A 11.0 capacitor in a heart defibrillator unit is charged fully by a 10500 power

Question

A 11.0 capacitor in a heart defibrillator unit is charged fully by a 10500 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 is the resistance of the body between the two paddles. Data indicate that it takes 73.1 for the voltage to drop to 19.5 .

1-How much time does it take for the capacitor to lose 86 of its stored energy?
2-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

Capacitance C = 11 * 10 ^ -6 F Maximum voltage V = 10500 V Time t= 73.1 ms = 0.0731 s Voltage after time t is V ’ = 19.5 V We know V ‘ = V [ 1- e –t/T ] V ' / V = [ 1- e –t/T ] 1.857 * 10 ^ -3 = [ 1- e –t/T] [ e –t/T ] = 1-( 1.857* 10^-3) =0.998 -t/T =-1.858 * 10 ^ -3 time constant T = t / ( 1.858 * 10 ^ 3 ) = 33324.97 ms (b). we know T = RC from this resistance R = T / C = 3.575 * 10 ^ 6 ohm = 3575 k ohm

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