# Calculating resistance from cross-section

valarking

## Homework Statement

http://www.vkgfx.com/physics/p5.gif [Broken]

R = rho*l/A

## The Attempt at a Solution

Initially I thought this would be easy. I could just write the surface area A of the trapezoid by its geometric area formula and multiply it by h. I thought about it though and that doesn't really get the resistance from side a to side b.
So my idea now is to design an integral with x representing the width (from a to b) like this:
$$R =\rho*\frac{l}{A}$$
$$= \rho{l}\int_a^b{\frac{dA}{A}}$$
$$= \rho{l}\int_a^b{\frac{h}{hx}{dx}}$$
$$= \rho{l}\int_a^b{\frac{dx}{x}}$$

which obviously comes out to $$(\rho{l})*(ln(\frac{b}{a}))$$

I sort of went out on a limb here and I'm not sure if it's correct, does anyone here know if that's the right way to approach this problem? Basically I'm saying that due to the way resistors add in series, I'm adding infinitely many infinitesimal resistor cross sections.

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## Answers and Replies

Mentor
Kind of along the right path, but not with the right variables. The initial integral should be over dl along the x axis. At each distance x between a (x=0) and b (x=b), there will be an area slice through the figure. That area is determined by the equations for the top and bottom sloping lines, and the constant width. At each x, you have an A which is just determined by the line slope equations. Integrate that area slice multiplied by its thickness dl, and integrate along the x axis.

Make sense?

valarking
Somewhat. I think you may be looking at the cross sections slightly differently, as you mention the constant width, but it looks like length and height are constant, and the only thing that changes is the width (from a to b).
I see what you are saying about using dl for the integral, so I will try to construct a new one that way.
Thanks for the help, I will post my further attempt as soon as I can construct that integral.

valarking
Here's my attempt along your suggestions:

$$A = h\left(\frac{(b-a)x}{l}+a\right)$$

So I'm looking at:
$$R = \int_0^l{\frac{\rho{l}}{h\left(\frac{(b-a)x}{l}+a\right)}dx}$$

This eventually works down (after some lengthy calculus) to:
$$\frac{\rho{l^2}}{h(b-a)}ln\left(\frac{b}{a}\right)$$

Mentor
The first equation does not look quite right to me (it might be, but it's not intuitive to me). When x=0, the area is ha. When x=b, the area is hb...

valarking
well take for example, if a and b were 4 and 6, and l was 8
at x = 0, we get 4h
at x = 8, we get 6h

the way i have it set up, it's simply an integral that assumes the x-axis is along l and integrates along that.

Mentor
You could be right. Do you know if you got the right answer to the problem? Or do you have to just turn it in? I think yoiu have the right understanding of how the integration goes now.

Homework Helper
The first equation does not look quite right to me (it might be, but it's not intuitive to me). When x=0, the area is ha. When x=b, the area is hb...

When x = l. area = hb. And that is correct.

valarking
You could be right. Do you know if you got the right answer to the problem? Or do you have to just turn it in? I think yoiu have the right understanding of how the integration goes now.

It's not online so I won't know until I turn it in.

Thanks for all the help.

imtheunknown
There shouldn't be an l in your integral for R. You are integrating along the length, dx, from 0 to l, not ldx from 0 to l. If you checked the units in your answer, something your professor should have taught you , you would have noticed that you Ohm*m, not Ohms.

valarking
How do you figure I got Ohm*m?
The numerator of the fraction has Ohm*m^2
The denominator has m^2
And the units of the logarithm's contents cancel out

edit:
Oh, I think you're looking at my answer in the original post. That one's definitely wrong.

imtheunknown
Resistivity has units of Ohm*m. So your new answer has units (Ohm*m)*m^2/(m*m)

Mentor
When x = l. area = hb. And that is correct.

Yes, thanks for the correction. x never equals b here. Doh.

valarking
Resistivity has units of Ohm*m. So your new answer has units (Ohm*m)*m^2/(m*m)

ah, i see. redoing the integral simply gives an answer with rho*l in the numerator rather than rho*l^2

thanks