resistant

315, try comprehending my last post.   

http://web2.0calc.com/questions/resistant#r15

 I put a lot of time into displaying it and Alan has checked that it is correct. Try studying it. :)

NOTE: R will never be proportional to L

I wanted to say more but I could not make my wording exact enough, what i have said is correct though.

i brought a tissue ;-;

good! now use it! :)lol

This is how I interpret it. (Elucidating comments from Alan are always appreciated)

This is called radial current flow. It is calculated via an infinite number of “shells” radiating out from the point of origin. As the current flows out from the center, each shell has a progressively greater surface area and a lower corresponding resistance. The shells’ surface areas, and its corresponding resistances, are summed to calculate the total resistance. However, because the area, and the rate of area, increases as the current moves outward, the cross-sectional area formula is integrated to effectively make the area uniform over predefined distances.

For DC circuits this phenomena seems trivial. For AC circuits it becomes important because the current flow moves closer to the surface of a conductor as the frequency increases. Where it really becomes interesting is in the conduction of signals in biology, specifically nerve cells that are mediated by sodium, potassium, and calcium channels. Understanding this is a brain strain (and a nerve pinch), and how anyone ever figured it out in the first place is mind boggling. 

_7UP_

I have no understanding of the initial  part but

Given that P and L are constants then

$$\\ \displaystyle\int_{a}^{b}\frac{P}{2\pi rL}\;dr\\\\
=\frac{P}{2\pi L}\displaystyle\int_{a}^{b}\frac{1}{r}\;dr\\\\
=\frac{P}{2\pi L}\;[ln(r)]_a^b\\\\
=\frac{P}{2\pi L}\;[ln(b)-ln(a)]\\\\
=\frac{P}{2\pi L}\;\left[ln\left(\frac{b}{a}\right)\right]\\\\$$

Excellent explanation SevenUp!  All I would add is that ρ is the electrical resistivity (it has SI units of ohm-metres).

And Melody's integration is, of course, correct (though the usual symbol for resistivity is the Greek letter ρ, rather than an upper case P!)

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Thanks Alan, do you know how to get ρ to display in LaTex?   I do not know what it is called :/

I suppose I could just google greek letters. :/      $$\rho$$      got it!    

http://www.ibiblio.org/koine/greek/lessons/alphabet.html

Thanks Alan and 7up,

I am trying to work out the first bit of this in terms of the mathematics.

I understand what 7up said.

I think I can see that a shell  has the volume of    dA= 2pi L dR       Is that right?

I can see that    $$R=\int dR$$

From what 7up said it seems logical that Resistance is inversely proportional to volume as the radius increases.

and i am guessing that $$\rho$$    is a contant for the specific material that the cylinder is made from.

So    

 $$\\R(of a shell)=dR=\frac{\rho}{dA}=\frac{\rho}{2\pi r L}\;dr\\\\
R=\int dR=\int_a^b\frac{\rho}{A}\;dr=\int_a^b\;\frac{\rho}{2\pi r L}\;dr\\\\$$

Is what I have written correct?  I am confused :/  

"I think I can see that a shell  has the volume of    dA= 2pi L dR       Is that right?"

Be careful not to confuse your r (a radial distance) with your R (electrical resistance).

The surface area of the shell is  2pi*r* L (the volume would be 2pi*r*L*dr) - see below.

ρ is an intrinsic property of the material (unlike resistance, which depends of the size of the material).

The relationship between ρ and R is  ρ = RA÷l where A is cross-sectional area and l is length in the direction of electrical current flow. In this problem l is not L, it is r.  Or, rather, for the shell it is dr.  The relevant cross sectional area of the shell (bearing in mind that current is flowing radially) is 2pi*r*L (this is stated incorrectly in the problem, though the correct value is used in the integral - perhaps this was the cause of xvxvxv's confusion), so the small increase in resistance across the shell is dR = ρl/A

or dR = ρdr/(2pi*r*L).

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Thanks Alan, 

I will take a little time to absorb this.

this      dA= 2pi L dR     was a typo,  I did mean dr       

the rest i will need to think about.    

