A heart defibrillator passes 11.5 A through a patient\'s torso for 5.00 ms in an

Question

A heart defibrillator passes 11.5 A through a patient's torso for 5.00 ms in an attempt to restore normal beating. How much charge passed through the torso? Number C What voltage was applied if 535 J of energy was dissipated by the current in the patient's body? Number V What was the effective resistance of the patient's torso? Number Ohm Find the average resulting temperature increase of the 8.75 kg of affected tissue. Assume that the human body has a specific heat of about 3.50 times 10^3 J/(kgmiddotmiddot degree C). Number degree C

Explanation / Answer

a) Q = It

Q = 11.5 x 0.005 s = 0.0575 C.

b) W = VIt

V = 535 / 0.0575 = 9304.35 Volt

c) V = RI

R = V/I = 9304.35 / 11.5 = 809.07 ohm

d) T = 535 / 3500*8.75 = 0.0175

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