If say, one has a small solonoid coil wound with 0.015mm enamel wire for 230V, drawing about 100mA, and one were to unwind all the windings and carefully lay them out all over the floor, will the wire survive if you applied 230V now? Not being a coil anymore, somehow I think the wire will just melt. If so, this brings me to my question. Is there then a maximum length OF COIL WIRE that may be used outside the coil before reaching a tie-off tag? Simply put, what would happen if my tie-of tags were say 1/2 meter away from the coil? Any ideas on this? I have no application for this, just curious, and obviously not an electrical person.

 

This shouldn't be a problem because the coil's inductance is still there to limit the current. With the wire laying out on the ground there is very little inductance so the current would be what ever the resistance of the wire and the applied voltage dictate. The current is the same in
all parts of a series circuit.

Regards

cww

 

Responding to Jan Huyser's Apr 11 query:

Your gut feel about the consequence is correct for an AC solenoid. And, it just wouldn't work if the solenoid was DC. But, before getting to the nitty-gritty comparison of the two scenario's you presented please clarify two apparent inconsistencies:

1) In countries that adhere to metricalization (although the USA agreed in 1866, repeat 1866) there are still many that oppose it) wire size is usually given in terms of area, i.e., mm-sq or mmq. Do you mean 0.015 mm in diameter or the area is 0.015 mmq?

2) For the current and wire size you chose, current density far exceeds the German norm of about 3 Amps/mmq. Even if the correct value is 0.015 mmq, then the current density is about 6.7 A/mmq, or about twice that used in German practice.

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

If this is an AC solenoid, then the inductive reactance is a significant part of the total impedance. The current will be higher if the coil is unwound, or if the iron is removed from the core. As to whether the wire will melt, that is more difficult to answer, as completely unwinding the coil will also improve the cooling.

Unwinding some turns from the coil is not a good idea. If you need longer tail wires, then solder some extra wire onto the end of the coil.

 

Responding to Curt Wollet's Apr 12 reply:

Curt, for analysis presume that the unwound coil yields a long straight wire. Furthermore, doubled back the wire on itself so that there are now 2 parallel wires. For this case the formula covering a single-phase 2-wire transmission line is applicable. The formula yields an inductance value proportional to both length and to the separation distance.

Because of the (presumably) great length, minimum inductance is not zero, even if they touch. Then, inductance increases, although not linearly, as the separation between them increases.

If Jan responds to my question, I intend to compare the inductance derived above to that of wire configured as a coil.

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

Hello Phil

I wasn't going to get into transmission lines with someone looking for an answer at the electronics 101 level. My point is, that the wire wound in a coil has much greater inductance and therefore inductive reactance than the same length of wire unwound. That is why the coil doesn't fry
while the unwound wire might with the same applied voltage. But, provided you don't remove a significant number of turns, unwinding the ends a bit to provide leads won't cause things to burn up any more than soldering longer wires on. That is how I interpret the question.

Regards

cww

 

Hi to all and thanks for all the replies. It has turned out to be much more interesting than I thought. The wire I proposed is meant to be 0.015mm in dia, voltage 230v AC, and I got these figures from measuring an existing coil. The crux of my question is however, how long can the lead-out wires be, from the coil to the termination tags, before being soldered to heavy duty wire? Will it burn, or make no difference? All theoretical of course. Thanks for the patience all..
Jan

 

It won't make any difference as the current in every part of the wire is the same, at least for our purposes, at line frequency. If fact the leads should run cooler by virtue of being able to readily dissipate heat. Any constraints would be mechanical.

Regards
cww

 

Responding to Jan Husyser's Apr 19 concluding comment:

Please excuse my tenacity (others would call it stubbornness), but the numbers don't make sense (to me.)

Ignoring the thickness of the enamel coating, then, the diameter of 0.015 mm (15 um) is less than the thickness of a human hair (if one accepts that a hair is about 100 um in dia). Furthermore, the cross sectional area is about 1.767E-4 mmq (or 0.349 cmil). I'm very doubtfull
that such a small wire, if copper, can carry such a current.

Of course I understand that you want to conclude further discussion, but a very simple test will reveal not only the coil resistance but its inductance as well.

Interested? Onsite or off?

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])