The immediate cause of many deaths is ventricular defibrillation, an uncoordinat
ID: 2084166 • Letter: T
Question
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 milliseconds. 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, are 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-mu F 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 ohm. (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? (d) What average power is delivered to the patient during the 1.2 ms?Explanation / Answer
a) Given C
C=50micro farads
V=6K volts
The energy stored in the capacitor is given by
E=1/2*C*V2
=0.5*50*10-6*6*103*6*103
= 900 joules
b) The charge on the capacitor is given by
Q = C*V
= 50*10-6 *6*1000
=300 micro coloumb
C) the time constant is given by t=R*C=240*50*10-6
t=12 milli seconds
the voltage across capacitor for 1.2 milli seconds is given by
vc=vs(1-e(-t/RC))
where vs=6k volts and here t=1.2 milliseconds and RC=12 milliseconds
vc = 0.571KV
The energy dissipated is calculated as Ed =1/2*C*vc*vc
Ed = 8.15 joules
d) average power dissipated for 1.2 ms is = energy / time
Ed / 1.2
= 8.15 / 1.2*10-03
= 6.79 K watts
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