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Defibulator physics problem

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data

    A defibrillator is used to restart a person's heart after it stops beating. Energy is delivered to the heart by discharging a capacitor through the body tissues near the heart. If the capacitance of the defibrillator is 9 ìF and the energy delivered is to be 300 J, to what potential difference must the capacitor be charged?

    2. Relevant equations

    U = 1/2C(delta V)^2.


    3. The attempt at a solution

    Could someone get me started on this problem?
     
  2. jcsd
  3. Feb 25, 2010 #2

    ideasrule

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    Re: Defibulators

    If you have U = 1/2C(delta V)^2 and you know U and C, why can't you calculate delta V?
     
  4. Feb 25, 2010 #3
    Re: Defibulators

    Hmm. I never had this problem before..

    I would use U = 1/2C(Delta V) ^2

    U / Delta V^2 = 1 / 2C

    1 / Delta V^2 = U / 2C

    Delta V^2 = 2C / U

    Delta V = sqrt( 2C / U )

    But what are C and U?

    C = 900J and U = 9iF??

    so V = sqrt( 2*900J / 9iF ) = 14.14 J/iF?
     
  5. Feb 25, 2010 #4
    Re: Defibulators

    That just doesnt seem right to me. I am not even sure if the equation is the right one to use.
     
  6. Feb 26, 2010 #5

    collinsmark

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    Re: Defibulators

    :bugeye:

    Perhaps you should try it this way,

    [tex] U = \frac{1}{2}CV^2 [/tex]

    [tex] 2U = CV^2 [/tex] (Multiplied both sides by 2)

    [tex] \frac{2U}{C} = V^2 [/tex] (Divided both sides by C)

    [tex] \sqrt{\frac{2U}{C}} = V [/tex] (took the square root)

    Wait, 900 J? I thought it was 300 J.

    In terms of the "iF" units, Are you sure it's not just "F"? Capacitors are measured in Farads. The abbreviation for a Farad is simply F. You can break down the Farad unit to be Coulombs/Volt.

    9 F is a fairly large capacitor. However, it is not out of the question. Using today's ultra-capacitor technology, you could hold a 9 F ultra-capacitor in the palm of your hand.

    By the way, when you finish calculating the voltage, it may not seem like very much. I'm assuming that there is a voltage converter involved in the defibrillator, not applicable to this problem (we're only dealing with the capacitor itself, not the entire defibrillator).

    [Edit]: It's still possible to store the same energy in a much smaller valued capacitor, albeit at a higher voltage. It could very well be that the author of the problem meant uF (as in micro Farad), since the 'u' is right next to the 'i' on the keyboard. But if it was up to me, I'd design the thing with a 9 F ultra-capacitor. Ultra-capacitors are cool. :cool:
     
    Last edited: Feb 26, 2010
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