# Resistance of a Spherical Wire

• albert1993

#### albert1993

So I am doing elementary physics, as in high school elementary, and learned that the equation for the resistance of a wire can be put in the form: $$\rho\frac{\ell}{S}$$

so what about a wire that is in the form of a sphere? (let's say something like that can exist :), or you can prove to me that it can not ;) )

I don't know about the math part, but a spherical mass of metal is not a wire.

Think of the sphere as an infinite number of discs connected in series. Applying
$$\rho\frac{\ell}{S}$$
to these discs, and using integration to sum the results, gives you an equivalent resistance of $$\frac{2\rho}{R\pi}$$. (Can somebody check whether this is right? I had two infinities that seemed to cancel out; I don't have enough knowledge about calculus to determine if they actually do.)

I did the this on a piece of scrap paper, but I was unable to get my infinities (ln(0)) to cancel out (I thought I might've screwed up, but I looked up the integral and it likewise doesn't cancel out).

The boundary conditions (what's the area of the differential slice right at the surface of the sphere?) are what mess things up.

However, for half a sphere with a slice of depth A chopped off the curved surface forming two parallel flats, I got a resistance (between these two flats) of:
$$\frac{\rho}{2\pi R}ln\left(\frac{2R-A}{2R+A}\right)$$

For the same hemisphere, with a slab thickness of B (such that B+A = R, the radius of the sphere), the resistance is:
$$\frac{\rho}{2\pi R}ln\left(\frac{R-B}{R+B}\right)$$

So if you had a sphere with two parallel sides chopped off, and you measured the resistance between these parallel faces, you could add two of my above half-spheres together.

Last edited:
Your spherical answer depends on the area of the contact (boundary) points as MAT implies.
In a uniform wire, it's assumed we connect across the entire unfiorm area at each end...in a sphere its impossible to do that: make a POINT contact on opposite sides of the sphere and virtualy the entire resistance will reside there...the resistance of the spherical volume is negligible in comparison.

I did the this on a piece of scrap paper, but I was unable to get my infinities (ln(0)) to cancel out (I thought I might've screwed up, but I looked up the integral and it likewise doesn't cancel out).

The boundary conditions (what's the area of the differential slice right at the surface of the sphere?) are what mess things up.

However, for half a sphere with a slice of depth A chopped off the curved surface forming two parallel flats, I got a resistance (between these two flats) of:
$$\frac{\rho}{2\pi R}ln\left(\frac{2R-A}{2R+A}\right)$$

For the same hemisphere, with a slab thickness of B (such that B+A = R, the radius of the sphere), the resistance is:
$$\frac{\rho}{2\pi R}ln\left(\frac{R-B}{R+B}\right)$$

So if you had a sphere with two parallel sides chopped off, and you measured the resistance between these parallel faces, you could add two of my above half-spheres together.

Wow, I used the wrong definite integral for csc x. I don't think that $$\frac{\rho}{2\pi R}ln\left(\frac{2R-A}{2R+A}\right)$$ can be right either, though, because that equation suggests R should approach 0 as A approaches 0; it should actually approach infinity.

I postulate the sphere would have less resistance than a wire of the same mass.
The electricity would have less distance to travel in a sphere from point A to point B.

Take a 16 AWG wire 122.3 feet long is 1 ohm.
Take the same mass and make a sphere.
The electricity is no longer traveling 122.3 feet, only centimeters.
So I think will be less than 1 ohm.

Wire Resistances Copper

Wow, I used the wrong definite integral for csc x. I don't think that $$\frac{\rho}{2\pi R}ln\left(\frac{2R-A}{2R+A}\right)$$ can be right either, though, because that equation suggests R should approach 0 as A approaches 0; it should actually approach infinity.

Worse than that, look what happens when you plug in values for A that are fractional values of R (e.g. 0.5R, 0.25R, etc.) Everything turns out negative, although the absolute magnitude increases as the depth of the slab increases--which at least is a good sign.

For completeness, I'll give my corrected full derivation here.

A circle, centered at the origin, has the following equation:
$$x^{2}+y^{2}=R^{2}$$

Solving for y for just the half that is above the origin (the height of our slice at any given point), we have:
$$y=+\left(R^{2}-x^{2}\right)^{1/2}$$

For a circular of radius y, the area is:
$$\pi y^{2}=\pi \left(R^{2}-x^{2}\right)$$

A circular slab, at any point along x, with thickness dx has a volume of:
$$\pi \left(R^{2}-x^{2}\right) dx$$

Assuming we have homogeneous resistivity $\rho$, the differential resistance (which we'll call dQ because we're already using R) of any of these slabs at any point such that $-R<x<R$ is:
$$dQ=\frac{\rho \ell}{A}=\frac{\rho}{\pi \left(R^{2}-x^{2}\right)}dx$$

Instead of finding the resistance of the whole sphere, we'll just take the resistance of half the sphere (to the point x=0) and we'll slice off a slab of thickness A (obviously, $\left|A\right|<R$) from the rounded face. Consequently, we integrate along the interval $-R+A<x<0$.

Equivalently, if we sliced off a slab of thickness A, we'd be left with a slab of thickness B (again, $\left|B\right|<R$). In that case, we'd integrate in the interval between $-B<x<0$.

The Wikipedia gives the following integral (for $\left|x\right| < \left|a\right|$):
$$\int\frac{dx}{x^{2}-a^{2}}=\frac{1}{2a} ln \frac{a-x}{a+x}$$
http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions

So, we find the resistance Q by integrating the differential resistance dQ:
$$Q=\int dQ=-\frac{\rho}{\pi} \int \frac{1}{\left(x^{2}-R^{2}\right)}dx$$

With our expression for a slab of thickness A being sliced off the hemisphere, we do the following integral:
$$Q=-\frac{\rho}{\pi} \int^{0}_{-R+A} \frac{1}{\left(x^{2}-R^{2}\right)}dx$$
$$=-\frac{\rho}{\pi} \frac{1}{2R} ln \left.\frac{R-x}{R+x} \right|^{0}_{x=-R+A}$$
$$=-\frac{\rho}{2\pi R}\left(0 - ln \frac{R-(-R+A)}{R+(-R+A)} \right)$$
$$=\frac{\rho}{2\pi R} ln \frac{2R-A}{A}$$

With our expression for a remaining slab of thickness B, we do the following integral:
$$Q=-\frac{\rho}{\pi} \int^{0}_{-B} \frac{1}{\left(x^{2}-R^{2}\right)}dx$$
$$=-\frac{\rho}{\pi} \frac{1}{2R} ln \left.\frac{R-x}{R+x} \right|^{0}_{x=-B}$$
$$=-\frac{\rho}{2\pi R}\left(0 - ln \frac{R-(-B)}{R+(-B)} \right)$$
$$=\frac{\rho}{2\pi R} ln \frac{R+B}{R-B}$$

EDIT: Which exhibit roughly the correct behaviour as B and A tend to 0 respectively, and whose resistances increase as B increases, and A decreases (i.e. thicker slab = higher resistance)

Last edited:
I don't know where the rest of you are going with this, but Naty1 is right. If the connection area at each pole goes to zero, the resistance goes to infinity. So an area where the current enters is an element of the analysis.

More than that, the current density in each cross section is not uniform nor can be assumed axial. I'm not touching this one. I'm sure it's non analytical.

Last edited: