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?
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?
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?
yes
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):
...you'll probably have enough info to figure it out.
cr0sh:
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 = voltagePlease 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):
...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.
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?
http://www.faa.gov/documentLibrary/media/Advisory_Circular/Chapter_11.pdf
Jump to page 15 for the simple table of wire sizes and fuse sizes.
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.
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)?