# Resistance of ring geometry

1. Dec 7, 2009

### mmiguel1

I just had this problem on my Electromagnetics final. I want to know if I got this right and I can't find the problem with google.
1. The problem statement, all variables and given/known data
Consider a conducting quarter-ring.
It can be envisioned as one piece of a hollow cylinder that has been cut into fourths down its length.
In cylindrical coordinates the geometry of this quarter ring is: $$r \in [a,b]$$; $$\phi \in [0,\pi/2]$$; and $$z \in [0,h]$$;
The material of interest has conductivity $$\sigma$$
What is the resistance from the $$\phi=0$$ end to the $$\phi=\pi/2$$ end?

2. Relevant equations
For a conductor with linear geometry down it's length:
$$R = \frac{L}{\sigma A}$$
Where L = length and A = cross-sectional area;
$$G = \frac{1}{R}$$

3. The attempt at a solution
I considered a slice of the conductor corresponding to a differential angle $$d \phi$$.
The slice is described by:
$$r \in [a,b]$$; $$\phi \in [\phi_0,\phi_0+d\phi]$$; and $$z \in [0,h]$$;
Let the resistance of this slice be dR. Since the total conductor is a series combination of these slices, and since series resistances add,
$$R = \int dR$$

Within the slice of interest, dR, consider a sub-slice located at radius r with a < r < b.
The sub-slice is described by:
$$r \in [r_0,r_0+dr]$$; $$\phi \in [\phi_0,\phi_0+d\phi]$$; and $$z \in [0,h]$$;
Let us call the resistance of this sub-slice dR'. Our original slice, dR, is a parallel combination of many dR' sub-slices. It is therefore more convenient to calculate the conductances for this part of the problem. Define $$dG = \frac{1}{dR}$$ and $$dG' = \frac{1}{dR'}$$. Since parallel conductances add, we have
$$dG = \int dG'$$
As the differential values $$d\phi$$ and $$dr$$ tend to zero, the subslice of interest is linearized along its length.
We can therefore say
$$dG' = \frac{\sigma A}{L} = \frac{\sigma h dr}{r d \phi}$$
Where $$A = h dr$$ and $$L = r d \phi$$
$$dG = \int_a^b \frac{\sigma h}{d \phi} \frac{1}{r} dr$$
$$dG = \frac{\sigma h}{d \phi} \ln \left (\frac{b}{a} \right )$$
$$dR = \frac{d \phi}{\sigma h \ln \left ( \frac{b}{a} \right ) }$$
$$R = \int_0^{\frac{\pi}{2}} \frac{d \phi}{\sigma h \ln \left ( \frac{b}{a} \right ) }$$
$$R = \frac{\pi}{2 \sigma h \ln \left ( \frac{b}{a} \right ) }$$

I was debating this question with my friends after the final. Each of us got different answers. I think my method is correct but I would like some verification.
Thanks!

2. Dec 7, 2009

### Redbelly98

Staff Emeritus
Your derivation looks fine, and your result is certainly true for the case a≈b as well as a=0.

Here is something to check with your friends: If a=0, then the two ends will be in contact, giving a resistance of 0. Your formula correctly reproduces that, since you are dividing by the ln term which approaches infinity as a→0.