Resistance of Cylindrical Conductor

[aaas]

Homework Statement

A circular disk of radius r and thickness d is made of material with resistivity p . Show that the resistance between points a and b (Fig. attached)is independent of r and is given by R=πp /2d

Homework Equations

R= p L / A
where L and A are length and section area respectively .

The Attempt at a Solution

I tried to use differential equations , but I always end up with ∞ or some indeterminate forms .

 

[m.medhat]

i think there is a mistake in this question . the resistance dependent of r .

 

[aaas]

i think there is a mistake in this question . the resistance dependent of r .

Maybe !
But it is taken from serway's book .
The trick here is that the current direction is perpendicular on the cylinder axis !

 

[Phrak]

I tried to use differential equations , but I always end up with ∞ or some indeterminate forms .

I don't know what to tell you. The resistance of any point origin of current working into a volume is infinite.

Perhaps you've misinterpreted the problem and the current originates from a line segment along the thickness of the disk.

 

[gabbagabbahey]

The resistance of any point origin of current working into a volume is infinite.

I have no idea what you mean here. The resistance between two points in a solid is usually well defined.

 

[gabbagabbahey]

Homework Statement

A circular disk of radius r and thickness d is made of material with resistivity p . Show that the resistance between points a and b (Fig. attached)is independent of r and is given by R=πp /2d

Homework Equations

R= p L / A
where L and A are length and section area respectively .

The Attempt at a Solution

I tried to use differential equations , but I always end up with ∞ or some indeterminate forms .

There's no need for any DEs here. Draw a straight line connecting your two points...its length is what you use as "L" in your relevant equation. The cross-section perpendicular to this line segment changes as move along the line, so you will need to divide the line into infinitesimal pieces, express the cross-section for each piece and integrate $dR=\rho \frac{dL}{A}$ over the entire line. I suggest you use cylindrical polar coordinates for the integration.

 

[Phrak]

I have no idea what you mean here. The resistance between two points in a solid is usually well defined.

It isn't.

A hollow ball of volumentric resistivity rho has source and sink on its inner and outer surfaces.
inner radius = r₁ and outer radius = r₂.

$$R = \frac{\rho}{4\pi}\left( \frac{1}{r_1} - \frac{1}{r_2} \right)$$

Clearly, any other solid angle of current flux will encounter greater resistance.

As $$r_1 \rightarrow 0$$ resistance tends to infinity.

All resistance through a volume from ideal point sources or sinks is undefined.

 

[gabbagabbahey]

It isn't.

A hollow ball of volumentric resistivity rho has source and sink on its inner and outer surfaces.
inner radius = r₁ and outer radius = r₂.

$$R = \frac{\rho}{4\pi}\left( \frac{1}{r_1} - \frac{1}{r_2} \right)$$

Clearly, any other solid angle of current flux will encounter greater resistance.

As $$r_1 \rightarrow 0$$ resistance tends to infinity.

Taking the limit as $r_1\to\infty$ makes no sense whatsoever...there would be no source. Instead, consider a solid sphere with a diametrically opposed source and sink of finite solid angle and take the limit as the solid angle approaches zero.

All resistance through a volume from ideal point sources or sinks is undefined.

I don't understand why you think this...if you have a very thin superconducting wire connected to an ohmmeter and across a solid resistor/conductor, would you really expect to measure infinite resistance?

 

[Phrak]

Taking the limit as $r_1\to\infty$ makes no sense whatsoever...there would be no source.

That's $r_1\to 0$, of course. This is exactly as the problem is presented and therefore it makes sense to say the resistance is undefined.

Instead, consider a solid sphere with a diametrically opposed source and sink of finite solid angle and take the limit as the solid angle approaches zero.

I don't understand why you think this...if you have a very thin superconducting wire connected to an ohmmeter and across a solid resistor/conductor, would you really expect to measure infinite resistance?

In the real world ohmmeter probes are not mathematical points. But the contact area has everything to do with measured resistance of bulk measurements--unless you are talking to an EE, who are generally clueless.

