3. (8 points) Automated External Defibrillator (AED) The immediate cause of many

Question

3. (8 points) Automated External Defibrillator (AED) The immediate cause of many deaths is ventricular defibrillation, an uncoordinated quivering of the heart, as opposed to proper beating. An electric shock to the chest can cause momentary paralysis of the heart muscle, after which the heart will sometimes start coordinated beating again. A defibrillator is a device that applies a strong electric shock to the chest over a time of a few seconds. The device contains a capacitor of a few microfarads, charged to several thousand volts. Electrodes called paddles, about 8 cm across and coated with conducting paste, ar held against the chest on both sides of the heart. Their handles are insulated to prevent injury to operator, who calls, "Clear and pushes a button on one paddle to discharge the capacitor through the patient's chest. (Source: Serway, R.A. and Vuille, C., 2007, Essentials of College Physics, Brooks/Cole, Belmont CA, p. 444, Chap. 16 A defibrillator passes a brief burst of current through the heart to restore normal beating. In one such defibrillator, a 50.0-uF capacitor is charged to 6.0 kV. Paddles are used to make an electric connection to the patient's chest. A pulse of current lasting 1.2 ms partially discharges the capacitor through the patient. The electrical resistance of the patient (from paddle to paddle) is 240 (a) What is the initial energy stored in the capacitor? (b) How much charge is on each plate of the fully charged capacitor? (c) How much energy is dissipated in the patient during the 1.2 ms? (Hint: Use the known values of R and C to calculate the time constant t.) (d) what average power is delivered to the patient during the 1.2 ms?

Explanation / Answer

a)

initial energy stored in the capacitor = 0.50 * C * V^2

initial energy stored in the capacitor = 0.50 * 50 *10^-6 * 6000^2

initial energy stored in the capacitor = 900 J

b)

charge on each plate = C * V

charge on each plate = 50 *10^-6 * 6000

charge on each plate = 0.30 C

c)
time constant , T = 50 *10^-6 * 240

T = 0.012 s = 12 ms

final charge is Q

Q = 0.30 * (e^(-1.2/12))

Q = 0.271

energy remaining = 0.271^2 * 0.50/(50 *10^-6) = 736 J

energy dissipated = 900 - 736 = 164 J

d)

average power = energy/time

average power = 164/0.0012

average power = 1.37 *10^5 W

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