Makes sense, even though no American would use the word arsehole in everyday conversation. It’s British English specifically, so we don’t say that here.
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Rather than leave what is becoming a tedious OT thread up I’ve deleted my original posts.
Cole’s Law: thinly sliced cabbage, vinegar, salt and mayonnaise.
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That’s the shit!
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Too much current. Which, given Omh’s Law, means too much voltage across the terminals.
You could run a 12v pot’ through a car battery, which is capable of delivering a massive amount of current, with no burnout whatsoever. Not so if you upped the voltage though.
It does depend however, what you’re using it for. If it’s to control a 12v motor which draws a heavy current, then it’d indeed burn out. If you were using it to control a small low-current motor then there’d be no problem, even connected through a car battery.
Any voltage higher than its rating though, would burn it out regardless of the current.
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In order to “burn out” the pot (or “release the smoke” 
), the components of the pot must get hot. In resistive heating, this is I^2 x R; there is no voltage term. Pass enough current (I) through the resistance element (R), and eventually you exceed the allowable heat limit and something starts to smoke; in a simple pot like the picture shows, the smoke will be from the conformal coating on the resistive element.
Of course, voltage is related, via Ohm’s law (I=V/R)
In your 12V car battery example and a 10Kohm pot, connecting to the outside terminals results in 0.0012amps, and therefore 0.014 watts (W). As long as the power rating of the pot is > 0.014 W, no smoke. The pot in the picture is likely able to handle more than 1 W (can’t be certain, but, having seen a lot of pots in my time…)
Now, attach the 12V to one of the outside terminals and the center terminal (the “wiper” is usually on the center terminal), with the dial turned so the wiper is at the max R end (so 10Kohm); same current, same wattage, same heat, no smoke.
But, as you turn the dial and move the wiper to lower and lower resistance, current will increase (I=V/R) until I^2xR heating is too large. Assuming a 1W 10K linear taper pot, at 1/2 dial setting (halfway between 10Kohms and 0ohm = 5Kohm), were still OK (approx 0.03 W); at 10% from the 0ohm end, we’re still OK (0.14 W).
At 144ohms, power is 1W. The pot will probably be noticeably warm, but should not fail. In reality, depending on manufacturing tolerances, quality of the part, duty cycle. etc, it could fail or have its working life reduced.
Of course, all the above assumes no internal battery resistance, standard temperature, pressure & humidity, wiring & connections with no resistance, no change in resistance to the resistive element in the pot as heat rises, etc.
The voltage rating on a simple pot pictured is “dielectric strength”, which is more about how much voltage can be safely applied before the insulating properties of the components are compromised. Above that voltage, you might get arcing between the closely spaced components inside the pot, or breakdown of the insulation on the wire of a wire-wound resistive element, and while these problems will likely release the smoke, it’s not due to heating as we’re discussing.
Apologies to all for the long-winded and off-topic ranting… I guess this is a pet peeve of mine, stemming from a long-ago electronics teacher that insisted “It’s the current that kills!” in human electrocution. Of course it is! But the current is related to the voltage, and the current is related to the resistance…
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No apology necessary. Very insightful. thank you.
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