A defibrillator passes a brief burst of current through the heart to restore nor
ID: 1529689 • Letter: A
Question
A defibrillator passes a brief burst of current through the heart to restore normal beating. In one such defibrillator, a 47.0µF capacitor is charged to 6.4 kV. Paddles are used to make an electrical connection to the patient's chest. A pulse of current lasting 1.0 ms partially discharges the capacitor through the patient. The electrical resistance of the patient (from paddle to paddle) is 236 .
(a) What is the initial energy stored in the capacitor?
____ J
(b) What is the initial current through the patient?
____ A
(c) How much energy is dissipated in the patient during the 1.0 ms?
_____ J
(d) If it takes 2.2 s to recharge the capacitor, compare the average power supplied by the power source with the average power delivered to the patient.
Psource/Ppatient = ____
(e) Referring to your answer to part (d), explain one reason a capacitor is used in a defibrillator
PLEASE SHOW WORK.
Explanation / Answer
1)
Energy stored = CV^2/2 = 47 x 10^-6 F*(6.4 x 10^3 v)^2 / 2 = 962.56 J.
2)
Current = Qo / CR = CV /CR = V/R = 6.4Kv / 236 = 27.12 A
3)
Charge remaining after t s., Q = CV e^ -[1/CR]t
Q = 0.3008*e^ - 90.15t . When t = 1ms,
Q = ln 0.3008 - 0.09015 l
Q = 1.011 C
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