Find current drawn from the cell

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Homework Help Overview

The problem involves a cylindrical conductor characterized by its length and inner and outer radii, with a specific resistance. A cell with an electromotive force is connected across the lateral faces of the conductor, and the original poster attempts to find the current drawn from the cell.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the method of integration proposed by the original poster and question its necessity, suggesting that the cross-sectional area could be determined directly. There is also a focus on the nature of the connections to the conductor, with some participants clarifying that the current may be radial rather than axial.

Discussion Status

The discussion is ongoing, with participants providing insights and questioning the assumptions made by the original poster. Some guidance has been offered regarding the setup of the integral and the interpretation of the problem, but there is no explicit consensus on the approach to take.

Contextual Notes

There is a lack of clarity regarding the application of connections to the cylindrical surfaces versus the flat surfaces, which may affect the interpretation of the problem. Additionally, the original poster mentions obtaining a "different answer," but it is unclear what this refers to.

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Homework Statement


A cylindrical conductor of length l and inner radius R1 and outer radius R2 has specific resistance ρ A cell of emf e is connected across the two lateral faces of the conductor. Find the current drawn from the cell.


The Attempt at a Solution


Consider a thin shell of radius dx from the centre of the cylinder.

dR = \dfrac{\rho l}{2\pi x dx} \\<br /> <br /> di=\frac{e}{dR}

Now, if I integrate the above equation I get a different answer.
 

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Why are you integrating? You can find the cross-sectional area of the cylinder directly.
 
It's not very clear but I think the connections are applied to the cylindrical surfaces and not to the flat surfaces at the end. So the current will be radial rather than axial. Usually the "lateral area" of a cylinder is used to designate the area of the curved surface not of the flat bases.

If this is NOT the case, then it is a trivial problem and you don't need to integrate.

And what do you mean by a "different answer"? Different in respect to what?
 
nasu said:
It's not very clear but I think the connections are applied to the cylindrical surfaces and not to the flat surfaces at the end. So the current will be radial rather than axial. Usually the "lateral area" of a cylinder is used to designate the area of the curved surface not of the flat bases.
I do believe you are correct.

In that case, when setting up the integral, the OP should be careful about what is the length and what is the area (with respect to the resistivity) when setting up the integral. The equation he used is not correct.
 

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