Copper wire question

[magnetman12003]

Hi All,

I have a large covered coil of single strand AWG copper magnet wire that weighs 2.5 pounds.
Can someone tell me about how long the wire inside the coil should be if I can measure 80 ohms resistance between the two ends of the wire.

I also have another covered coil the same weight that measures 12.2 ohms resistance between the ends of both wires??

What might be the AWG wire size and lenght? I dont have a wire gauge.

All help appreciated.

 

[ronv]

Can't tell without the wire size.

 

[ChrisP58]

Do you have a micrometer or calipers? Like Ron said, you really can't tell the length from the resistance without knowing the diameter.

 

[Pommie]

I would guess 25 gauge (0.45mm) and 2500 feet (774m) based on a metric calculation. I don't do imperial.

In metric,

Resistance = Rl/A = 1.7E-8 * 774 / (PI * 0.00045^2 /4) = 80Ω

Weight = density*volume = 8930 * 774 * PI * 0.00045^2 /4 = 1.124 kg = 2.47lb.

The other one is 29 AWG and 991 ft.

HTH

Edit, the above assumes no core or bobbin weight. BTW, if you want the (very rough) excel sheet let me know.

Mike.

 

[KeepItSimpleStupid]

Do you have index cards? Assume they are between 0.008 and 0.010" thick.

Measure the diameter in # of index cards.

Or, pick an object that others might have. e.g. Dollar bills or about the thickness of a penny, a dime, a nickel etc.

Drills also work.

 

[Bob Scott]

I will repost this reply after corrections.

Darn. BobW, below, beat me to it.

 

[Pommie]

R=0.00294 and L=1,316" gives a volume of 0.036 square inches. That is very heavy copper.

Edit, I see you're recalculating. It'll be interesting to see if the imperial calculation comes up with the same answer as mine.

Mike.

 

[KeepItSimpleStupid]

Is a "covered coil of wire" an insulated wire. Magnet wire is typically varnish insulated and thus would very minimally contribute to the weight.

You do have lbs of copper or is this lbs of copper on a spool? Wire tables do have lbs/1000 feet, but you need to know the wire gauge.

The other equation is R=ρL/A which states that
The resistance of a block of homogeneous material of uniform dimensions and length L is the constant p * the Length divided by the cross sectional area. ρ is a material constant in example units of (ohm-length); e.g. ohm-cm

Electricians use a variation of this formula unknowingly for Aluminum and Copper and generally use twice the length because of 2 conductors being run some distance. Since it's specific, it uses data direct from the wire table. This https://www.powerstream.com/Wire_Size.htm table is more comprehensive but misses the cross-sectional area and this https://en.wikipedia.org/wiki/American_wire_gauge one.

That basically tells us that R is proportional to length, but we knew that already.

Confused here:
Givens:
Coil 1 80 ohms 2.5 lbs
Coil 2: 12.3 ohms 2.5 lbs

Insulated?: Y/N

 

[Pommie]

There is only one wire diameter that will have a resistance of 80Ω and a weight of 2.5lb and that is 25 AWG. See above for explanation.

I'm assuming the OP allowed for any bobbin weight.

Mike.

 

[BobW]

I agree with Pommie. The weight and resistance is all the information that is necessary.

The density of copper is 8.96e3 kg/m^3
The resistivity is 1.68e-8 ohm-m

The weight is 2.5 lb = 1.13 kg
The resistance is 80 ohms.

The resistance of a piece of wire is equal to the length divided by cross sectional area times resistivity:
R=L/A*r

The weight is equal to the cross sectional area times the length times the density:
W=L*A*d

You have two equations in two unknowns which can be combined to give:
L=sqrt(W*R/(d*r))
where:
L=Length in meters
W=weight in kilograms
R=resistance in ohms
d=density in kg/m^3 = 8960
r=resistivity in ohm-m = 1.68e-8

Substituting the known info into this formula, you get a length of 776.78 meters, and then using the length to calculate the wire cross section:
A=L*r/R
you get 1.63e-7 m^2 which is equivalent to a diameter of 0.46mm or wire gauge 25 AWG.

For the second coil with the same weight and resistance of 12.2 ohms you get a length of 303 meters and a wire diameter of 0.73mm or #21AWG. This one disagrees with Pommie's calculation, but if the resistance is less for the same weight, then the wire size must be larger, and hence the length must be less.

Edit:
BTW, the formula to convert wire diameter to AWG is:
AWG=36-8.624889* Ln(200*D)
where D is diameter in inches, or
AWG=36-8.624889* Ln(7.874*D)
where D is diameter in mm

 

[Pommie]

For the second coil with the same weight and resistance of 12.2 ohms you get a length of 303 meters and a wire diameter of 0.73mm or #21AWG. This one disagrees with Pommie's calculation, but if the resistance is less for the same weight, then the wire size must be larger, and hence the length must be less.

I agree. Not sure how I got the above (wrong) answer. I'll check my spreadsheet when I get home later.

Edit, my spreadsheet was correct and so it looks like I had a senior moment when I looked up the AWG number.

Mike.

 

[Pommie]

For anyone interested, here is the spreadsheet I used to work this out. Whoops, xls files aren't allowed. Zipped instead.

Mike.