When the battery connected:
An AAAA cell has 200-350 mohms internal resistance. A 9v battery has 6 of them in series (many of them are literally that, others have their cells as a stack of plastic buckets). The nose ring is a short run of wire, it’s idunno a 0.2 ohm heater?
I think the septum is going to get pretty toasty.
I just tested this (for science!) with a 9V battery and an iron nail of roughly nose-ring diameter. Both the nail and the battery get unpleasantly hot after several seconds. I don’t think they’d get hot enough to burn you, though. (Don’t take my word, though, please!) I believe the internal resistance of the battery does also increase with the temperature, so it effectively somewhat self regulates itself.
Common nose ring materials like Titanium and Stainless Steel are 4× and 7× more resistant than iron, which means they should dissipate more power than the nail, and thus get hotter. I was calculating something around 3 milliohms for a titanium 16 gauge 10mm wire, and 0.7 milliohms for an iron wire.
Regardless of material, at 1000 milliohms internal resistance, i think the battery itself is doing most of the heat dissipation. (But also over a much bigger surface area!)
How long did you keep the nail on the battery for?
A 9V battery can be used as a foam cutter.
Styrofoam and most art foams melt at about 200°C
About 10-20s, I left it on until it didn’t seem to be getting much hotter. I also didn’t want the battery to overheat and fail catastrophically. I think because the “wire” is such a large gauge, there’s not enough current for it to get seriously hot. In a foam cutter, you’re passing all that current through a much smaller cross-sectional area.
Edit: just to confirm, I did a little math. A 10cm steel wire with a tenth of the diameter would have a resistance of 5 ohms. That means that instead of 1% of the total heat dissipating in the thick wire, 80% of the heat is dissipating in the wire in foam cutter’s case, and there’s more total resistance, so more heat dissipation as well.
This is because:
A = π r²
R = ρ × L / A
So resistance is proportional to the material resistivity (ρ), the length (L), and the inverse square of the radius (r⁻²). That is to say, decreasing the radius by a factor of 10 increases resistance by a factor of 100.
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Wakey wakey, septum bakey
Would the person feel anything? Presumably the electricity would flow through the metal as path of least resistance?
Why would you spread incorrect information so confidently? You’re absolutely wrong. A 9 volt will get hot in a few seconds when shorted.
They prob wouldn’t get shocked but the ring and battery will get very hot very quick as there i pretty much no resistance wich means there is a lot of current flowing.
Someone else in the comments claimed to have calculated a resistance of 0.2 ohms.
I=U÷R
9V ÷ 0.2Ω = 45A
45 amps is a lot of current to flow through a nose ring. I don’t know how much that is in heat but i’d expect you to get burned from it
No, 9v batteries have almost no shock to them. Put one on your tongue or lick your finger and touch the terminals. You’ll feel nothing.
Dry skin is hundreds of thousands of ohms. Even wet skin has pretty good resistance. When you touch a 9v to your tongue, you’re starting to mess around with lower resistance flesh, it is definitely not a comfortable thing to do.
The metal ring doesn’t do anything to move more electricity into your body, but it soaks every electron the battery can push and turns it into heat. Best I can figure it would amount to a few watts, which would be toasty if you were holding it between your fingers. The septum is a thin piece of flesh, I think it would sauté pretty quickly.
I mean, if you lick a 9v you definitely feel something. It’s not painful, but it’s not natural or pleasant either.
You’re doing it wrong. You gotta place a metal sphere close to it connected to a Van de Graaff generator.
