When replacing the resistance wire in an electric kiln, most manufacturers recommend wrapping the resistance wire back on itself at the ends where it connects to the electrics. Although I haven't tested this, in theory wrapping the wire back on itself should increase the resistance slightly.

Does wrapping the wire this way increase or decrease the temperature of the stretch of wire that has been wrapped back on itself? I.e. Do they recommend this to protect the electrical wiring and relay from the temperature of the resistance wire, or is it for some other reason?


enter image description here Image source.


3 Answers 3


I read the question as looping the wire back at the terminal, like this, image from here. I might be wrong though, and an illustration from the OP may be needed to correct me.

enter image description here

There are two ways that this can reduce the temperature of the end stretch of wire

  • To the extent that the reversed bit makes good electrical contact with the wire it's twisted with, it reduces the resistance, reducing the I2R heating in that section. If the wire is not twisted tightly, or if the wire has an oxidised or otherwise contaminated surface, electrical contact will not be good, and there may be no temperature reduction due to this.
  • The reversed bit makes reasonable thermal contact with the wire, and increases its mean surface area, making it able to dissipate the heat produced at a lower temperature. Whether this effect is significant compared to the large heatsink of the terminal it will be screwed to is another matter.

Now we have seen the illustration of the kiln wire configuration, the coiled section will get so much hotter than the straight, terminal, section, that I would expected the further temperature reduction due to twisting back would be practically irrelevant. However, kilns can get verrrry hot, so maybe you do have to take those extra steps to get reliable terminal contact?

  • \$\begingroup\$ Exactly so. Kiln temperature can easily exceed 1000 degrees C (and the wire has to be hotter), so keeping the terminals to some reasonable temperature is necessary for a reliable connection, even at 230VAC. \$\endgroup\$ Jun 16 at 10:05

We could model the twisted and untwisted sections like this:


simulate this circuit – Schematic created using CircuitLab

Each \$R\$ is some arbitrary length of wire, each length having the same resistance. The blue box represents a section of wire pair twisted together. The orange section is a length of single, untwisted wire.

Ignore for the moment the cross connections, such as the connection between A and B.

Current \$I\$ splits into two halves, shared equally between the two paths of the twisted section. Since \$P=I^2R\$, the power dissipated in each blue \$R\$ section is one quarter of the power dissipated in orange \$R\$. This implies that each path of the entire twisted section is receiving 25% of the heating power being received by the same length of untwisted section. The twisted pair together receive 50%. The twisted section is going to heat far less than untwisted section, with clear benefits for whatever mechanical connection is joining this wire to the power source at node C.

By twisting bare, un-insulated sections of wire together, obviously there will be points where the wires touch, represented by the cross connections such as the one between nodes A and B. Little or no current will pass via those points of contact, because the potential at each corresponding node on opposite sides (for example \$V_A\$ and \$V_B\$) will be very similar. Symmetry of the system ensures that potential changes at the same rate along both paths, and these cross connections have no consequence.


or is it for some other reason?

They probably recommend this so that the self-inductance of the element wire is minimized and thus electromagnetic interference generated is also minimized. You have to consider that the element wire might be controlled via thyristors or triacs and, these will generate very sharp changes in voltage (and current) that could easily interfere with other equipment.

Non-inductive bifilar winding: -

enter image description here

Does wrapping the wire this way increase or decrease the temperature of the stretch of wire that has been wrapped back on itself?

It may change the temperature slightly but, I suspect, that this is not the main reason for the recommendation.

  • \$\begingroup\$ Bifilar or Ayerton-Perry winding would be unlikely candidates within heater coil, given the wire is bare and uninsulated. I think the lowest-induction configuration I've seen is just a rectangular zig-zag shape; there doesn't seem to be any concern with making whole turns either. (That said, an Ayerton-Perry configuration, on an otherwise spiral solenoid shaped winding, would do. That is, doubling back on itself, in the largest curvature following the kiln walls -- not within the smallest wire spirals themselves.) (But I digress, and this is probably just confusing without a diagram...) \$\endgroup\$ Jun 15 at 9:57

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