Question

A 11.6 μF 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 76.3 ms for the voltage to drop to 20.5 V .

time constant= 12.0 ms

a) what is resistance R=? K (ohm)?

b) how much time does it take for capactor to lose 86% of its stored energy? ( ms)

c) U 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)

Answer 1

Solyo C = 11.6 4F T = 12ms V 12000 V. ca) Time Constant , c = RC R= 듣 122103 11.6 X10-6 R = R = 1.03 xlops (b) In Discharging case, charge on capcitor is given as q=% ce tlRc take square both sides q² = que ce tre j² q? (stored Energy in capacitor) by 20 both sides q? So we divide 2 2C (et/Rc Uo (ě 24/Rc oj ec) U - 100-86 Vo = 100 Uo (e/Rc) - (724/e) 14 100 (:2=12ms) Vo = U. 14 Joo -2tle e 2t - Into) tento = ta (12. in , mas => t = 11.8 ms

(C) After the by many time constants, capacitor is completely discharged. and 9t deler delivered to tell Energy to the chest/ heart, which is stored the capacitor. So. Total stored Energy delivered Energy U= 1 68² U= 3(116X1000) (-2000) U = 835.2 J