Why is resistance of similar cases calculated differently?

[Shivang kohlii]

Homework Statement

1.In the WE-11 resistance is calculated by integrating and that too by taking length as dr and area as the CSA of the small cylinder ..
Shouldn't length be l and area by 2πrdr..?
I also don't understand why can't we simply use
Dl/A formula
, Where A =π((2R)^2 - R^2)

2) in second question the initial resistance is calculated by the method I think Is correct .. so then why are two different methods being used for same situation??

Homework Equations

R= DL/A
D= specific resistance
L= length of resistor
A= area

 

[mjc123]

In the first case they ask for the resistance between the inner and outer surfaces, i.e. in the radial direction. In the second case they ask for the resistance in the length direction. (Well actually they don't specify that, but that is what they calculate.)

First rule of exams - always read the question carefully, and answer what they ask!

 

[Shivang kohlii]

In the first case they ask for the resistance between the inner and outer surfaces, i.e. in the radial direction. In the second case they ask for the resistance in the length direction. (Well actually they don't specify that, but that is what they calculate.)

First rule of exams - always read the question carefully, and answer what they ask!

But I still don't get why integration is required .. it's not like specific resistivity is changing... And shouldn't length by taken as 2R-R = R only then??

 

[mjc123]

Because the area changes with radius - surface of the cylinder is 2πrL. So you have to model it as a series of cylindrical shells of area 2πrL and thickness dr, and integrate from R to 2R.

 

[RPinPA]

I also don't understand why can't we simply use
Dl/A formula

Because of the direction the current is flowing. That's what you would use if the current was flowing vertically, from bottom to top or top to bottom. In that case the cross section is a constant over the entire path, and it's equal to $\pi [(2R)^2 - R^2]$ as you say. And the appropriate length of the current path is the height $l$ of the cylinder.

But they tell you that the current is flowing from the inner face to the outer face. And the cross section area gets bigger and bigger as the current proceeds. It's not constant. So first it flows through a little cylindrical slice on the inside. The area of that slice is $2\pi R l$. The length of the current flow through that slice is $dr$.

Then it flows through a slightly bigger cylinder, of area $2\pi r l$ and length $dr$, with $r = R + dr$. And then the next little slice outward, and the next, until you get the outer ring of radius 2R. These slices are in series, and the total resistance is the sum of the resistance as the current flows through all of them. So that's an integration.

You could find the resistance for current flowing vertically by integration as well. Your slices would be horizontal. Each slice would have cross section area $A = \pi [(2R)^2 - R^2]$, a constant. Each slice would have length $dh$ for $h$ going from 0 to $l$. And when you integrate $D dh/A$ from $h = 0$ to $l$ and $D/A$ constant, you get the usual expression $D l/A$. So integration isn't wrong, it's just unnecessary.