# Resistivity

Discussion in 'Math' started by Lightfire, Aug 24, 2011.

1. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Hello,

I found this formula for calculating the resistivity of a certain thing.

So as quoted above, to get the resistivity of a certain thing, you need to multiply the length to resistance (measured in Ω) and the product will be divided by a cross section area.

Now, how could I measure the resistivity of a wood? Most says it's a millions Ω but I need clarification.

Secondly, cross section area? I don't know the formula to get the cross section area. Please tell it to me.

Thirdly, is there any standard for the unit of length? Isn't it in meter or?

Fourthly, I hope for your answers. I really need. This, Whoop! What is the correct resistivity of a wood?

Postscript: My teacher taught us, told us rather, that the Greek symbol Ω is alpha. I couldn't tell her that it is omega as she might give me the floor to teach.

Yours truly,
Lightfire

2. ### wmodavis Well-Known Member

Oct 23, 2010
739
150
There is no "correct resistivity of a wood".

In fact the resistivity of wood varies with wood species, moisture content of the wood and with temperature. Wood moisture meters are used to approximate the actual moisture content of the wood and the pin type are basically specialized ohm meters. They measure the resistance between two pins inserted into the wood fibers. The resistance value is then converted to moisture content and displayed. The meters have different conversion factors depending on the wood species and temperature.

3. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Okay.

So how about for the other questions?

4. ### wmodavis Well-Known Member

Oct 23, 2010
739
150
The same way you measure the resistivity of a wire or anything else. Measure the resistance, determine the cross section area and the length. Then use the formula to calculate resistivity. But that calculated resistivity only holds for the specific species, temperature and moisture content at the time of measurement since wood is hygroscopic and its resistivity varies widely.

5. ### wmodavis Well-Known Member

Oct 23, 2010
739
150
"Pin or resistance meters read the moisture content of the wettest area that contacts the uninsulated part of the pin/electrode. Effective measurement range of resistance or pin meters is from 7 to 30% MC. These meters are specialized ohm meters, and below 7% MC, the electrical resistance is too high for reliable readings. For example, at 7% MC the resistance for red oak is about 15,000 megohms, and for hard maple 72,000 megohms. Above 30% MC, the electrical resistance has too much variability and too little change for reliable readings. At 30% MC the resistance for red oak is about 0.50 megohms and for hard maple it is about 0.60 megohms." (R. S. Boone and E. M. Wengert)

6. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
I don't know how to get the cross-sectional area. Help, please.

7. ### wmodavis Well-Known Member

Oct 23, 2010
739
150
Measure it. If it has a square cross section the area is height x width. If it is circular measure the diameter and calculate using the area of a circle formula etc.

May 25, 2011
148
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9. ### BillO Distinguished Member

Nov 24, 2008
994
137
My guess is you are not going to be able to do this effectively for wood. First, wood is not a homogenous substance. There is considerable variance in any one piece of wood. Resistivity will vary greatly depending on where and how it is measured. Second, each different species of wood will have a very different bulk resistivity (given all other things being equal). Third, the mineral and salt content of were the wood was grown will greatly affect its resistivity. Fourth, humidity and moisture level will also have a huge effect. Fifth, but not final, the age and general density of the wood sample will have a large effect. There are just too many variables.

What is it you are actually after? Your ultimate goal?

10. ### u-will-neva-no Member

Mar 22, 2011
230
2
When you measure the cross sectional area, you have to look at which way your passing your test current through. Obviously it does not matter if you have something like a cube but for any asymmetrical shape this would apply.

11. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
$R=\frac{\rho L}{A}$

$\rho$ resistivity
L length
A cross sectional area

I am getting very confused with
How do I find the resistivity if I am still calculating it? Eh?

.

Now, I am confused with radius squared. How do I find the radius squared? Is there any formula for it?

12. ### Wendy Moderator

Mar 24, 2008
21,155
2,855
r² = r X r

Simple.

= 3.14159

But I think the correct forumula is

2$image=http://forum.allaboutcircuits.com/mimetex.cgi?%5Cpi&hash=3ba5bfe3e9df73360fc1350ce4e8ef06$

Lightfire likes this.
13. ### strantor AAC Fanatic!

Oct 3, 2010
4,899
2,875
Bill I think you are running the circumference and the area formulas for circles together into one. area of a circle is $image=http://forum.allaboutcircuits.com/mimetex.cgi?%5Cpi&hash=3ba5bfe3e9df73360fc1350ce4e8ef06$r². circumference is 2$image=http://forum.allaboutcircuits.com/mimetex.cgi?%5Cpi&hash=3ba5bfe3e9df73360fc1350ce4e8ef06$r.
http://www.mathsisfun.com/area.html

have you learned what radius is in school yet? it's the distance from the center of a circle to the outside. for example, if you have a circle 4cm across, the radius is 2cm. so the area of that circle is 2cm*2cm*3.14159 = 12.57cm²

Lightfire likes this.
14. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Ah, I see. That's why every return I had when I am googling are circle because of that.

Um, what if the thing i want to measure its resistance is a square. how? rather, a triangle?

pls. help

15. ### strantor AAC Fanatic!

Oct 3, 2010
4,899
2,875
The link I posted has the area of most common shapes
http://www.mathsisfun.com/area.html

Oct 5, 2010
690
21

Oct 3, 2010
4,899
2,875
no problem

Oct 5, 2010
690
21
19. ### strantor AAC Fanatic!

Oct 3, 2010
4,899
2,875
So, since you specify 1.59 X 10^-8, I assume you are talking about silver.
for example,
$R=\frac{\rho L}{A}$

we have a 100m silver conductor that is 1cm wide (.5cm radius)
.5cm * .5cm * 3.14159 = .7854
$R=\frac{.0000000159 * 100}{.7854}$

resistance would be about 2.024 μΩ.

someone please shout if I got that wrong. Im not good at math!

Lightfire likes this.

Mar 24, 2008
21,155
2,855