A 11.2 F capacitor in a heart defibrillator unit is charged fully by a 11000 V p

Question

A 11.2 F capacitor in a heart defibrillator unit is charged fully by a 11000 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 74.4 ms for the voltage to drop to 21.0 V .

Part A: Find the time constant.

Part B: Determine the resistance, R.

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

Discharging voltage is given by :

V = Vo*e^(-t/RC)

here, Vo = 11000 V

V = 21 V

C = 11.2*10^-6 F0

So, 21 = 11000*e^(-0.0744/(R*11.2*10^-6))

So, R = 1061 ohm <-------- answer(b)

time constant , T = RC = 1061*11.2*10^-6 = 0.0119 s <------- answer(a)

c)

U = 0.5*CV^2

So, U'/U = (V'/V)^2

So, (0.11*U/U) = (V'/V)^2

So, V' = 0.332*V

So, 0.332*Vo = Vo*e^(-t/(11.2*10^-6*1061))

So, t = 0.0131 s <-------- answer

d)

U = 0.5*CV^2

= 0.5*(11.2*10^-6)*(11000^2)

= 677.6 J <------ answer

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