Resistance of a sphere and cone

In summary: E. Shannon, an American mathematician and engineer, developed the mathematical theory of information in 1948. In his paper, "A Mathematical Theory of Communication," Shannon proposed that communication is a process of converting information from one entity to another. He defined communication as "the process of transferring symbols from one entity to another, where the two entities are not the same." In summary, Claude Shannon proposed that communication is a process of converting information from one entity to another. He defined communication as "the process of transferring symbols from one entity to another, where the two entities are not the same."
  • #1
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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  • #2
What happened to the squaring operation in your denominator?
 
  • #3
There is nothing wrong wat?
 
  • #4
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.
 
  • #5
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
 

1. What is the resistance of a sphere and cone?

The resistance of a sphere and cone refers to the force required to move these shapes through a fluid, such as air or water.

2. How is the resistance of a sphere and cone calculated?

The resistance of a sphere and cone can be calculated using the drag equation, which takes into account the shape, size, and velocity of the object as well as the properties of the fluid it is moving through.

3. How does the resistance of a sphere compare to that of a cone?

The resistance of a sphere is typically higher than that of a cone due to its larger cross-sectional area and lack of streamlined shape.

4. What factors affect the resistance of a sphere and cone?

The resistance of a sphere and cone can be affected by factors such as the shape, size, velocity, and density of the object, as well as the viscosity and density of the fluid it is moving through.

5. How can the resistance of a sphere and cone be reduced?

The resistance of a sphere and cone can be reduced by altering their shape to be more streamlined, increasing their velocity, or changing the properties of the fluid they are moving through, such as using a thinner or less dense fluid.

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