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Homework Help: Find the capacitance

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data
    The flash unit in a camera uses a special circuit to “step up” the V = 3.1 V from the batteries to V' = 300 V, which charges a capacitor The capacitor is then discharged through a flashlamp. The discharge takes t = 12 μs, and the average power dissipated in the flashlamp is P = 12 W. What is the capacitance of the capacitor?

    2. Relevant equations
    Energy = power * time
    energy stored in a capacitor = CV2/2

    3. The attempt at a solution
    Energy dissipated by the flashlamp = Pt
    This energy comes from the decreasing energy stored in the capacitor = CV’2/2 – CV2/2
    so Pt = CV’2/2 – CV2/2
    C = 2Pt/(V’2 – V2) = 3.2*10-9 F

    Anything wrong?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 16, 2009 #2


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    Homework Helper

    Your basic reasoning appears to be sound.
  4. Jun 17, 2009 #3
    but the website says ir is incorrect!
    I don't know why.
  5. Jun 17, 2009 #4


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    Homework Helper

    What's "ir"? I thought you were finding the capacitance.

    Anyways, if the capacitor completely discharges (which it does), then in your equation:

    CV’2/2 – CV2/2

    V would be zero, not the battery's voltage.
  6. Jun 17, 2009 #5
    The question is on www.masteringphysics.com
    You're right, V = 0, but it won't affect the numerical result (only 2 significant figures needed).
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