A 10.6 microF capacitor in a heart defibrillator unit is charged fully by a 1200
ID: 2050305 • Letter: A
Question
A 10.6 microF 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 74.6 ms for the voltage to drop to 22.0 V.a)Find the time constant.
b)Determine the resistance, R.
c)How much time does it take for the capacitor to lose 82 % of its stored energy?
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
we have (a) u(t)=12000*exp(-t/RC) u(t)=22 for t=73ms => RC=0.012 =time constant (b)resistance R=RC/C=1092.9O (c) initial energy E=1/2*C*120002 after that: E'=0.13*E=1/2*C*u(t)2 => u(t)=4326.7 =>t=0.0118 s (d) energy delivered: E=1/2*C*120002=763.2 J (including thermal energy ofresistor)
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