Is resistance of a wire increases with wounding it around a cylinder?

[waqarrashid33]

is resistance of a wire increases with wounding it around a cylinder?

 

[Born2bwire]

No, but the adjacent loops of the wire give rise to mutual and self inductance. This gives rise to an inductive reactance in the wire's total impedance. Though of course the wire itself has a resitance per unit length, so the additional coils will result in more resistance by virtue of the fact that you have a much longer length of wire.

 

[mabs239]

Born's answer is more accurate.

My version:

Resistence would remain same if you apply DC voltage
"Resistence" is increased with same wire when it is wound for AC voltage. It will reduce the AC current. However in that case it is known as reactance, not resistence.

 

[Longstreet]

It *could* increase physical resistance if the wire gets stretched at all during the winding process. Stretching both decreases cross-sectional area AND increases length, both of which would act to increase resistance.

 

[DaveC426913]

All other things (such as length of wire and thickness) being equal, winding it around a cylinder will increase resistance. The current running through the wire creates a magnetic field which then reacts to impede the current.

This can be shown quite easily by coiling an industrial extension cord. If you try to operate a 14V drill using a cord that's coiled, you might only get 10 or 12 volts (and the cord will get quite hot).

 

[sophiecentaur]

Low voltage tools draw a lot of current if they are at all powerful and the resistance of the (long) supply wire is very relevant and will produce a voltage drop unless very thick cable is used (a modest little 140W motor will need 10A with a 14V supply; easy to lose a few volts there).
There is no net magnetic field if the two wires (supply and return) are coiled together because the fields will cancel as the current flow will be clockwise in one case and anticlockwise in the other.
Anyway, are they not all DC operated? The reactance is zero for DC.

 

[ericgrau]

The wire's DC resistance stays the same, but generating the magnetic field created by the coils resists changes in current because it takes energy to change the strength of the magnetic field. So when the current is changing the total impedance (wire resistance + coil impedance) will increase. For constant current systems like batteries supplying a constant amount of power, this makes no difference. For AC (alternating current) circuits, audio sound waves, etc. it makes a difference depending on how fast the current is changing.

 

[Bob S]

The wire's DC resistance stays the same, but generating the magnetic field created by the coils resists changes in current because it takes energy to change the strength of the magnetic field. So when the current is changing the total impedance (wire resistance + coil impedance) will increase... For AC (alternating current) circuits, audio sound waves, etc. it makes a difference depending on how fast the current is changing.

If you coil up a wire that has both the current supply and the current return conductors in the same jacket, there is no increase in inductance, because the magnetic field is between the two opposite-current conductors, and very small field outside. This is the same for both straight and coiled power cords. The inductance is minimum when the two conductors are twisted together in the same jacket.
Bob S

 

[mabs239]

All other things (such as length of wire and thickness) being equal, winding it around a cylinder will increase resistance. The current running through the wire creates a magnetic field which then reacts to impede the current.

This can be shown quite easily by coiling an industrial extension cord. If you try to operate a 14V drill using a cord that's coiled, you might only get 10 or 12 volts (and the cord will get quite hot).

I think the resistence will remain same. The more voltage drop and heated chord will be due to increase in the current because power factor will be lower in this case.

 

[Born2bwire]

I think the resistence will remain same. The more voltage drop and heated chord will be due to increase in the current because power factor will be lower in this case.

Yes I agree, the resistance would only change, barring any structural stress added by the creation of the coils as previously mentioned, if we ended up using a longer run of wire than we otherwise would not have done. But the inductance created by the coils adds a reactive impedance to the wire. This means that we cannot deliver as much real power as we used to. While the reactance does not dissipate the power, it does tie up energy into reactive power. The loss in deliverable real power, also described by the power factor as mentioned above, is what results in the lowered performance of the drill.

 

[sophiecentaur]

But the inductance is more or less Zero - as stated in at least two posts, along with the reason. Also, the capacitance between the two conductors will be no more than, perhaps 1000pF - also negligable.

