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I didn't even know you need AWG rating to get a coil's resistance. But this is may sound naive, can I use the physics formula: $$R=\rho\frac{L}{A}$$.

\$\rho\$ is resistivity of copper, which I looked up to be \$0.000 000 017 W m\$. I know the length, \$L\$, is 100m. And the diameter is \$.5mm\$. So I know the cross-sectional area, \$A\$.

Now, my question is how accurate is the result?

Here is the website where I bought the coil: http://www.ebay.com/itm/142313954017

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    \$\begingroup\$ You measure it with a multimeter. \$\endgroup\$ – PlasmaHH Jun 23 '17 at 13:03
  • \$\begingroup\$ I dont have one \$\endgroup\$ – most venerable sir Jun 23 '17 at 13:13
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    \$\begingroup\$ That generally is a problematic thing when doing anything with electronics. \$\endgroup\$ – PlasmaHH Jun 23 '17 at 13:14
  • \$\begingroup\$ Which multimeter should I buy \$\endgroup\$ – most venerable sir Jun 23 '17 at 13:14
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    \$\begingroup\$ any that turns out to be good after doing some research on your own on the topic. You might want to start with what one is and how to use it and how not to kill yourself. If you are the kind that loves videos, I recommend watching all multimeter related eevblog videos on youtube. \$\endgroup\$ – PlasmaHH Jun 23 '17 at 13:16
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And the diameter is .5mm

I'd look it up on google and find a table like this: -

enter image description here

Your diameter of 0.5 mm lies somewhere between 24 and 25 AWG and this means a 100 metre length will have about 9 ohms.

Now, my question is how accurate is the result?

It will vary with temperature - hotter means higher resistance. At 20 degC copper has a resistivity of 1.68E-8 ohm-metres so using your formula I get a 0.5mm diameter wire of 100 metre length to be 8.57 ohms.

But it also depends on impurities in the copper: -

enter image description here

Only the "pure copper" trace aligns with the calculations above because there are no impurities. Now I'm not saying your vendor has added any nickel into the alloy but if you do eventually get an ohm meter and measure it, this might be an explanation.

Above picture taken from here.

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  • \$\begingroup\$ Please read the eBay link. Diameter is 0.2 mm. \$\endgroup\$ – WhatRoughBeast Jun 23 '17 at 16:14
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    \$\begingroup\$ @WhatRoughBeast yes it is so, why did the op write 0.5mm? Anyway, the method is the same whether 0.2 mm or 0.5 mm diameter. \$\endgroup\$ – Andy aka Jun 23 '17 at 16:28
  • \$\begingroup\$ "why did the op write 0.5mm?" Why indeed. I can't insert a "roll your eyes" in comment. \$\endgroup\$ – WhatRoughBeast Jun 23 '17 at 16:39

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