stupid dangerous question

[bryan93644]

hey, im not going to rush out and try this tomorrow, but i really want to know if anybody knows what happens at higher voltages. I have a 1,500 volt power supply, and i am also curious about 120 just rectified by four diodes. Im not saying i think this is a great idea- but i hope someone can help me understand what 120 or 1,500 volts would do, and why, so i better understand anodizing. thanks!

[Fibergeek]

120 Volts would imply for PAR = 2.5 Ohms, a current density of:

I = V / R = 120V / 2.5R = 48 Amps per sq. ft.

This would either boil the electrolyte or draw an arc, before the 15 or 20 Amp mains circuit breaker would trip. You place yourself in danger of electrocution by attempting to use rectified 120 VAC without transformer isolation. Don't do it.

The implied current density would be 600 Amps per sq. ft. at 1500 V.

Will your 1500 V supply put out 600 Amps? This would make it a 900,000 Watt power supply, and it would weigh 3 or 4 tons. A normal size 1500 V supply; say 500 Watts, will simply short out and shut down if you did this. You might also draw an arc before shut down. If its an old power supply, which isn't as well protected as the newer ones, you will likely destroy it.

(edited to improve completeness)

[bryan93644]

i can keep the draw on the 1500 between four and six milliamps.

[Fibergeek]

That makes it 9 watts max.

[bryan93644]

which would be lame
so, wattage per surface area is an issue, but a wide range of voltages will work with the right corrosponding current? is there a way to calculate the ideal voltage?

[Fibergeek]

No.
You have this turned around a bit.

As is the case with everything electrical, current (amps) does the work. You need enough voltage to overcome the resistance (Ohms) and push the current to the desired level. That is all that voltage does.

Watts is Volts times Amps, and is what power is. If you have too much voltage, the power dissipation goes up, and heats up the electrolyte, and several other bad things also occur.