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Ultracapacitors - How long until the potential difference drops to 6V

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Compact "ultracapacitors" are with capacitance values up to several thousand farads are now commercially available. One application for ultracapacitors is in providing power for electrical circuits when other sources (such as a battery) are turned off.

    Assume a 1200-F ultracapacitor is initially charged to 12.0 V by a battery and is then disconnected from the battery. If charges is then drawn off the plates of this capacitor at a rate of 1.0mC/s. say, to power backup memory of some electrical gadget, how long (in days) will it takes for the potential difference across the capacitor to drop to 6.0 V?

    2. Relevant equations

    V(t)=V(0)e^(-t/t-constant)
    t-constant=RC
    I=Q/t
    R=V/I

    3. The attempt at a solution

    I=0.001/1=0.001
    R=12/0.001=12000
    t-constant=12000*1200=14400000
    ln(6/12)*14400000=-t
    t=9981319.4
    9981319.4 seconds = 115.52 days
     
  2. jcsd
  3. Mar 21, 2009 #2

    LowlyPion

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    Re: Ultracapacitance

    Isn't it a little simpler than that?

    Qo = Vo*C

    The charge is drawn at a constant rate. At Vo/2 then isn't there Qo/2 .

    Isn't then ΔQ/I = (Qo - 1/2Qo)/I = Δt ?
     
  4. Mar 21, 2009 #3
    Re: Ultracapacitance

    Thank you.
     
  5. Mar 21, 2009 #4
    Re: Ultracapacitance

    Where exactly was I going wrong above then?
     
  6. Mar 22, 2009 #5

    LowlyPion

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    Re: Ultracapacitance

    As it reads to me there is a constant current drain. You have so much water in the bucket and you are taking it out at a constant rate.
    i.e the I is constant and not a function of t.

    The equation you are using is a solution for the differential equation of

    C*dV/dt + V/R = 0

    But V/R in this equation is the current and is a function of t.

    Except in this problem V/R is a constant Io.

    Substituting Q/C for V and V/R with Io the equation reduces to a simpler statement that dQ/dt = -Io
     
  7. Mar 22, 2009 #6
    Re: Ultracapacitance

    If two of these capacitors are placed in a series, how long (in days) would it take? In parallel?

    I can calculate the two equivalent resistances for this question, but are they even really needed?
     
  8. Mar 22, 2009 #7

    LowlyPion

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    Re: Ultracapacitance

    I'd say what you really want to calculate is the equivalent capacitance. That tells you how big a bucket of charge you have to draw from.
     
  9. Mar 22, 2009 #8
    Re: Ultracapacitance

    That's what I meant to say. Once I have the two equivalent capacitances, what equations do I plug them into?
     
  10. Mar 22, 2009 #9

    LowlyPion

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    Re: Ultracapacitance

    Has Q = V*C changed to determine total charge as before?
     
  11. Mar 22, 2009 #10
    Re: Ultracapacitance

    Yes, it has changed. I realize that, but what equation am I missing that relates all of these elements to time? In my notes from class, I recorded that, when something is discharging... V(t)=V(0)e(-t/t-constant)

    Would I find a new initial voltage through Q=VC and plug it into that equation I just mentioned?
     
  12. Mar 22, 2009 #11
    Re: Ultracapacitance

    Okay. So I got the correct answer by finding the new Q and plugging it into that equation you gave me for the first part of the question. However, I still do not know where you derived that equation from. My class is strictly algebra-based and we have not even mentioned differentials yet.

    ΔQ/I = (Qo - 1/2Qo)/I = Δt
     
  13. Mar 22, 2009 #12

    LowlyPion

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    Re: Ultracapacitance

    Since I is defined to be

    I = ΔQ/Δt ...

    And by Q = V*C you know that when V(t) = Vo/2 , that Qo necessarily = Qo/2 ...
     
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