Multi concept Exercise 18.56 A 11.4 muF capacitor in a heart defibrillator unit

Question

Multi concept Exercise 18.56 A 11.4 muF capacitor in a heart defibrillator unit is charged fully by a 12000 V power supply. Each capacitor plate is connected to the chest of a patient by wires and flat "paddles," one on either side of the heart. The energy stored in the capacitor is delivered through an RC circuit, where R is the resistance of the body between the two paddles. Data indicate that it takes 75.7 ms for the voltage to drop to 18.5 V . Part C How much time does it take for the capacitor to lose 89 % of its stored energy? Part D If the paddles are left in place for many time constants, how much energy is delivered to the chest/heart area of the patient?

Explanation / Answer

v = Vo*(e^-(t/Rc))

18.5 = 12000*e^-(0.0757/(R*11.4*10^-6))

R = 1025.55 ohms

energy , U = 0.5*C*Vo^2

U = 0.5*c*v^2

U = 0.5*c*vo^2*e^-2t/Rc

U = Uo*e^-2t/RC

U = uo-0.89 Uo

1-0.89 = e^-(2t/(1025.55*11*10^-6))

t = 12.4 ms

+++++

D)

U = 0.5*c*vo^2 = 0.5*11.4*10^-6*12000^2 = 820.8 J

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