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Understanding Resistance

  1. Oct 24, 2009 #1
    Hi

    have a question:

    When I connect a PM DC generator to an ultracapacitor and begin charging the ultracapacitor is there more resistance from the outset with the resistance dropping proportional to the voltage?

    Thanks
    Colin
     
  2. jcsd
  3. Oct 24, 2009 #2
    I don't think so. The capacitor will have a very small resistance, and the current is probably limited in some other way, by a series resistor or the current limiting in your power supply.
     
  4. Oct 24, 2009 #3
    You should connect a PM DC generator (is it unregulated?) to an ultra capacitor with a series diode rectifier, or else the generator will become a motor when you stop cranking. When you start cranking, the ultra cap will start drawing current, even when the generator voltage output is very low. As the ultra cap charges, the threshold voltage for torque resistance will rise, up until when the ultra cap is fully charged.
    Bob S
     
  5. Oct 24, 2009 #4

    vk6kro

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    You mean "resistance" as in how hard it is to turn the handle of the generator?

    The capacitor will draw more current when it is first being charged and this makes the generator harder to turn.
    As the capacitor charges up to the voltage being put out by the generator, the current decreases and the generator gets easier to turn.

    Some hand cranked DC generators can generate 100 volts or more which may be more than your capacitor can handle. If your generator looks old, this might be the case.
     
  6. Oct 25, 2009 #5
    Thanks everyone for making that very clear. Have a couple of other questions if oyu don't mind regarding charging capacitors and batteries from DC generators:

    So I have a DC generator connected to a 15v 52F ultracapacitor. There is a diode in place. If the generators has in inherent 1.7ohm resistance and the capacitor is at 10v. Does this mean that voltage produced by the generator would need to be the sum of the (voltage needed for internal generator resistance) + (10v to charge the capacitor) + (losses)? in order to charge up the capacitor? Or have I got it completely wrong!

    Hope the question is clear.

    And again thanks for helping my understand.
    Colin
     
  7. Oct 25, 2009 #6

    vk6kro

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    Science Advisor

    So I have a DC generator connected to a 15v 52F ultracapacitor. There is a diode in place. If the generators has in inherent 1.7ohm resistance and the capacitor is at 10v. Does this mean that voltage produced by the generator would need to be the sum of the (voltage needed for internal generator resistance) + (10v to charge the capacitor) + (losses)? in order to charge up the capacitor? Or have I got it completely wrong!

    The capacitor will charge if the generator is producing more than the capacitor voltage plus the diode voltage. The diode voltage will be about 0.6 volts to 1 volt depending on the current.

    The internal resistance of the generator limits the current that can flow into the diode/ capacitor combination.

    So, if the capacitor is already at 10 volts and the generator is producing 14 volts, there will be 10 volts across the capacitor, 0.6 volts across the diode and 3.4 volts across the internal 1.7 ohm resistor. (This adds up to 14 volts.)

    This last bit tells you the current will be about 2 amps, ( 3.4 V /1.7 ohms = 2 amps) so the diode needs to be a big one.
    When the capacitor was completely discharged, the current would have been about 7.9 amps. (14 -0.6) / 1.7 = 7.9 amps
     
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