A defibrillator passes a brief burst of current through the heart to restore nor

Question

A defibrillator passes a brief burst of current through the heart to restore normal beating. In one such defibrillator, a 40.5 µF capacitor is charged to 6.1 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 278 .

(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.

p source/ p patient =

Explanation / Answer

(a)

energy stored in the capacitor is

E = 1/2 * CV^2 = 0.5 ( 40.5 * 10^-6)( 6.1 * 10^3 )^2 = 753.5 J

(b)

I = V/R = 6.1 * 10^3 V/278 = 21.94 A

(c)P = E/t = 753.5 J/1 * 10^-3 s = 753500 W

(d)

P_source = E/t =  753.5 J/2.2 s = 342.5 W

P_patinet = E/t = 753.5 J/1 * 10^-3 s = 753500 W

p source/ p patient =4.54* 10^-4

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