| Author |
Message |
   
Anonymous
| | Posted on Wednesday, January 12, 2005 - 6:55 pm: |      |
now im thinking about building a van de graaff to power an electric motor, but will that extra voltage give the motor any more "umph" and will the VDG be able to give a constant current? |
   
Simon Quellen Field (sfield)
New member Username: sfield
Post Number: 110 Registered: 12-2004
| | Posted on Wednesday, January 12, 2005 - 7:05 pm: |      |
No. You lose an enourmous amount of power by converting since the VDG is inefficient, and high voltage motors are inefficient. The electric motor that is running the VDG will always get more power than the electric motor run by the VDG. |
   
Anonymous
| | Posted on Wednesday, January 12, 2005 - 8:52 pm: |      |
makes sense, but aside from the VDG, Ive seen these "voltage multipliers" on the internet involving capacitors and resistors and I have no idea how they work since there is a lot of mumbo-jumbo thats too advanced for me... ever heard of them and how they work? |
   
Simon Quellen Field (sfield)
New member Username: sfield
Post Number: 114 Registered: 12-2004
| | Posted on Thursday, January 13, 2005 - 1:37 pm: |      |
People often confuse voltage with power. Voltage is just the pressure on the electrons. Power is voltage times current. Current is how many electrons per second are going through the wire. A voltage multiplier is like a nozzle on a hose. The pressure is increased, but the power is almost the same, because the water is coming out of a smaller hole. I say "almost the same" because there are always losses (usually due to heat) when converting. To understand a voltage multiplier made from capacitors and diodes, consider two capacitors in parallel, being charged from a battery. They both get charged to 1.5 volts from the 1.5 volt battery. Now remove them, and connect them in series. Now you have 3 volts. However, the power is the same, since two capacitors in parallel can deliver twice the current as two capacitors in series. |
   
Anonymous
| | Posted on Thursday, January 13, 2005 - 10:16 pm: |      |
ahhh, so since P = V x I, it is just swapping the current for voltage so it can give a better show (with the sparks and all)? and if so, is there actually any way to mulitply power? |
   
Simon Quellen Field (sfield)
New member Username: sfield
Post Number: 117 Registered: 12-2004
| | Posted on Friday, January 14, 2005 - 12:28 am: |      |
Sure. Just do the same work in less time. Charge a capacitor slowly, then discharge it all at once. Note that the total energy is the same (minus the normal losses in real systems). You can't create energy, you can only change the type. |
   
lysdexia Unregistered guest
| | Posted on Saturday, March 5, 2005 - 10:30 pm: |      |
Voltage is like pressure, but it certainly isn't pressure! Voltage is another creature entirely; I would call it "demand" because it's short of a mechanically-meaningful property. E/q Fs/q mas/q = kQs/S2 Scientists and engineers are careless and call voltage a pressure, force (electromotive), or potential when it's nothing of the sort. They're also careless and dumb when they cancel both spatial variables in the potential expression. :P |
   
Simon Quellen Field (sfield)
Senior Member Username: sfield
Post Number: 274 Registered: 12-2004
| | Posted on Tuesday, April 12, 2005 - 10:22 pm: |      |
Voltage is pressure. When you put two electrons together, they repel one another. When you charge two metal plates with electrons, they repel each other. The pressure you feel as they repel one another is voltage. It is a direct result of the electrons being crowded together. When the electrons are less crowded, there is less pressure, and thus less voltage. This is exactly the same kind of pressure that gas molecules undergo when they are crowded together. |