The resistance of an element shaped like a half-cylinder

In summary, the author is asking if it is correct to use the formula S = π/2*(r2^2-r1^2) to calculate the resistance of an element when the current flows from A,B,C. The author does not provide any additional information.
  • #1
polibuda
52
9
Homework Statement
Determine the resistance of element put on the picture, if the current flows from A,B,C.
Relevant Equations
Given informations: r1, r2, h and ϱ(resistivity)
1606668864020.png

So I started doing this with formula:
1606668803136.png

S is equal to π/2*(r2^2-r1^2)
l is equal to h
Is it correct ?
 
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  • #2
polibuda said:
Homework Statement:: Determine the resistance of element put on the picture, if the current flows from A,B,C.
Relevant Equations:: Given informations: r1, r2, h and ϱ(resistivity)

View attachment 273370
So I started doing this with formula:
View attachment 273369
S is equal to π/2*(r2^2-r1^2)
l is equal to h
Is it correct ?

What does "if the current flows from A,B,C." mean?
 
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  • #3
berkeman said:
What does "if the current flows from A,B,C." mean?
In the picture it is shown like currents flows to this element from A direction, B direction and C direction.
 
  • #4
polibuda said:
In the picture it is shown like currents flows to this element from A direction, B direction and C direction.
Currents have to flow in closed loops. The figure shows currents entering 3 places, but where do they leave? And are they all supposed to be flowing at the same time, or are these 3 separate calculations that they want you to do?
 
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  • #5
berkeman said:
Currents have to flow in closed loops. The figure shows currents entering 3 places, but where do they leave? And are they all supposed to be flowing at the same time, or are these 3 separate calculations that they want you to do?
That informations are not given in my task.
 
  • #6
berkeman said:
The figure shows currents entering 3 places, but where do they leave?
Yes. Are you (@polibuda ) trying to replace the object with an equivalent set of three resistors between the points AB and C? The formula you quote refers to a piece of metal which is uniform between the end faces. It need not be a cylinder but the formula doesn't actually apply to resistances between any of the three connecting points. It would only apply between C and the face at the other end.
 
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  • #7
Is it possible to determine resistance with two radii, length and resistivity of this element? So should I determine two more resistances and sum them with first?
 
  • #8
There is no answer to that until you choose the contact points / surfaces. You must appreciate the essential difference between a simple cylinder and the shape you are discussing.
 
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  • #9
Sorry, I'm just not sure we can help. This question is posed in such a confusing way, even if you know about solving this sort of question, that doesn't mean you can decipher the intent.

I will give you some of my thoughts:

1) It does seem clear to me that they want three different answers: A, B, and C directions.

2) If you interpret this as current coming from a point contact with the conductor at points A, B, and C, then you will also need to know where they are flowing to. If I specify current from A to B (both point contacts), that is a different answer than from point A to C. You could just solve all of these (A-B, A-C, B-C), but the points aren't really specified well enough for that. Also, frankly, that calculation is nasty; more a simulation exercise than a closed form answer.

3) So my second guess is that they are assuming that the electrons flow in the (uniform) direction of A, B, or C. I guess as if there is an equipotential planar surface that cuts through the object normal to the given direction (otherwise the electrons would curve). That is understandable in the C direction since the start and end surfaces are normal to the direction given. But it doesn't make any sense to me in the other directions since the boundary surfaces aren't planar, and aren't consistent with an equipotential plane. I suppose there is a way to do this if you specify that this shape is actually imbedded in a material with the same resistivity so the electrons don't see any change when crossing the boundaries (no dissimilar metals allowed either, if you are a stickler for precision). This surely isn't what they meant, it's a bazaar interpretation to fix a simple, but confusing description.

4) Maybe my best guess: Perhaps you should assume that the electrons flow from a boundary surface to the "opposite" boundary. So the C direction is axial flow (along the cylinders), the B direction is radial flow (outside to inside cylinders), and the A direction is circumferential flow (around the cylinder). This way you can have an equipotential surface at each entry and exit boundary. Also the electrons won't need to cross the perpendicular boundary surfaces.

So, sorry, I'm not at all sure this is of any help at all. I think maybe you should be confused. Can you ask for clarity from your instructor?
 
