Resistance of a half ring of a given resistivity

[Tronx]

Lets suppose a half ring whit internal radius a and external radius b and height t. We can calculate the resistance of the half ring considering that it is made of a number of parallel resistors each one with resistance given by: R = ρ L/S where L = π·r and S = t·dr. The resistance of each differential resistor is: R = ρ (πr) / (t·dr).

Then as the resistors are connected in parallel, we have:
1/R = t / (ρπ) integrate (dr/r) from a to b = [t ln(b/a)]/ (ρπ)
Finally, the total resistance of the half ring is:
R = ρπ/[t·ln(b/a)]

It would be possible to arrive at the same result by considering there are a number of serial resistors each one with surface t·(b-a). The problem is that the length L of each these resistors is not constant and it is a·dθ at the inner part and it is b·dθ at the outer part.

In this case, how could we perform the integral to arrive at the same result when the calculation is made by considering a parallel connection?​

 

[anuttarasammyak]

We can assume
$$R=\rho \frac{l}{A}$$
where
$$A=t(b-a)$$
and l is a certain value between
$$a\pi <l < b\pi$$

Disintegrating the ring to infinitesimal thin rings of radius r within which same amount of current flows
$$l =\frac{\int_a^b \pi r dr}{b-a} = \frac{\pi}{2}(b+a)$$
I get average of a and b multiplied by $\pi$ as l. Do you think it is reasonable ?

 

[DaveE]

"Assume the current is uniformly distributed on a cross section of the half ring"...
What? Why? I don't understand. So, do they mean in the limit as a→b?

Sorry It's late, maybe I'll reply tomorrow.
Also, I recall this problem being beat to death somewhere else on PF.

 

[Tronx]

We can assume
$$R=\rho \frac{l}{A}$$
where
$$A=t(b-a)$$
and l is a certain value between
$$a\pi <l < b\pi$$

Disintegrating the ring to infinitesimal thin rings of radius r within which same amount of current flows
$$l =\frac{\int_a^b \pi r dr}{b-a} = \frac{\pi}{2}(b+a)$$
I get average of a and b multiplied by $\pi$ as l. Do you think it is reasonable ?

Thank you for your response.

I don't know if it is reasonable or not, but I'm sure that the result to this problem is :
R = ρπ/[t·ln(b/a)]

And the result obtained considering one ensemble of parallel resistors MUST BE the same when considering one ensemble of serial resistors. That's the question.

 

[Tronx]

"Assume the current is uniformly distributed on a cross section of the half ring"...
What? Why? I don't understand. So, do they mean in the limit as a→b?

Sorry It's late, maybe I'll reply tomorrow.
Also, I recall this problem being beat to death somewhere else on PF.

Hello Dave:

These are the links in PF related with this problem:
https://www.physicsforums.com/threa...ctor-with-a-rectangular-cross-section.934323/
https://www.physicsforums.com/threa...n-element-shaped-like-a-half-cylinder.996616/

In none of then was given a good result.

It is possible to find this problem in Tipler (4 edition) problem 26-34 (for example). I have the solutions manual too and, as I said, the result is:
R = ρπ/[t·ln(b/a)]

 

[anuttarasammyak]

Thank you for your response.

I don't know if it is reasonable or not, but I'm sure that the result to this problem is :
R = ρπ/[t·ln(b/a)]

And the result obtained considering one ensemble of parallel resistors MUST BE the same when considering one ensemble of serial resistors. That's the question.

Formula for ensemble of parallel resistance is
$$\frac{1}{R}=\sum_i\frac{1}{R_i}$$
Applying it to the case
$$\frac{1}{R}=\int_a^b \frac{t}{\rho r\pi}dr=\frac{t}{\rho \pi}\ln \frac{b}{a}$$
I prefer it but I am afraid it would not go with the problem statement "Assume the current is uniformly distributed ... ", because inner part of the ring cross section should have larger current density.

 

[berkeman]

Yeah, IMO that's a pretty non-physical assumption. Not the OP's fault, and it does simplify the problem significantly.

 

[DaveE]

Yeah, IMO that's a pretty non-physical assumption. Not the OP's fault, and it does simplify the problem significantly.

I think it's just described really poorly. What they mean is that when you integrate over r, each half cylinder of height t and depth dr has constant current density flowing circumferentially. Yet each dr slice will have a different current than the other slices. i.e. current density doesn't vary over the t dimension. As @anuttarasammyak did above.

 

[Tronx]

I think it's just described really poorly. What they mean is that when you integrate over r, each half cylinder of height t and depth dr has constant current density flowing circumferentially. Yet each dr slice will have a different current than the other slices. i.e. current density doesn't vary over the t dimension. As @anuttarasammyak did above.

I think when the current is not constant over a differential resistor it is not possible to apply that R = ρ L/S for each resistor.
I mean, when we consider the half ring is made like an ensemble of parallel strips of length πr then the current into one strip is constant (although different to other adjacent strip). But when we consider the half ring is made like an ensemble of serial slices with surface t·(b-a) then the current is not constant into this slice (is bigger when the radius is shorter). Then we can`t consider the half ring as a ensemble of serial resistors.

It is the same situation with this other problem from Serway 6E. The result is obtained considering the wedge as a superposition of serial resistors slices and doing an integration over L dimension (in each slice the current is constant):

But it is not possible to arrive at the same result considering the wedge as a collection of parallel resistors strips in w direction because the current is not constant inside one strip. The initial height of one strip is Y1 and the final height is Y2 and then the current is not constant on each strip.

 

[SammyS]

I think when the current is not constant over a differential resistor it is not possible to apply that R = ρ L/S for each resistor.
I mean, when we consider the half ring is made like an ensemble of parallel strips of length πr then the current into one strip is constant (although different to other adjacent strip). But when we consider the half ring is made like an ensemble of serial slices with surface t·(b-a) then the current is not constant into this slice (is bigger when the radius is shorter). Then we can`t consider the half ring as a ensemble of serial resistors.

View attachment 299275

It is the same situation with this other problem from Serway 6E. The result is obtained considering the wedge as a superposition of serial resistors slices and doing an integration over L dimension (in each slice the current is constant):

But it is not possible to arrive at the same result considering the wedge as a collection of parallel resistors strips in w direction because the current is not constant inside one strip. The initial height of one strip is Y1 and the final height is Y2 and then the current is not constant on each strip.

That really is not the same assumption. The Serway problem states that the current is the same for every slice. Furthermore, the "resistors" are in series, not in parallel.

It does appear that Serway has a typo in that integrand.

The integrand should be $\displaystyle \ \frac 1 {\ y_1 ~ w + \left(\frac{y_2-y_1}{L}x \right) w \ }$ .

 

[Tronx]

That really is not the same assumption. The Serway problem states that the current is the same for every slice. Furthermore, the "resistors" are in series, not in parallel.

It does appear that Serway has a typo in that integrand.

The integrand should be $\displaystyle \ \frac 1 {\ y_1 ~ w + \left(\frac{y_2-y_1}{L}x \right) w \ }$ .

Ha,ha.. the typo in the integrand is mine, not from Serway.
And of course, the resistors are in serial (not in parallel). That's what I said...

Instead of this, if we try to do the integration considering parallel strips from face A to face B, it is not possible to do the integration because the current is not constant over one of each strip.

I think this is the reason but I'm nor sure. And don't know if there is another kind of assumption we could make to arrive at the same results but considering parallel strips.

In the original problem of the half ring, it happens the contrary, the integration is made considering parallel strips and we cannot make the integration if we choose serial slices.