A 10.0 F capacitor in a heart defibrillator unit is charged fully by a 10000 V p
ID: 1424063 • Letter: A
Question
A 10.0 F capacitor in a heart defibrillator unit is charged fully by a 10000 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 71.0 ms for the voltage to drop to 22.0 V .
A) Determine the time constant (ms)
B)Determine the resistance (ohm)
C)How much time does it take for the capacitor to lose 87 % of its stored energy? (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? (J)
Explanation / Answer
1) V = Vo e^(-t / T) => 22 = 10000 e^- (0.071 / T)
=> 0.0022 = e^- (0.071 / T)
take (ln) for both sides u obtain :
-6.12 = - 0.071 / T
hence T = 0.0116 s = 11.6 ms
2) T = RC , => R = 0.0116 / (10 x 10^-6) = 1160 ohm = 1.16 Kohm
3) the initial stored energy = 1/2 C (Vo)^2 = 1/2 (10 x 10^-6) (10000^2) = 500 J
so its 87% = 0.87 x 500 = 435 J = 1/2 C V^2
=> V = 9327.4 = 10000 e^(- t / 0.0116)
solve for t :
t = 0.808 ms
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