Find current drawn from the cell

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To find the current drawn from a cylindrical conductor connected to a cell, the resistance must be calculated based on the geometry and resistivity of the material. The discussion highlights the importance of understanding that the current flows radially through the cylindrical surface rather than axially. There is confusion regarding the integration approach used to derive the current, as it may not be necessary if the connections are correctly understood. The participants emphasize the need to accurately define the length and area in relation to resistivity when setting up the integral. Overall, clarity on the conductor's configuration is crucial for solving the problem correctly.
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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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