(896) Problem 7: A heart defibrillator being used on a patient has an RC time co
ID: 777972 • Letter: #
Question
(896) Problem 7: A heart defibrillator being used on a patient has an RC time constant of 10.5 ms due to the resistance of the patient's body and the capacitance of the defibrillator Randomized Variables t-10.5 ms V-11.5 kV 50% Part a) If the defibrillator has an 8.5 capacitance, what is the resistance of the path through the patient in k? (You may neglect the capacitance of the patient's body and the resistance of the defibrillator.) Grade Summary Deductions Potential 100% cos0 cotanOasin s0 atan)acotan)sinh) cosh tanhOcotanhO Degrees ORadians tanOt ( acoso Submissions Attempts remaining: 10 Sin % per attempt) detailed view 0 END (0 CLEAR Submit Hint I give up! Hints: 1 % deduction per hint. Hints remaining: 2 Feedback: 0%-deduction per feedback. 50% Part (b) If the initial voltage is 11.5 kV, how long does it take to decline to 6.00 × 102 V in ms?Explanation / Answer
(a) Using the equation = RC we can calculate the resistance:
R = /C = 1.05*10^-2 s/8.5*10^-6F
= 1.235*10^3 = 1.23K
(b) Using the equation
V = Voe^-/RC we can calculate the time it takes for the
voltage to drop from 11.5 kV to 600 V :
= -RC ln(V/Vo)
= -(1.235*10^3)(8.5*10^-6F)*ln(600V/1.15*10^4V) = 3.10*10^-2 s = 31.0ms
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.