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Homework Help: Capacitance of capacitor

  1. Feb 8, 2010 #1
    1. The problem statement, all variables and given/known data
    A capacitor, previously charged by a 1000 V source, is discharged through a photographic flash unit. The capacitor delivered an average power of 1000 W and the flash lasts 0.040 s. What is the estimated capacitance of the capacitor

    a. 40 x 10-6 F
    b. 20 x 10-6 F
    c. 60 x 10-6 F
    d. 80 x 10-6 F
    e. 80 x 10-3 F


    2. Relevant equations
    Q = CV
    W = 1/2 QV
    P = VI
    Q = I t


    3. The attempt at a solution
    I got two different answers....

    1.
    P = W/t = CV2 / 2t

    C = 2 Pt / V2 = 80 x 10-6 F (d)


    2.
    P = VI = VQ/t = CV2 / t

    C = 40 x 10-6 F (a)

    Something's definitely wrong with my concept but I don't know what...

    Thanks
     
  2. jcsd
  3. Feb 8, 2010 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    First, find the energy stored in the capacitor. (hint: an average power of 1000 J/sec for .04 seconds provides a total energy of ____ J.).

    Then use your formula for stored energy in the capacitor (E = CV^2/2) to determine the capacitance.

    Your first answer was correct. In your second attempt, you were assuming that the voltage is a constant 1000 V as the capacitor discharges. It decreases as charge decreases.

    AM
     
  4. Feb 8, 2010 #3
    Hi AM

    Thanks a lot !! :smile:
     
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