 

[Integral]

But the inductance is more or less Zero - as stated in at least two posts, along with the reason. Also, the capacitance between the two conductors will be no more than, perhaps 1000pF - also negligable.

I am not sure what you are reading. I have seen no discussion about magnitudes of of the inductance or capacitance, it simply does not matter, the effects exist.

 

[sophiecentaur]

Integral
"I am not sure what you are reading ...it simply does not matter, the effects exist "

Posts by Born2bqire, Dave c42 and ericgrau, to mention just three.
These posts refer to the reactive effect of a changing magnetic field associated with the supply wires. That is commonly known as Inductance. The 'coil' to which the posts refer seems to be bi-filar wound (look them up if you are not familiar with the terms) - because the flow and return coils have the same number of turns and carry the same current - producing zero net magnetic field. The 'Integral' is zero! There will be, as has been said, vanishingly small reactance to 50 (/60Hz) AC. Perhaps only one of the conductors is wound into a coil? That might alter matters.
Failing that, any effect that the OP is mentioning must be due to something other than magnetism. If it is not the copper conductor getting hot - and this is unlikely to be happening or the insulation would have melted - then I suggest that there may be a connection within the cable which gets warm or stretches and forms a bad contact.

Btw, how much current is flowing to the device? We would really need to know some more details before coming to any serious conclusion but the 'magnetic' idea is a very unlikely cause of lost volts. With low power devices, it is possible to lose some power due to resistance in the supply without it showing up as a big temperature increase.

 

[mabs239]

Sophiecentaur,

You raised very good point and I missed it altogather in first reading. OP used the word 'wire', so your conccept requieres some stretch.

Dave's observational setup is a good one to show the effect, but I am not sure what shape is the "industrial extension cord". If it has separate wires then the inductive effect could be even doubled.

 

[sophiecentaur]

Sophiecentaur,

You raised very good point and I missed it altogather in first reading. OP used the word 'wire', so your conccept requieres some stretch.

Dave's observational setup is a good one to show the effect, but I am not sure what shape is the "industrial extension cord". If it has separate wires then the inductive effect could be even doubled.

If you subscribe to what JC Maxwell wrote about then you will realize that two wires wound into a coil and carrying equal and opposite currents generate Zero Magnetic Field. That is to say, there will be no inductance. Don't argue that one with me - argue it with the first textbook you care to pick up!

 

[Born2bwire]

If you subscribe to what JC Maxwell wrote about then you will realize that two wires wound into a coil and carrying equal and opposite currents generate Zero Magnetic Field. That is to say, there will be no inductance. Don't argue that one with me - argue it with the first textbook you care to pick up!

But that isn't what the OP is asking about. The OP was talking about winding the wire around a cylinder, like making a solenoid. There is no statement from the OP about the return path coinciding with the loops as well.

 

[sophiecentaur]

I was just following the thread as it was progressing. Threads often move beyond the content of the OP, which is what makes them interseting sometimes. Should we always start a new thread when things digress a bit?

afaik, most 'industrial extension cords' consist of two conductors. That was the post to which I was responding. Furthermore, a later post compounded the problem by suggesting that having two wires would double the effect.

If you want the basic answer, involving the simplest scenarions, then, it's got to be:
1. Making a coil has no effect on resistance.
2. A coil has Inductance, which produces a Reactance to AC. But coiling 'a wire' in air will produce a low inductance and, hence, a very low / negligible reactance to 50Hz AC (unless you use thousands of turns).
3. For a 'real situation', involving a supply lead to an drill, the reactance will be zero (v.low) because there are two conductors in a supply lead, each having the same number of turns in the coil and the currents are equal and opposite.

 

[mabs239]

afaik, most 'industrial extension cords' consist of three conductors.

If you subscribe to what JC Maxwell wrote about then you will realize that two wires wound into a coil and carrying equal currents generate double Magnetic Field as copared to just one wire.

Your two wires might as well be treated as one wire by winding the second wire in reverse direction.

