Go Down

Topic: how to calclulate the thickness of a wire according to the current? (Read 1 time) previous topic - next topic

KobiAflalo

if for example in a wire there is a 3A current, how to calculate the area or AWG of this wire?

i know the formula

but not so sure how to find the area by a given current....

i think about the current but is there other factors i need to think about?
begginer Board Designer.

tmd3

The question isn't clear.  3 amps could be persuaded to run in a tiny wire, or a fat one.  The current doesn't tell you anything about the size of the wire.

Are you asking how to decide what size wire you need to carry a particular current?

KobiAflalo

The question isn't clear.  3 amps could be persuaded to run in a tiny wire, or a fat one.  The current doesn't tell you anything about the size of the wire.

Are you asking how to decide what size wire you need to carry a particular current?
yes
begginer Board Designer.

keeper63

Hmm - based on refs in this:

https://wiki.xtronics.com/index.php/Wire-Gauge_Ampacity

First we'll rearrange things to solve for A:

A = (p / R) * l

Then note - Ohm's law - V = I * R or R = V / I - so sub that in place of R:

A = (p / (V / I)) * l

Which I think can be re-arranged (math ain't my strong point!) to:

A = (p * I * l) / V

Where:

A = cross-sectional area of the specimen (measured in square meters, m²)
p = static resistivity (measured in ohm meters, Ω-m)
I = current (amps)
l = length of the piece of material (measured in meters, m)
V = voltage

Please cross-check my math, of course - it's probably wrong! At any rate, I think between the link above, and the following (or something similar):

http://www.powerstream.com/Wire_Size.htm

...you'll probably have enough info to figure it out.
I will not respond to Arduino help PM's from random forum users; if you have such a question, start a new topic thread.

Archibald

Hmm - based on refs in this:

https://wiki.xtronics.com/index.php/Wire-Gauge_Ampacity

First we'll rearrange things to solve for A:

A = (p / R) * l

Then note - Ohm's law - V = I * R or R = V / I - so sub that in place of R:

A = (p / (V / I)) * l

Which I think can be re-arranged (math ain't my strong point!) to:

A = (p * I * l) / V

Where:

A = cross-sectional area of the specimen (measured in square meters, m²)
p = static resistivity (measured in ohm meters, Ω-m)
I = current (amps)
l = length of the piece of material (measured in meters, m)
V = voltage

Please cross-check my math, of course - it's probably wrong! At any rate, I think between the link above, and the following (or something similar):

http://www.powerstream.com/Wire_Size.htm

...you'll probably have enough info to figure it out.
Your formula:
A = (p / R) * l
should be:
A = (ρ / R) * L   where L is the length of the conductor.

The remainder of your post is therefore incorrect.

tmd3

OK, so you're asking how to decide what size wire you need to carry a particular current.  I think that subject could fill a medium-length book.  It will depend on your requirements. 

You don't want the wire's insulation to melt.  You don't want the wire to get so hot that adjacent materials will melt, or soften, or ignite.  If you're building something that you will handle, you don't want the wire to become uncomfortably hot.  You don't want the wire to drop too much voltage from one end to the other, and that will depend on the length of the wire, the current, its frequency, and its power factor. 

Before we recommend some course of action that will make your house burst into flame, tell us more.  What are you building?  How long are the wires?  Are you enclosing them in something like conduit, fishing them through walls, or otherwise limiting their ability to lose heat to their surroundings?

MorganS

"The problem is in the code you didn't post."

Paulcs

22 AWG.

See: AWG sizes


If it is long or warm environments or inside conduit, you might use a 20 AWG.  It never hurts to use a little better wire than you need.  Also, 22 AWG is fairly light weight if there are any mechanical issues, flexing, even cutting and stripping should be careful not to nick it up too much.

jcallen

You would need to insert temperature rise into the formula.

How much current can the wire carry without getting hot enough to melt the insulation (or the wire)?

Go Up