Now to make this clear to me 

What is " cross sectional area " 

Is it a circle which its area is $${\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{r}}}^{{\mathtt{2}}}$$ 

No.  The cross-sectional area is the area perpendicular to the direction of current flow, through which the current is flowing.  If the current were flowing axially through a solid cylinder of radius r, then the cross-sectional area would, indeed, be pi*r², and it would be constant all the way along the cylinder.  However, here the current is flowing radially outward , so the cross-sectional area gets bigger as you go further out from the centre.

. 

So here the cross sectional area is the surface area of the cylinder ?  Which is 

2$${\mathtt{\pi}}$$rL

Try and find the odd one out.

Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!  Alan!

Mwahahaha there isn't one.

I am very pleased with what I have nutted out (and explained) here, but Alan could you please examine it with a fine toothed combe to check that everything I have said is correct.    

-------------------------

This question is, in a way, related to finding the volume of a hollow tube. 

The normal way to do this is to find the area of the big circle and subtract the area of the little circle. and then times it by the length of the pipe (L)    

 $$\\V=(\pi b^2-\pi a^2)*h\\
=\pi*(b^2-a^2)*h$$

BUT if the cylinder is made up of an infinite nmber of infintesimally narrow shells there is another way to think about it.

The circumference of a shell is   $$2\pi r$$ and the length of the cylinder is L  if you flattern this  out you will have a rectangle,  the area of the rectangle is  $$2\pi rL$$   

The thinkness of the flatterened of this rectangle is $$dr$$ $$$$(dr is the difference in the radii of the outer circle to the inner circle)

the integral sign    $$\int$$     is literally an S it stands to the sum of all the infintesimally small parts.  (I would like to word this bit better   )

So, if you want the volume of the pipe it will be        $$V=L*\int_{inner\; radius}^{outer \;radius} \;2\pi r \;dr$$

NOTE that

  $$\\\int_{inner \;radius}^{outer \;radius}\;(2\pi r) dr \\\\
= [\pi r^2]_{inner \;radius}^{outer \;radius}\\\\
=[\pi (outer \;radius)^2]-[\pi (inner \;radius)^2]\\\\
= $Area of the annulus$$$

-------------------------------------------------------------------------

NOW  - this question is not really about volume - it is about the surface are of an infinite number of infintesimally thin shells  (together they make up the volume but think of them as shells)

The resistance is inversely proportional to the surface area that it connects with.

The surface area for any given radius is   $$2\pi r *L$$   

$$\\R\;\alpha \;\dfrac{1}{2\pi r*L}\\\\
$BUT r is all values from a to b \qquadSO$ \\\\
R\;\alpha \;\displaystyle\int_a^b\dfrac{1}{2\pi r*L}\;dr\\\\
$Adding in the constant which intrinsically belongs to the material we get$\\\\
R= \;\displaystyle\int_a^b\dfrac{\rho}{2\pi r*L}\;dr\\\\$$

NOW is that COMPLETELY CORRECT ?      I had to think really hard to put that together.     

(I hope I didn't include any more typos and I really hope I know what I am talking about)      

You are mean anon - I was trying to find the odd one out !!

You are correct Melody and xvxvxv.

.

Thanks you Alan.

"The cross-sectional area is the area perpendicular to the direction of current flow, through which the current is flowing.  If the current were flowing axially through a solid cylinder of radius r, then the cross-sectional area would, indeed, be pi*r², and it would be constant all the way along the cylinder.  However, here the current is flowing radially outward , so the cross-sectional area gets bigger as you go further out from the centre."

I do not understand the difference between these 2 highlighted terms.  

(It is probably because I have absolutely no understanding of electricity at all)

Axially through simply refers to a situation in which the current is flowing in the direction of the long red arrow (see diagram below).  Radially outward refers to a situation in which the current is flowing in the direction of the short blue arrow.

The direction current flows will be determined by the potential difference in a particular case.  If there is a potential difference between the ends then current will flow axially; if there is a potential difference between an inner and an outer curved surface (as in the original problem) then current will flow radially.

.

Now what is wrong here? 

Thanks Alan :)

@ xvxvxv

 R is not proportional to the length L of the cylinder in this problem.

Also, if you are going to differentiate R with respect to A in your expression you are best to change the A from the denominator to A^-1 in the numerator.

.

Here R  is not proportional to L because I flow radially  

Right? 

Yes, R is not proportional to L, which measures the length of the cylinder, because the current flows radially.

.

Now its very clear I understood this 

Thank you all.