Now for the other points:
1. I agree but including the points in post4
2. You can't say that. It depends upon the length of the wire, cross-sectional area of the final coil and the materal through which magnetic flux flows. (I see you mentioned two of the three factors)
3. This is a very special case. You do have other ways of winding where reactance is not necessarily zero.

 

[sophiecentaur]

afaik, most 'industrial extension cords' consist of three conductors.

If you subscribe to what JC Maxwell wrote about then you will realize that two wires wound into a coil and carrying equal currents generate double Magnetic Field as copared to just one wire.

Your two wires might as well be treated as one wire by winding the second wire in reverse direction.

Now for the other points:
1. I agree but including the points in post4
2. You can't say that. It depends upon the length of the wire, cross-sectional area of the final coil and the materal through which magnetic flux flows. (I see you mentioned two of the three factors)
3. This is a very special case. You do have other ways of winding where reactance is not necessarily zero.

If there is, indeed, any significant current flowing in the Earth wire, then you have a problem and loss of volts is the least of your worries. I think the Earth wire statement is a yellow and green striped herring - and I think you do too.

I suppose what this all boils down to is whether we are having a sensible discussion about the initial query - and I think we agree that Resistance is not affected except, possibly, by non specified extra practical factors.

All the subsequent interchange about reactance (which is the only thing that can affect the Impedance of the cable) is hardly relevant to mains operation. Reactance will be caused by only a few pF of Capacity between the conductors and the inductance of the single loop, comprised of the flow and return conductors, irrespective of their parallel route through space. Bearing in mind that they are very close together and enclose a very small area and, moreover, are twisted, The resulting Inductive Reactance must be vanishingly small - any field might just possibly be 'detectable' by a Hall Effect Probe. But this is Angels on a pinhead stuff and nothing to do with any practicable cable route is going to produce a practically measurable change in Impedance. Possibly a route around Birmingham with a few hundred turns might do it for you but these posts are discussing supply cables - nothing more. I thought there was mention of a 14V drill, in fact, supplied via a cable which which "gets hot". That must be resistive loss because reactance is lossless! and the reason you notice the heating effect when rolled up is that the heat can't get out. I have never actuallt come across a low voltage tool which you could actually operate 'corded, in any case. Aren't they all cordless? (Apart from little minidrills, of course)
The idea that two coils equals twice the inductance / field, whatever their arrangement, is just ludicrous and I think most people reading this will know that. Look at how a RCCB works - it relies totally on what I have been saying.

 

[Per Oni]

V good post sophiecentaur.

I don’t understand DaveC426913’s post #5.

 

[Redbelly98]

I don’t understand DaveC426913’s post #5.

DaveC is referring to the inductance that is created when you wind a wire around a solenoid. The inductance is effectively in series with the resistance, increasing the magnitude of the impedance for an AC signal.

But the OP specifically said resistance, not impedance, and the wire's resistance would not be changed by the simple act of winding the wire.

This whole discussion could benefit from clarification and involvement by the OP.

 

[mabs239]

A mistake on my part, doubling the number of turns of the cylendrically wound coil would actually increase the inductance four times. Resistence would remain practically unchanged.

 

[Per Oni]

DaveC is referring to the inductance that is created when you wind a wire around a solenoid. The inductance is effectively in series with the resistance, increasing the magnitude of the impedance for an AC signal.

But the OP specifically said resistance, not impedance, and the wire's resistance would not be changed by the simple act of winding the wire.

This whole discussion could benefit from clarification and involvement by the OP.

Following on from previous posts the impedance is going to be small. The I think the reason Dave's industrial cord gets a bit hotter is that there's a lot of wire confined to a small space with not much ventilation.

 

[Redbelly98]

Following on from previous posts the impedance is going to be small. The I think the reason Dave's industrial cord gets a bit hotter is that there's a lot of wire confined to a small space with not much ventilation.

I had missed that part of Dave's post (#5) earlier. It has me puzzled, why would a 14V drill, presumably battery-powered and cordless, have a cord attached to it in the first place?