A 12.0 F capacitor in a heart defibrillator unit is charged fully by a 10500 V p

Question

A 12.0 F capacitor in a heart defibrillator unit is charged fully by a 10500 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 75.0 ms for the voltage to drop to 21.5 V.

A) Find the time constant. (Express in ms)

B) Determine the resistance, R. (Express in k)

C) How much time does it take for the capacitor to lose 84% of its stored energy? (Express in ms)

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? (Express in J)

Explanation / Answer

Part A: for a discharging capacitor

final voltage = initial voltage * e^(-t/T)

21.5 = 10500 * e^(-75/T)

21.5/10500 = e^(-75/T)

take ln of both sides

-6.1911 = -75 / T

T = 12.114 ms is the time constant

Part B: R = T / C = 12.114 / 12.0 = 1.0095 kohm

Part C: to lose 84% of its stored energy... this means its final energy is only 0.16 times its initial. So the final voltage is only sqrt(0.16) = 0.40 times the initial voltage. And now...

final voltage = initial voltage e^(-t/T)

0.4 = e^(-t/T)

take ln...

-0.91629 = -t/T

t = 0.91629 T = 0.91629 * 12.114 = 11.10 ms

Part D: the energy delivered is the initial energy stored in the capacitor...

energy = (1/2) CV^2 = (1/2) * 12.0x10^-6 * 10500^2 =

= 661.5 Joules

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