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Van de graaff generator as a current source

  1. Nov 19, 2014 #1
    I know that van de graaff generator is a high voltage generator, i was wondering though if we can use it as a current source, if we get any resistor, connect the negative terminal to the generator "if the generator generates negative charge " then ground the positive terminal, then any charge that appears on the sphere will be immediately pulled to ground, creating a current, using simple formulae we can derive the current, is that feasible or not?
     
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  3. Nov 19, 2014 #2

    phinds

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    Sure, it's feasible as far as I can see but it is a HUGE waste of electricity to do that since the amount of current you'll get is tiny vs the amount you have to use to get it.
     
  4. Nov 19, 2014 #3
    Yea i know, i calculated it, and the efficiency was extremely low
     
  5. Nov 19, 2014 #4
    Just wanted to know if it would work, gonna use it to measure how much time does my generator need to reach maximum voltage
     
  6. Nov 19, 2014 #5

    phinds

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    I don't get how you could do that.
     
  7. Nov 19, 2014 #6
    My plan is by grounding the generator and measuring the current in the wires i can use it to know the time to reach the maximum voltage, can't write the formula now because i am using the mobile app
     
  8. Nov 19, 2014 #7
    Assuming I=const the formula is simply t=Vmax⋅C/I
     
  9. Nov 19, 2014 #8
    I don't know yet if I is constant, but theoretically, it should be
     
  10. Nov 19, 2014 #9

    phinds

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    I still don't get it. If you ground the wires, or run the output of the generator through a resistor, you won't GET any voltage on the sphere. I think you must be talking about something that I'm not following.
     
  11. Nov 20, 2014 #10
    That's the whole idea, i dont need any voltage, by grounding the sphere, the generator became a current source, so i measure the current in the wire, or my case i will connect a high resistance and measure the voltage across it because the current will be low and hard to measure.
    The maximum voltage of a van de graaff generator is 3*10^6*radius(in meter), and the voltage on the sphere is kQ/r where K is coulomb's constant
    Q=I*t and we're done
     
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