I call challenge accepted. In fact, I think it would be far more useful (and entertaining) if we all just put the meters and the slide rules down and set about trying to recreate the incident.
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Potentiometer Fire, Smoke and Sparks
- Thread starter Rebop
- Start date
I call challenge accepted. In fact, I think it would be far more useful (and entertaining) if we all just put the meters and the slide rules down and set about trying to recreate the incident.
Easy. Gimme a guitar, a car battery and a set of jumper cables.
Maybe the customer used a fire-wire cable?
The only guitar I've seen on fire was with Jimi Hendrix - and this was no pot issue ...
The only guitar I've seen on fire was with Jimi Hendrix - and this was no pot issue ...
Ok, just for fun, I've filmed a pot smoking from being connected to a 9V battery.
http://youtu.be/Rxq6DzDVdcg
And yes, I'm aware that this is a pot smoking video!
http://youtu.be/Rxq6DzDVdcg
And yes, I'm aware that this is a pot smoking video!
I am just re-telling what I was told. He said one outlet had the hot and neutral reversed. When his guitar touched the mic (amp and mixer on different circuits) his strings turned red hot. Sure sounds possible to me.
Ahh, well that's different. Didn't know the guitar strings touched a metal object. I'm sure you're just re telling the story-didn't mean to imply you made it up.
Still not sure I believe it.
Still not sure I believe it.
Sure it is safe, there is no way the number you get can hurt you in any way. And, if you multiply any voltage by any current the number you get will have the units of Watts. But if you want to claim the number represents the amount of power the battery can supply to a load the voltage and current have to be measured under the same conditions. Making them at the same time guarantees that will be true.
Your peak voltage*peak current is 8.9*4.9 = 44W. A reasonable assumption is that the Ammeter resistance is near zero and that 8.9/4.9 = 1.8Ohm is the battery internal resistance when you first connect the meter. The maximum load power will occur when the load equals the internal resistance. Under those conditions the total power would be 8.9*8.9/3.6 = 22W and only half that, 11W, will appear in the external load, the rest will be dissipated as heat in the battery. It is clearly enough to smoke a pot but not nearly 44W.
Ken
Even I am not sure I believe it. The only reason I DO remember it was, the guy bought some mic cables from us and a batch we had in stock all had pin 1 connected to the shell on BOTH ends, and we thought either we'd be sued or else someone would get killed.
