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Defibrillator: charge, capacitance, power and current problem

  1. Feb 17, 2009 #1
    1. The problem statement, all variables and given/known data
    A heart dysfunction that can cause death is ventricular fibrillation. This is an
    uncoordinated quivering of the heart as opposed to regular beating. An electric shock to
    the chest can cause momentary paralysis of the heart muscle, after which the heart will
    sometimes start organized beating again. A defibrillator is a device that applies a strong
    electric shock to the chest over a time of a few milliseconds. The device contains a
    capacitor of several microfarads, charged to several thousand volts. Electrodes called
    paddles, about 8 cm across and coated with conducting paste, are held against the chest
    on both sides of the heart. Their handles are insulated to prevent injury to the operator,
    who calls “Clear!” and pushes a button on one paddle to discharge the capacitor through
    the patient's chest.


    Assume that the capacitor in the defibrillator is 20.0 μF and is charged to 4,000 Volts.

    a. How much charge is stored in the capacitor before it is discharged?
    b. How much energy is released when the capacitor is discharged?
    c. If the capacitor completely discharges in 2.0 ms, what is the average current
    delivered by the defibrillator?
    d. What is the average power delivered?


    2. Relevant equations
    C= Q/V
    Energy stored= 1/2 C*V^2
    I = Q\t
    P=I*V


    3. The attempt at a solution

    a. How much charge is stored in the capacitor before it is discharged?
    C= Q/V
    Q = C*V = 20E-6 * 4000 = .08 C

    b. How much energy is released when the capacitor is discharged?
    Energy stored= 1/2 C*V^2 = 1/2 (20E-6)(4000^2 = 160 J

    c. If the capacitor completely discharges in 2.0 ms, what is the average current
    delivered by the defibrillator?
    I = Q/t = .08/.002 = 40 A

    d. What is the average power delivered?
    P = I*V = 40 (4000)= 1.6X10^5W


    I dont know if i did this problem right can someone help me
     
  2. jcsd
  3. Feb 18, 2009 #2

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    a, b, and c look good.

    A problem with d:
    V is 4000v only at the beginning of the discharge, and drops to 0V by the end of the discharge. So P=I*V=1.6X10^5W is only true at the beginning; the average power will be less.

    Do you know any other equations that deal with power?
     
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