Chapter 20: Q20.6-92PE (page 736)
A heart defibrillator passes \(10.0{\rm{ }}A\) through a patient’s torso for \(5.00{\rm{ }}ms\) in an attempt to restore normal beating.
(a) How much charge passed?
(b) What voltage was applied if \(500{\rm{ }}J\) of energy was dissipated?
(c) What was the path’s resistance?
(d) Find the temperature increase caused in the \(8.00{\rm{ }}kg\) of affected tissue.
Short Answer
- Passing through patient's torso the charge is obtained as: \(q{\rm{ }} = {\rm{ }}50.0{\rm{ }}mC\).
- To the patient the voltage applied is obtained as: \(\Delta V{\rm{ }} = {\rm{ }}10.0{\rm{ }}kV\).
- The resistance of the path is obtained as: \(R{\rm{ }} = {\rm{ }}1.00{\rm{ }}k\Omega \).
- Caused in the affected tissue the temperature increased is obtained as: \(\Delta T{\rm{ }} = {\rm{ }}1.79{\rm{ }} \times {\rm{ }}{10^{ - 2{\rm{ }}o}}{\rm{ }}C\).