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Energy in a Capacitator

  1. Mar 15, 2016 #1
    1. The problem statement, all variables and given/known data
    A 2.8 µF capacitor is fully charged to a value of 0.009 C. A 1.5 × 10^6 Ω resistor is connected to this capacitor so that it begins to discharge. One second after the capacitor has begun to discharge, how much energy is stored in the capacitor? Answer in units of J.

    2. Relevant equations
    Q=CV

    Vo*(e^(-t/RC))=Vf

    1/2C(Vf)^2 = E

    3. The attempt at a solution
    .009C / 2.8*10^-6 F = 3214.28571429 V

    3214.28571429V * e^-1 = 1182.4696323 V

    (1/2)* (2.8*10^-6 F) * (1182.4696323)^2 = 1.95752820396 J

    What I did was was divide .009 C by 2.8 * 10^-6 F to get the initial voltage then multiply by e^-1 to get the voltage in the capacitor. I then took that value, squared it, and multiplied it by 1/2 and 2.8 *10^-6 for the value of the capacitance. My final answer was 1.95752 J. I don't know what I did wrong.
     
  2. jcsd
  3. Mar 15, 2016 #2

    mfb

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    Staff: Mentor

    Where does that value come from?
     
  4. Mar 15, 2016 #3
    Well in the solution to the problem, the description says "the RC time constant is one second so that after one second the charge has fallen to Q(1s) = C((∆Vinit)e^ −1) ." We always work it like that in class because seconds is interchangeable with RC units.
     
  5. Mar 15, 2016 #4

    mfb

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    Staff: Mentor

    I don't know what the solution says, but 2.8*1.5 is never anything like 1.
    2.8 µF*1.5 MΩ = 4.2 seconds.
     
  6. Mar 15, 2016 #5
    I'm sorry, I'm not really understanding why you did that.
     
  7. Mar 15, 2016 #6
    Okay never mind I understand why you did that, but I was wondering why that would work because in the sample problem the values were 10µF, 100kΩ, and 1 s and they still used e^-1.
     
  8. Mar 15, 2016 #7
    NEVER MIND I'M STUPID k= 10^3 not 10^4! THANKS SO MUCH!
     
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