Resistance of a sphere and cone

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    Cone Resistance Sphere
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Discussion Overview

The discussion revolves around the resistance of solid geometric shapes, specifically a sphere and a cone, when modeled as resistors. Participants explore the implications of integrating resistance formulas under different conditions, including the effects of contact points and integration limits.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes using the formula R=(rho)L/A for calculating resistance but encounters issues with integration due to singularities at contact points.
  • Another participant questions the handling of the squaring operation in the denominator of the resistance formula.
  • A different participant suggests that the limits of integration must be adjusted to avoid infinite resistance, indicating that integration from 0 to 2r is inappropriate.
  • One participant asserts that the resistance is infinite at points of contact, such as the poles of the sphere or the vertex of the cone, due to zero cross-sectional area at those points.
  • Another participant emphasizes the necessity of having non-zero area at the surfaces to achieve finite resistance, suggesting that evaluating integration from the center to a point less than the radius is essential.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to integrating resistance for these shapes, with no consensus on how to resolve the issues of singularities and infinite resistance. The discussion remains unresolved regarding the appropriate limits and methods for integration.

Contextual Notes

Participants highlight limitations related to the assumptions of contact points and the implications of integrating over certain ranges, which may lead to infinite resistance. There is also uncertainty regarding the correct interpretation of integration limits.

lee_chongeu
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Lets assume we have a resistor material, with a perfect solid spherical shape and no defect, we connect it from south pole to north pole, by using the general formula of R=(rho)L/A where rho is the resistivity and L is the length of the resistor, and A is the cross sectional area. I found that i cannot integrate it, because i get a Log negative. I realize that the problem is at the point of contact between the wire and the spherical resistor's cross sectional area is close to zero.

Those are my question:
i) How to solve such conflict?
ii) How to integrate such resistor if i connect the wire 90 degree instead of pole to pole(180 degrees). What is the L length of the resistor? will it be the circumference of the sphere?
iii) I found the same problem occur when i change the shape to cone shape.

I've enclose my working
 

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What happened to the squaring operation in your denominator?
 
There is nothing wrong wat?
 
U have taken limits of integration as -r to +r. That u have to do if u start integrating from the center in which case R-L will become L. For ur integration, take limits as 0 to 2r. But even then the resistance will be infinite after integration. U can only find the resistance of a sphere or a cone between two limits say a and b not between 0 to 2r. Ur method is correct but the statement of question is wrong.
 
The problem is that of singularities. The resistance is literally infinite. At the top or bottom of the sphere, where r = +/-R, or at the vertex of the cone, there is conductivity at only a point, the tangent point. A point has zero cross sectional area, hence infinite resistance. If the integration is evaluated from zero (center) to less than +R or -R, say +/- 1.9*R, then there is an area which is non-zero at the surface. In order to have finite resistance, the surfaces at each end must have non-zero area. A curved surface with one tangent point conducting will give infinite resistance. Does this help?

Claude
 

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