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If you take a straight piece of wire that is insulated and coil it around a non ferrous object like piece of wood would the resistance change?

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    \$\begingroup\$ Of course not. A ferrous or non-ferrous former doesn't matter, either. \$\endgroup\$ – Leon Heller Jun 18 '15 at 21:19
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The simple answer is no. Winding wire on a non-ferrous form will not change its resistance.

Of course, it all depends on the details. If the wire is uninsulated, and the form is conductive (copper, silver, gold, platinum, etc) the form will short out the wire turns and reduce the total resistance. Or were you thinking of insulated wire? You didn't say.

Ah, you say, but it is insulated. To be precise, it's insulated with enamel, and called magnet wire.

Well, now it gets tricky. Let's assume your reference point is a straight wire suspended between two contacts in still air. As you run current through the wire, it will heat up, even if only slightly, and for most wire materials the resistance will change. If the wire is now wound around a form of some material, the form will change the rate at which power is dissipated to the environment, and thus the resistance. A thermal insulator, such as foam, will increase the temperature of the wire. A good thermal conductor, such as silver, will tend to reduce the temperature changes, and even this effect depends on the physical details of the form to determine the long-term temperature effects. Size and shape will matter, and the result can, in principle, go either way.

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No the resistance only cares about the electron path through the material. Regardless of the shape or if it is wrapped around something it will not change.

On the other hand the inductance, which is related to the energy stored in the magnetic field created around the wire as current passes through it, will change with geometry and the material it is wrapped around.

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  • \$\begingroup\$ I guess my thinking was the B field induced by the DC current maybe some how change the resistance of the wire. I guess if it was an AC signal than the time varying magnetic fields would induce current on the other coils? \$\endgroup\$ – SamFisher83 Jun 18 '15 at 22:00
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Everyone so far seems to have forgotten (or not heard of) proximity effect.

Ignoring the inductance of a coiled wire and only concentrating on losses i.e. the resistive part of the wire then, at DC there will be no change. However, as frequency increases proximity effect between the coils will certainly increase the resistance of a conductor.

This is very much akin to skin effect where the rapidly changing magnetic field of the wire (due to the alternating current it carries) causes the electrons to flow in the outer circumference of the wire thus all that nice copper in the centre carries very little current. Current not using the full cross sectional area of the conductor will have a higher resistance to flow.

Remember I'm not talking about inductance.

So, proximity effect is where two coils are close by and the magnetic field of one wire causes the current flow in the next wire to be even more constricted than by skin-effect alone: -

enter image description here

The above is with current in opposite directions. With currents in the same direction the effects are opposite to that shown above.

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Contrary to popular belief, the resistance will change - if only by a miniscule amount - because of the wire being work-hardened by winding it around the form.

From: http://www.copper.org/publications/pub_list/pdf/a1360.pdf

enter image description here

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  • \$\begingroup\$ What do you mean by "work-hardened"? Maybe I just don't get it because English is not my first language. \$\endgroup\$ – Rev1.0 Jun 19 '15 at 17:00
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    \$\begingroup\$ Sorry, I should have included some supporting documentation in my answer. It's there now, and if you Google "work-hardened" or "cold-worked", there are zillions of hits explaining both terms, which are equivalent. \$\endgroup\$ – EM Fields Jun 19 '15 at 21:06
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This is one of those questions that starts the real meat of EE. The DC resistance will not change if you coil the wire, but the overall impedance will change. A components impedance consists of the DC resistance, inductive reactance and capacitive reactance, all of which get "sumed" up to form a frequency dependent electrical impedance. I would rather not get into the down and dirty details, but this wiki article sums it up pretty nicely! Electrical impedance

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Andy aka gave an awesome answer the question was so vague. Sounds like you are trying to make an inductor.. If so, I suggest you read up on inductors. If you are asking about your ac extension cord and whether or not it matters to leave the cord coiled up while trying to use it, you can test that as well. You will discover you are turning your extension cord into an inductor and depending on how many feet you have coiled up, you will notice a significant drop in the voltage, increase in impedance, and an increase in power consumption. Even with household appliances, you should never coil up your excess cord. If you do, straighten them out and enjoy the money you will be saving on your electric bill.

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  • \$\begingroup\$ Actually, the increased impedance of the cord acting as an inductor would decrease power consumption. You are correct that it would decrease voltage to the load, though. :-) \$\endgroup\$ – Robherc KV5ROB Feb 21 '16 at 11:59
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    \$\begingroup\$ @CHall: Welcome to SE. The inductance of a coiled up extension cord will be very low due to cancellation caused by the opposing direction of the feed and return currents. Even if there were some inductance that would reduce the voltage and current (and reduce the power factor). The power demand would normally reduce - the exceptions being electronic power controllers which will adjust their switching to maintain the required power. The main problem with coiled extension leads is that of temperature rise due to lack of heat dissipation. You may edit your answer to fix any issues. \$\endgroup\$ – Transistor Feb 21 '16 at 12:44

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