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Relationship between volts and charge

  1. Jun 29, 2006 #1
    Hi,

    I have been given an assignment question detailing a circuit with a few resistors in as well as a capacitator.

    Further more the information given was the resitance of the resitors and that the battery had had a potential difference of 9 volts.

    The question's part a was workign out the current in the circuit - no problem

    The problem however comes in at part b with a question on what the capacitance of the capacitator is.

    Now from my understanding of capacitance you need to know the charge and use the forumal C=Q/V but I'm missing Q.

    I know V = joules per coulomb but I'm not seeing the relationship to find out what Q is.

    Any pointers would be greatly appreciated.

    Thanks

    Jacques
     
  2. jcsd
  3. Jun 29, 2006 #2

    berkeman

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    Staff: Mentor

    If the problem is referring to the DC situation after all voltages and currents have settled to their final values, then I agree that it sounds like you don't have enough information to solve for C. If they give you some time-related information, like you close a switch at time t=0 and by time t=5us the voltage across the capacitor has risen to half of its final value.... or something like that, then you can use the exponential rise property of an RC circuit to figure out C.

    Could you post the exact question, along with the circuit diagram?
     
  4. Jun 29, 2006 #3

    Kurdt

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    Staff Emeritus
    Science Advisor
    Gold Member

    current is the rate of flow of charge
     
  5. Jun 29, 2006 #4
    Thanks Berkeman,

    In fact they do give me a chart showing the how the current falls from an arbitary 10 amps to 2 amps in 2.5 miliseconds. the question based on the this asks for the time constant.
    This question then is followed by the capacitance question.

    thanks j
     
  6. Jun 29, 2006 #5
    Berkeman,

    Thanks for your input, I've got it now, I can read the time constant from the decay graph (time where 36.8% is left) and then I can work out the capacitance

    thanks, again

    j
     
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