more or less “significance”. Therefore, the current between these edges must be distributed equally. That is, the current strength in each edge is equal to:
The result is that the voltage drop across each of these three edges is the same and is equal to , where is the resistance of each edge. 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 are the same and equal. For symmetry reasons, the point potentials 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 find that the edges A.C., AD And A.E. T. Likewise the ribs FB, G.B. And K.B. 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 equivalent edges are AD And A.C.. The same current will flow through them. Moreover, equivalent are also KE And KF. The same current will flow through them. Let us repeat once again that the current between 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. This means that 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 a vertical section (directly between the points T And M) no current flows. Indeed, the situation is similar to a balanced measuring bridge. This means that this link can be excluded from the chain. After this, calculating the total resistance is not difficult:
The resistance of the upper link is equal to , the resistance of the lower link is . Then the total resistance is:
Resistance of a cube between adjacent vertices of the same face
This is the last possible variant connecting the cube to an electrical circuit. In this case, the equivalent edges through which the same current will flow are the edges A.C. And AD. And, accordingly, points will have identical potentials C And D, as well as points symmetric to them E And F:
We again connect points with equal potentials in pairs. 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 unite into a point 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 using standard methods. We replace each segment of two parallel-connected resistors with 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 one resistor with a resistance of , in parallel. The resistance of a circuit consisting of two parallel-connected resistors 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 equal to:
Well, the total resistance of two parallel connected resistors 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 whose edges were identical resistors. I also have a multimeter that I turned on in resistance mode. The resistance of a single resistor is 38.3 kOhm:
Electrical resistance of a cube
A cube-shaped frame made of metal wire is given. The electrical resistance of each edge of the cube is one ohm. What is the resistance of a cube when passing through electric current from one vertex to another if it is connected to a constant current source as shown in the figure?
We calculate the resistance of the circuit using the formulas for parallel and series connection of resistances, and we get the answer - the electrical resistance of the cube is 5/6 Ohms.
Interesting facts about the problem about the resistance of a cube of resistors
1. Solving the problem about the resistance of a cube in general view can be read on the Kvant magazine website or viewed here: “At the end of the forties, a problem about 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. The problem was very popular, and everyone quickly learned about it Very soon they began to ask her in exams and she became...
0 0
Let's consider a classic problem. Given a cube, the edges of which represent conductors with some identical resistance. This cube is included in an electrical circuit between all its possible points. Question: what is the resistance of the cube in each of these cases? In this article, a physics and mathematics tutor 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 confirming all the calculations.
So, the cube can be connected to the circuit in three different ways.
Resistance of a cube between opposite vertices
In this case, the current, having reached point A, is distributed between the three edges of the cube. Moreover, since all three edges are equivalent in terms of symmetry, no edge can be given more or less “significance”. Therefore, the current between these edges must be distributed equally. That is, strength...
0 0
Strange..
You 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 is a drawing: --
Simple reasoning will suffice. Enough with school knowledge of physics. Geometry is not needed here, so let’s move the cube onto a plane and first mark the characteristic points.
Attached is a drawing: --
Still, it is better to provide logical reasoning, and not just numbers at random. However, they didn’t guess right!
I suggest you look original ways solutions. You guessed it, but how did you decide? The answer is absolutely correct and the topic can be closed. The only thing is that the problem can be solved this way not only for identical R. Simply, if...
0 0
Let me comment on Teacher'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 in the section of the circuit external to the cube.
The figure shows currents flowing along the faces of a cube. From symmetry considerations it is clear that the currents flowing along the faces AB, AA" and AD are equal - let's denote this current I1; in the same way we find that the currents along the faces DC, DD", BC, BB", A"B", A"D " are equal to (I2)l; the currents along the facets CC, B"C" and D"C" are also equal to (I3).
We write down 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 ABCC contour 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) =...
0 0
Students? These are school tasks. Ohm's law, series and parallel connections of resistances, a problem about 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 problems that are at least 50 years old (I solved them from a collection older than Irodov’s first edition - 1979, as I understand it).
But it’s still strange to hear that “the problems are not Olympiad.” IMHO, the “olympics” of problems is determined not so much or even so much by their complexity, but largely by the fact that when solving it you have to guess (about something), after which the problem from very complex becomes very simple.
The average student will write a system of Kirgoff equations and solve it. And no one will prove to him that the decision is wrong.
A smart student will figure out symmetry and solve problems faster than the average student.
P.S. However, “average students” are also different.
P.P.S....
0 0
Using universal mathematical packages is unwise if you have circuit analysis programs. The results can be obtained both numerically and analytically (for linear circuits).
I’ll 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 a 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 conductivity in half (we double the resistance). Expand half of the cube. We then obtain a circuit with two internal nodes. We replace one triangle with one star, since the numbers are integers. Well, then some basic arithmetic. It may be possible and even easier to solve, vague doubts are gnawing...
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 will want with it...
0 0
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, downloading torrents: D
type_2: I wonder, what if he put a cast iron cube in there, painted like a Rubik’s cube? :)
Discussion of a Lego robot that solves a Rubik's cube in 6 seconds.
type_2: I wonder what if he put a cast iron cube painted into a Rubik’s cube in there? :)
punky: guess the country from the comments...
xxx: did you try on the new panties?
yyy: Nope)
yyy: Tomorrow...
0 0
Solving calculation problems electrical resistance using models
Sections: Physics
Objectives: educational: systematize students’ knowledge and skills in solving problems and calculating equivalent resistances using models, frames, etc.
Developmental: development of logical thinking skills, abstract thinking, skills to replace equivalence schemes, simplify the calculation of schemes.
Educational: fostering a sense of responsibility, independence, and the need for skills acquired in class in the future
Equipment: wire frame of a cube, tetrahedron, mesh of an endless chain of resistance.
DURING THE CLASSES
Update:
1. Teacher: “Let’s 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 sketches an elementary...
0 0