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

Question

Map d 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? Number What voltage was applied if 525 J of energy was dissipated by the current in the patient's body? Number What was the effective resistance of the patient's torso? Number Find the average resulting temperature increase of the 7.75 kg of affected tissue. Assume that the human body has a specific heat of about 3.50 × 103 J/(kg·). Number

Explanation / Answer

given

i = 10.5 A , dt = 5.0 ms = 5 X 10-3 sec

a )

to find the charge

using Q = i X dt

Q = 10.5 X 5 X 10-3

Q = 0.0525 C

b )

the energy is applied of W = 525 J

we have P = W / t

P = 525 / 5 X 10-3

P = 105000 watts

even we have P = V X i

V = P / i

V = 105000 / 10.5

V = 10000 volts

c )

to find resistance R

using ohms law

we have V = i X R

10000 = 10.5 X R

R = 10000 / 10.5

R = 952.38 ohms

d )

given mass 7.75 Kg = m

specific heat 3.50 X 103 J/Kg.0C = C

Q = 0.0525 C

using

Q = m X C X dt

0.0525 = 7.75 X 3.5 X 103 X dt

dt = 0.0525 / 7.75 X 3.5 X 103

dt = 0.0525 / 27125

dt = 1.93 X 10-6 sec

or

the average temperature increase is 1.93 microseconds

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