# Resistivity

#### Lightfire

Joined Oct 5, 2010
690
Hello,

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

Resistance is length X resistivity / Cross section area
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

#### wmodavis

Joined Oct 23, 2010
739
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.

#### Lightfire

Joined Oct 5, 2010
690
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.
Okay.

So how about for the other questions?

#### wmodavis

Joined Oct 23, 2010
739
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.

#### wmodavis

Joined Oct 23, 2010
739
"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)

#### Lightfire

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

#### wmodavis

Joined Oct 23, 2010
739
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.

#### TBayBoy

Joined May 25, 2011
148

#### BillO

Joined Nov 24, 2008
990
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?

#### u-will-neva-no

Joined Mar 22, 2011
230
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.

#### Lightfire

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

$$\rho$$ resistivity
L length
A cross sectional area

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

$$\pi$$ multiplied by radius squared
.

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

#### Wendy

Joined Mar 24, 2008
22,178
r² = r X r

Simple.

= 3.14159

But I think the correct forumula is

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

#### strantor

Joined Oct 3, 2010
5,369
But I think the correct forumula is

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

How do I find the radius squared? Is there any formula for it?
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

Joined Oct 5, 2010
690
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
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

#### strantor

Joined Oct 3, 2010
5,369
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
The link I posted has the area of most common shapes
http://www.mathsisfun.com/area.html

#### strantor

Joined Oct 3, 2010
5,369
no problem

#### strantor

Joined Oct 3, 2010
5,369
$$R=\frac{\rho L}{A}$$

$$\rho$$ resistivity
L length
A cross sectional area
Oops, just a question.

What does this mean : 1.59 x 10^-8
Can be found at http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/rstiv.html#c1 .

When I calculated it with a calculator, the answer is still 1.59. Yay.

pls help
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!

#### Wendy

Joined Mar 24, 2008
22,178