GFI Caused a Fire!

No. It was NOT a short. In most cases it is a loose, intermittenr, or high resistance connection that causes a fire - not a short.

In order for this to happen there MUST be a load of some sort on the circuit. A normal breaker will not respond to an "arc fault" of this type, and nor will a CFCI - an "arc fault detector" breaker is designed to trip under those conditions. Not sure how effective they are in real life - or if they are more likely to false trigger.

It had at least one four legged one.

If you have a lot of spare cash. They are put in bedroom circuits because that is where people are most likely to be asleep and unaware if a fire were to start there.

So you live in a "pretty new" house. Almost 30 years older than mine

- which is also a "relatively new" house.

Then they were not wired properly. Mine is over 40 years old and not a single wiring problem - but I AM replacing all switches and outlets with new COALR devices. Surprisingly, there is no such thing as a COALR GFCI device available in Canada (according to my electrical supplier) Need to use GFCI breakers

I know how you feel. My house was built in '79 and I think it's pretty new.

Old is 1930.,

Oh. Maybe I should just do my bedroom,

and let any weekend visitors fend for themselves. (I do have smoke alarms.)

You finally made the right point. I didnt' see it until now.

I think you'd be OK, it's houses that predate the 50's where I find all manner of rube wiring. In your case the only probable issue would be 3 wire circuits, which are two circuits which share a common neutral. AFCI breakers can't be connected to these circuits. If you open your panel and find any cables that have a red, black, and white wire, you won't be able to protect them with an AFCI.

Your house was built in 2028?

AFCI protection is currently required in practically all habitable rooms

Hi, Arc and overheating can occur on any loose connection wherever they maybe.

The homes I speak of passes inspection at the time using the guidelines at the time. Aluminum wiring connections will degrade over time due to the thermal and metallurgical properties of the wire. I've seen connections fail even when antioxidant compound was used. O_o

TDD

Thanks. I"m saving this for next summer, when I'll have time.

Aha! Now I understand what 'common ground' means. My former house was built in 1980, and had common grounds (3-wire) circuits. I understood how they worked, but I did not realize the danger involved. Thanks

me

These circuits, which are called multiwire branch circuits, three wire circuits, or Edison circuits, are common "neutral" in that two hot legs, of different potential share one neutral. Your typical residential electric service, which is 120/240 is an Edison circuit. When installed properly, they're no more dangerous than a two wire circuit

The aluminum simply relaxes over time and the connections start arcing. In areas near salt water, oxidation seemed to rapidly deteriorate the connections

I can add some comments about your last point from experience over the last

Frankly, I expected more nuisance trips based upon what I had been hearing and reading when the installation was new since, at that time, I was working with one of the NEC code panels and there was a lot of discussion about it. As I understand it, AFCIs react to an electrical arc by sensing the radio frequencies that the arc generates. But opening/closing a switch can generate such an arc (usually very small) and the same thing can happen when an incandescent lamp filament fails or electric motors with brushes or inertia switches power up.

Tomsic

An Edison circuit allows you to save running one neutral wire. Let's say I'm putting in new outlets, a long run from the panel. There are enough outlets that I would need to pull two romex cables, ie one hot, one neutral, one ground. Each cable would serve half those outlets.

With an Edison, I can pull one romex with two hots, one neutral, one ground. The one neutral is shared by the two groups of outlets. I've traded two cables for one and reduced the number of conductors by one, all of which save some $$.