# Find current drawn from the cell

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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.

$dR = \dfrac{\rho l}{2\pi x dx} \\ di=\frac{e}{dR}$

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

## The Attempt at a Solution

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tms
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?

tms
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.