A) A heart defibrillator passes 10.5 A through a patient\\\'s torso for 5.00 ms

Question

A) A heart defibrillator passes 10.5 A through a patient's torso for 5.00 ms in an attempt to restore normal beating. How much charge passed through the torso? Answer in C.

B) What voltage was applied if 525 J of energy was dissipated by the current in the patient's body? Answer in V.

C) What was the effective resistance of the patient's torso? Answer in Ohms.

D) 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 × 103 J/(kg·°C). Answer in Degrees Celsius.

Explanation / Answer

A) I = q / t

10.5 = q / (5 x 10^-3)

q = 0.0525 C

B) P = VI

and energy = Pt = VIt

525 = V* 10.5*5*10^-3

V = 10,000 V

C) V = IR

R = V/ I = 952.38 ohm

D) Q = m c dletaT

525 = 8.75 x 3.50 x 10^3 x deltaT

deltaT = 0.017 deg C

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