 

[gabbagabbahey]

That's $r_1\to 0$, of course. This is exactly as the problem is presented and therefore it makes sense to say the resistance is undefined.

My apologies, I was thinking of voltage for some reason.

Of course, the resistance between points is undefined since the total current through a point for any given current density will be zero.

Anyways, @aaas I'm sure you are meant to calculate the resistance between the two diametrically opposed line segments (of length "d"), and for that just use the method I described in my second post.

 

[Phrak]

My apologies, I was thinking of voltage for some reason.

Of course, the resistance between points is undefined since the total current through a point for any given current density will be zero.

Anyways, @aaas I'm sure you are meant to calculate the resistance between the two diametrically opposed line segments (of length "d"), and for that just use the method I described in my second post.

And the answer will be rubbish. This sort of homework problem really bothers me. Some typical students will come up with the acceptable wrong answer, as given, through gross oversimplification. The smarter students will be well aware that the current will not flow in parallel lines and fail to obtain a result requiring a numerical solution to a multivariable, nonanalytic differential equation.

 

[aaas]

Draw a straight line connecting your two points...its length is what you use as "L" in your relevant equation. The cross-section perpendicular to this line segment changes as move along the line, so you will need to divide the line into infinitesimal pieces, express the cross-section for each piece and integrate over the entire line. I suggest you use cylindrical polar coordinates for the integration.

Actually , this what I exactly did . but as the variable must be in the denominator the resulting integral will include In 0 which is not defined (approach to minus infinity )

 

[gabbagabbahey]

Actually , this what I exactly did . but as the variable must be in the denominator the resulting integral will include In 0 which is not defined (approach to minus infinity )

YOu shouldn't be getting any natural logarithms...why don't you show your calculations?

 

[aaas]

I divide the line into infinitesimal pieces
$dR=\rho \frac{dL}{A}$
$dR=\rho \frac{dx}{2\pi (R-X) d}$
$R = 2 \int \rho \frac{dx}{2\pi (R-x) d} , integrate for x from 0 to R$

 

[gabbagabbahey]

Your crossectional area is incorrect. Draw a picture.

 

[aaas]

Your crossectional area is incorrect. Draw a picture.

Isn't the cross sectional area is the surface area of dashed line circle ?

 

[gabbagabbahey]

No, the cross-section should be orthogonal to "dx", and will look like a rectangle with height "d"...what is its width at any given value of x?

 

[aaas]

The length of a chord if you know the radius and the perpendicular distance from the chord to the circle center. This is a simple application of Pythagoras' Theorem.

$$w= 2 \sqrt{(r-x)^2 - r^2} \\<br /> [\tex]<br /> where<br /> r is the radius of the circle<br /> r-x is the perpendicular distance from the chord to the circle center$$

 

[gabbagabbahey]

You have a sign error, it should be [tex]w=2\sqrt{r^2-(r-x)^2}[/itex]...anyways, that looks good. Now, in order to make the integration easiest, try the substitution $x=r(1-\cos\theta)$ (which is exactly what you would have had if you'd set the problem up in cylindrical polars)<br /> <br /> P.S. you need to use a "/" instead of "\" in your closing tex tag.[/tex]

 

[aaas]

thanks very much :!)
I got the same answer now which means that my problem was in considering the cross sectional area as a circle .
But I am a little confused , If you look at this example (b) ( hollow cylinder ) we considered the cross sectional area as surface area of cylinder . what is the difference ?
The current flow in both problems faces the same cross sectional area !
so why they don't be the same in our calculations ?
Is that because the voltage is applied only between points A and B in previous problem ?

 

[gabbagabbahey]

The difference is that in example (b), the potential difference is applied to the ends of the cylinder, so the current flows down its length and the cross-section that is perpendicular to it is the circle/lamina.

In this problem however, the current flows from one side of the cylinder to the other (not lengthwise), and the cross-section perpendicular to it is rectangular.

 

[aaas]

Thank you very much , it is very clear now

 

[chesswt]

I really want to know what is the exactly procedure, someone can explain me, how I can have the same answer πp/2d