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I need to replace a burned resistor in an electric kettle, but I'm not sure I identified it correctly. The bands are: Red - Grey - Grey - Gold - Black. My preliminary guess 28.8 Ohms no tolerance.

According to one of the answers in this post, a black band indicates non-inductive wire-wound resistor, supposedly its value calculates as a 4-band one. In this (1) case, we have 28 x 1e+08 = 2.8G Ohms 5%.

According to one of the answers in the next post, a black band could be the leading zero(although according to what I've read it never starts with a black marking, but it seems the latter post disproves that, eh?), making it a 5-band resistor. This (2) case doesn't seem to make much sense as the 2nd band is gold. In all unlikely hood, assuming I'm wrong, it's actually a brown, then we have 018 x 1e+08 = 180M Ohms 2%.

According to one of the answers in that post,

Sometimes a black band, centered over the end cap, is added to make it more obvious which is the right-hand end.

In this (3) case, we calculate it as a 4-band resistor, starting from a red band, and we have the same answer as in (1) case.

I calculated it on the premise it's a 5-band resistor, starting from a red band, assuming a black band implies no tolerance, as it was said here: 288 x 1e-01 = 28.8 Ohms w/o tolerance.

My main question: What's the value of the resistor?

My follow-up question: Can a resistor have no tolerance?

My off-topic question: I didn't want to do the research before posting a question, because it's pain in the ***, and now it is only reinforced. There's so much contradictory information, and it only made me more confused. Why do you guys encourage to conduct a research before posting, if the information is so unstructured and complex so that in each case we have different answer? Thank you.

P.S. Although it's burned, and because of the quality of the image, it may seem like the black band is burned gray one, but it is confirmed it's black. I made another picture, but because of the size, I couldn't upload it.

the resistor


The place it was connected to is on the left side of the circuit, nearby the marking "R1":

fig.1A fig.1B


Here's a resistor that was used to replace the original. It failed, and got burned too. I believe its value is 47Ω 5% tolerance enter image description here

The kettle info: polaris PWK 1783CAD

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  • \$\begingroup\$ Is the resistor physically damaged such that you can't measure it? Are there any similar resistors in the device that are not burned that you could measure? \$\endgroup\$ – Ron Beyer Sep 12 at 19:08
  • \$\begingroup\$ @Ron Beyer, no, there are none. Answering first question, yes. \$\endgroup\$ – Elizaveta Cherkasova Sep 12 at 19:23
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    \$\begingroup\$ So band 4 is clearly Gold. You cannot have Gold as band 2. Whether we choose to help or not depends upon the perceived commitment of the asker. We have no idea what you know, so it is easier to correct misconceptions, than to generate a perfect 100% answer. \$\endgroup\$ – StainlessSteelRat Sep 12 at 21:19
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Summary: I suspect it is a 20Ω 5% tolerance wirewound resistor - analysis below.


I need to replace a burned resistor in an electric kettle

It would help to know what function this resistor had in the kettle. It's obviously a power resistor, but where was it connected? What was it doing? It wasn't in series with the element, if I'm correct that its value is 20Ω.

My main question: What's the value of the resistor?

  • I believe the most likely interpretation, is as a 4-band colour code, with the black ring possibly indicating a non-inductive wirewound construction, but more likely just to mark the right-hand end.

    Notice how the black band is in the middle of the end cap, whereas the red band (other end) is closer to the body of the resistor, making the black band the "odd one out", as expected for being the "right-hand end" of the markings.

    Another indicator of the black band not being part of the value is, as kindly commented by StainlessSteelRat, band 2 cannot be gold. So the black end band is not part of the value.

  • So what is the value? I don't believe the (now) grey bands 2 & 3 were originally grey. I suspect they were originally black. That makes sense, because value 20 (red, black) is valid in the cheap & common E24 value range and since E24 range resistors are usually 5% tolerance, that makes sense with the gold band.

    A grey 3rd band on a 4-band resistor would mean a very large (x100M) multiplier, which seems unlikely in a power resistor as seen in the photo. Also red, grey, grey (2, 8, 8) on a 5-band resistor isn't a valid value in any of the E-series ranges. These and several other reasons suggest that the original value was not red, grey, grey.

    Therefore I suspect the original colour code (before heat affected the colour of bands 2 & 3) was actually:

    red (2), black (0), black (x1), gold (5%) i.e. 20Ω 5% tolerance

  • I think I can see the subtle ridges of a wirewound construction in the photo. But why would there be a non-wirewound resistor in a kettle, which usually has a wirewound heater? Why avoid using a normal, inductive, wirewound resistor?

    This goes back to the first point above - it would help to know what the resistor was doing, to guide the interpretation of the back band as either non-inductive wirewound, or just to mark the right-hand end.

My follow-up question: Can a resistor have no tolerance?

No.

A couple of other points which weren't in your question:

  • What is the power rating of that resistor?

    To decide that, you need to measure the existing resistor and try to match its size to other wirewound resistors, using one of the well-known distributor websites.

  • Why did that resistor fail?

    If you don't find & fix the cause, then any replacement is also likely to fail. It will help to start by finding what function that resistor has in the kettle.

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Dave Tweed Sep 18 at 12:04
  • \$\begingroup\$ @DaveTweed - Many thanks for triggering that, Dave. I wish I knew how to do that myself... \$\endgroup\$ – SamGibson Sep 18 at 12:05

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