Multiconcept Exercise 18.56 Part A A 10.0 F capacitor in a heart defibrillator u

Question

Multiconcept Exercise 18.56 Part A 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 76.1 ms for the voltage to drop to 18.0 V Find the time constant. TE ns Submit My Answers Give Up Part B Determine the resistance, R Submit My Answers Give Up Part C How much time does it take for the capacitor to lose 80 % of its stored energy? 80 ns Submit My Answers Give Up 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? Submit My Answers Give Up

Explanation / Answer

Formula for a RC Discharging Circuit. Solve for time constant.
v = V_i e^( -t / )
v / V_i = e^( -t / )
-t / = ln (v / V_i)
= -t / ( ln (v / V_i) ) = -76.1ms / ( ln (18.0V / 10,000V ) ) = 12 ms Ans.


Time constant.
= R C
R = / C = 12 ms / 10 F = 1.2 k Ans.

U = ½ C V² = ½ * 10F * (10kV)² = 500J

So capacitor loses 80% of energy. 20% is left over. We have to determine energy, work back to find voltage and then the time.

U_new = 20% of U = 0.2 * 500J = 100 J

U_new = ½ C V²
V = ( 2 U_new / C ) = ( 2 * 100J / 10F ) = 4472 V

v = V_i e^( -t / )
v / V_i = e^( -t / )
-t / = ln (v / V_i)
t = - ln (v / V_i) = -12ms * ln (4472V / 10,000V) = 9.66ms = 10ms Ans.

If paddles are left in place 500J of energy will be delivered to the patient

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