A heart defibrillator passes 12.5 A through a patient?s torso for 5.00 ms in an

Question

A heart defibrillator passes 12.5 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 7.25 kg of affected tissue. Assume that the human body has a specific heat of about 3.50 x 10^3 J/(kg degreeC). Number degree C

Explanation / Answer

Q = i*t = 12.5*5e-3 = 0.0625 C

E = dV*q

dV = W/Q = 535/0.0625 = 8560 v <-----answer

R = dV/i = 8560/12.5   684.8 ohms

Q = E = m*C*dT

535 = 7.25*3.5e3*dT

dT = 0.02 oC

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.