Disclaimer: what I am proposing is dangerous, and I know that. I would discourage anyone, regardless of expertise from undertaking such a pointless risk to their health and home.

After my recent (mis)adventure with the dryer, I now have a physically intact, technically functional heating element. I've been looking for an excuse to repurpose it for a DIY project, and then realized I've wanted to try making my own 2 stage Cockroft-Walton voltage doubler for a while. Seeing as the element itself runs on 240 volts, and the current running to my house is 120, I thought, "Maybe I could run a voltage multiplier off one of these transformers I have lying about, hook a potentiometer up to it with the wiper running to the input side of the element, and then run it to ground."

This is, inherently, a bad idea. I know. But can it be done? I have drawn here a pretty common depiction of the circuit, with the addition of the voltage divider and element attached as a load.

Upon re-examining it, it occurred to me that I probably couldn't return the 240 volt DC output to the same ground connection the circuit used without something very very bad happening, so I am a bit stumped there. How would I properly terminate this, seeing as I need some sort of ground to reference from? Other concerns I have are what AC input source should I use -- transformer, or main?

What effect will arbitrarily changing the output voltage of the circuit have on the circuit itself?

Is using a pot even feasible here?

Here is a horrible drawing of the circuit I am suggesting

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    \$\begingroup\$ With any practical value for those capacitors you will get next to no power out of it into 10 ohms, regardless of the pot setting. \$\endgroup\$ – Brian Drummond Dec 30 '19 at 19:31
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    \$\begingroup\$ You are not going to be able to draw anywhere near that much current through a circuit like this. \$\endgroup\$ – evildemonic Dec 30 '19 at 19:34
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    \$\begingroup\$ That would make it worse...you need more resistance in your load, not less. The current in this circuit will be limited by the CW V-doubler, not your load. \$\endgroup\$ – evildemonic Dec 30 '19 at 19:38
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    \$\begingroup\$ I don't think you can generate enough current to explode anything. Maybe enough to injure or kill somebody. To get an idea of how much voltage this circuit will drop under load: Vdrop = (load current / (360 * c )) * 42. C is in farads. So with a 0.1 amp load and 100µF caps: 116.67 V! Back to where you started at only a tenth of an amp and big caps. \$\endgroup\$ – evildemonic Dec 30 '19 at 19:55
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    \$\begingroup\$ "this element was designed to run on 240 volts AC at 30 amps...for all I know it could burn out at 240 volts DC." - no, DC or AC makes no difference to a resistor. electronics-tutorials.ws/accircuits/rms-voltage.html But where are you going to get >60A at 120V to power your circuit? \$\endgroup\$ – Bruce Abbott Dec 30 '19 at 20:06

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