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RC Circuit defibrillator Question

  1. Nov 11, 2013 #1
    1. The problem statement, all variables and given/known data

    A 140μF defibrillator capacitor is charged to 1500 V. When fired through a patient's chest, it loses 95% of its charge in 40 ms. What is the resistance of the patient's chest?

    2. Relevant equations



    3. The attempt at a solution
    According to "Mastering Physics," the solution that I have provided is incorrect. I don't know what I've done wrong.

    Given: C=140μF V=1500 V
    -loses 95% of its charge in 40 ms

    Needed: R=?

    Q=Qoe-t/τ
    (Q/Qo)=e-t/τ
    ln(Q/Qo)= -t/τ
    τ=RC
    ln(Q/Qo)=-t/RC
    Rearrange for R
    R=-t/(C(ln(Q/Qo))
    =-(40x10-3s)/((140 x10-6F)(ln(0.95)))
    =5.6kΩ
     

    Attached Files:

    Last edited: Nov 11, 2013
  2. jcsd
  3. Nov 12, 2013 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The solution looks correct, try to use more significant digits.

    ehild
     
  4. May 21, 2017 #3
    Sorry that I can't help you, but this is for any future students coming here for a similar problem. The OP did almost everything right. His only problem was that since the capacitor loses 95% of its charge, the final equation should include ln(0.05) for the 5% of the charge remaining, not ln(0.95). Hope this helps!
     
    Last edited: May 21, 2017
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