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  • #10
DaveE said:
current coming from a point contact
From a point contact the resistance would always be infinite.

I think it a reasonable guess that for A we want the resistance between the two rectangular upward facing surfaces. That one looks the trickiest.

For C, it would be between the two half annuli. This is the easiest, and @polibuda's approach looks right for that one.

For B, probably between the two half cylinder surfaces, being the only pair left.
 
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  • #11
Your question does not make sense.

Please post the exact words in the question or an image of the question.
 
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  • #12
Frodo said:
Your question does not make sense.

Please post the exact words in the question or an image of the question.
This task is translated from another language. Everything is correctly translated, but person, which wrote this tasks made a lot of mistakes. But my university has still used lists with his tasks (I don't know why). I sent email to my instructor and I'm waiting for his responding.
 
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  • #13
polibuda said:
This task is translated from another language. Everything is correctly translated, but person, which wrote this tasks made a lot of mistakes. But my university has still used lists with his tasks (I don't know why). I sent email to my instructor and I'm waiting for his responding.
Why not proceed with my interpretation in post #10? It looks to me like a fair bet.
The only thing that bothers me is the expected method for solving A. An easy approach is to treat the current flow as being in parallel semicircular arcs, giving an effective resistance proportional to ln(r2/r1); but I suspect that is not really valid, and a correct solution may involve solving Laplace's equation as a boundary value problem.
 
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  • #14
haruspex said:
Why not proceed with my interpretation in post #10? It looks to me like a fair bet.
The only thing that bothers me is the expected method for solving A. An easy approach is to treat the current flow as being in parallel semicircular arcs, giving an effective resistance proportional to ln(r2/r1); but I suspect that is not really valid, and a correct solution may involve solving Laplace's equation as a boundary value problem.
I'm sorry. I had to have sure, that is good interpretation. Well, I have the formulas to B and C, but I consider about A. My instructor, said that in A formula is logarithm (that is advise from him), so maybe you are right.
 
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  • #15
I consider how to determine the resistance for A case in this easy method.
For C formula is:
1607115798259.png

For B formula is:
1607115836365.png
 
  • #16
I don't get quite the same for B. Please post your working
 
  • #17
I made one mistake. That is good result.
1607160982411.png
 
  • #19
Could you advise me how to determine resistance in A case. I'm confused. I have no even dr in my formula.
1607250860492.png
 
  • #20
polibuda said:
Could you advise me how to determine resistance in A case. I'm confused. I have no even dr in my formula.
View attachment 273778
Consider an arc of thickness dr, at radius r from the centre of curvature. What is its resistance?
 
  • #21
Is it good?
1607293829640.png
 
Last edited:
  • #22
I don't see how you would get that. Please explain your reasoning.
 

1. What is the formula for calculating the resistance of a half-cylinder element?

The formula for calculating the resistance of a half-cylinder element is R = (ρ * L) / (A * π * r), where R is the resistance, ρ is the resistivity of the material, L is the length of the element, A is the cross-sectional area, and r is the radius of the half-cylinder.

2. How does the shape of a half-cylinder affect its resistance?

The shape of a half-cylinder affects its resistance because it increases the surface area of the element, which in turn increases the resistance. This is because the electrons have to travel a longer distance along the curved surface of the half-cylinder, resulting in more collisions and a higher resistance.

3. What factors influence the resistance of a half-cylinder element?

The resistance of a half-cylinder element is influenced by several factors, including the length of the element, the material it is made of, the cross-sectional area, and the temperature. Additionally, the shape of the half-cylinder and the presence of any imperfections or irregularities can also affect the resistance.

4. How does the resistivity of the material affect the resistance of a half-cylinder element?

The resistivity of the material plays a significant role in determining the resistance of a half-cylinder element. Materials with higher resistivity will have a higher resistance, while materials with lower resistivity will have a lower resistance. This is because materials with higher resistivity impede the flow of electrons more, resulting in a higher resistance.

5. Can the resistance of a half-cylinder element be changed?

Yes, the resistance of a half-cylinder element can be changed by altering the factors that influence it, such as the length, material, and temperature. For example, increasing the length or using a material with higher resistivity will result in a higher resistance, while decreasing the length or using a material with lower resistivity will result in a lower resistance.

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