A 10.8uF capacitor in a heart defibrillator unit is charged fully by a 10000V po
ID: 2053298 • Letter: A
Question
A 10.8uF capacitor in a heart defibrillator unit is charged fully by a 10000V 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, R where is the resistance of the body between the two paddles. Data indicate that it takes 72.8ms for the voltage to drop to 21.0V .
A)Find the time constant.
t= m/s
B)Determine the resistance
R= kohm's
C)How much time does it take for the capacitor to lose 81% of its stored energy?
Explanation / Answer
21 = 10000(e-t/RC)
t = 72.8*10^-3
calculating R = 1093.24 ohms
tou = time constant = RC = 1093.24 * 10.8*10^-6 = 0.0118
Energy = 1/2 C V^2
V^2 = V^2 o ((e-2t/RC))
0.81 = e-2t/RC
t = 1.24*10^-3 sec = 1.24ms
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