% of Assignment value) A heart defibrillator being used on a patient has an RC time constant of 8.5 ms due to the resistance of the patient's body and the capacitance of the defibrillator. Randomized Variables T = 8.5 ms C=8.5 μF V = 14 kV Part (a) If the defibrillator has an 8.5 uF capacitance, what is the resistance of the path through the patient in ks2? (You may neglect the capacitance of the patient's body and the resistance of the defibrillator.) R=1 Grade Summary Deductions 4% 96% Potential Submissions Attempt(s) Remaining: 2 4% Deduction per Attempt detailed view sin() cotan() cos() tan() T007 9 8 HOME asin() acos() E^^4 5 6 atan() acotan() sinh() * 1 2 3 1 cosh() tanh() cotanh() + - 0 END ● Degrees Radians BACKSPACE DEL CLEAR Submit Hint Feedback I give up! 2 Submission(s) Remaining Hints: 4% deduction per hint. Hints remaining: 2 Part (b) Feedback: 5% deduction per feedback. If the initial voltage is 14 kV, how long does it take to decline to 6.00 x 10² V in ms? 4% % of Assignment value) A heart defibrillator being used on a patient has an RC time constant of 8.5 ms due to the resistance of the patient's body and the capacitance of the defibrillator. Randomized Variables T = 8.5 ms C=8.5 μF V = 14 kV Part (a) If the defibrillator has an 8.5 uF capacitance, what is the resistance of the path through the patient in ks2? (You may neglect the capacitance of the patient's body and the resistance of the defibrillator.) R=1 Grade Summary Deductions 4% 96% Potential Submissions Attempt(s) Remaining: 2 4% Deduction per Attempt detailed view sin() cotan() cos() tan() T007 9 8 HOME asin() acos() E^^4 5 6 atan() acotan() sinh() * 1 2 3 1 cosh() tanh() cotanh() + - 0 END ● Degrees Radians BACKSPACE DEL CLEAR Submit Hint Feedback I give up! 2 Submission(s) Remaining Hints: 4% deduction per hint. Hints remaining: 2 Part (b) Feedback: 5% deduction per feedback. If the initial voltage is 14 kV, how long does it take to decline to 6.00 x 10² V in ms? 4%
% of Assignment value) A heart defibrillator being used on a patient has an RC time constant of 8.5 ms due to the resistance of the patient's body and the capacitance of the defibrillator. Randomized Variables T = 8.5 ms C=8.5 μF V = 14 kV Part (a) If the defibrillator has an 8.5 uF capacitance, what is the resistance of the path through the patient in ks2? (You may neglect the capacitance of the patient's body and the resistance of the defibrillator.) R=1 Grade Summary Deductions 4% 96% Potential Submissions Attempt(s) Remaining: 2 4% Deduction per Attempt detailed view sin() cotan() cos() tan() T007 9 8 HOME asin() acos() E^^4 5 6 atan() acotan() sinh() * 1 2 3 1 cosh() tanh() cotanh() + - 0 END ● Degrees Radians BACKSPACE DEL CLEAR Submit Hint Feedback I give up! 2 Submission(s) Remaining Hints: 4% deduction per hint. Hints remaining: 2 Part (b) Feedback: 5% deduction per feedback. If the initial voltage is 14 kV, how long does it take to decline to 6.00 x 10² V in ms? 4% % of Assignment value) A heart defibrillator being used on a patient has an RC time constant of 8.5 ms due to the resistance of the patient's body and the capacitance of the defibrillator. Randomized Variables T = 8.5 ms C=8.5 μF V = 14 kV Part (a) If the defibrillator has an 8.5 uF capacitance, what is the resistance of the path through the patient in ks2? (You may neglect the capacitance of the patient's body and the resistance of the defibrillator.) R=1 Grade Summary Deductions 4% 96% Potential Submissions Attempt(s) Remaining: 2 4% Deduction per Attempt detailed view sin() cotan() cos() tan() T007 9 8 HOME asin() acos() E^^4 5 6 atan() acotan() sinh() * 1 2 3 1 cosh() tanh() cotanh() + - 0 END ● Degrees Radians BACKSPACE DEL CLEAR Submit Hint Feedback I give up! 2 Submission(s) Remaining Hints: 4% deduction per hint. Hints remaining: 2 Part (b) Feedback: 5% deduction per feedback. If the initial voltage is 14 kV, how long does it take to decline to 6.00 x 10² V in ms? 4%
Chapter10: Direct-current Circuits
Section: Chapter Questions
Problem 54P: A heart defibrillator being used on a patient has an RC time constant of 10.0 ms due to the...
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Question
Transcribed Image Text:% of Assignment value)
A heart defibrillator being used on a patient has an RC time constant of 8.5 ms due to the resistance of the patient's body and the capacitance of the defibrillator.
Randomized Variables
T = 8.5 ms
C=8.5 μF
V = 14 kV
Part (a)
If the defibrillator has an 8.5 uF capacitance, what is the resistance of the path through the patient in ks2? (You may neglect the capacitance of the patient's body and
the resistance of the defibrillator.)
R=1
Grade Summary
Deductions
4%
96%
Potential
Submissions
Attempt(s) Remaining: 2
4% Deduction per Attempt
detailed view
sin()
cotan()
cos()
tan()
T007 9
8
HOME
asin()
acos()
E^^4
5 6
atan() acotan()
sinh()
*
1
2
3
1
cosh() tanh() cotanh()
+
-
0
END
● Degrees
Radians
BACKSPACE
DEL
CLEAR
Submit
Hint
Feedback
I give up!
2 Submission(s) Remaining
Hints: 4% deduction per hint. Hints remaining: 2
Part (b)
Feedback: 5% deduction per feedback.
If the initial voltage is 14 kV, how long does it take to decline to 6.00 x 10² V in ms?
4%
Transcribed Image Text:% of Assignment value)
A heart defibrillator being used on a patient has an RC time constant of 8.5 ms due to the resistance of the patient's body and the capacitance of the defibrillator.
Randomized Variables
T = 8.5 ms
C=8.5 μF
V = 14 kV
Part (a)
If the defibrillator has an 8.5 uF capacitance, what is the resistance of the path through the patient in ks2? (You may neglect the capacitance of the patient's body and
the resistance of the defibrillator.)
R=1
Grade Summary
Deductions
4%
96%
Potential
Submissions
Attempt(s) Remaining: 2
4% Deduction per Attempt
detailed view
sin()
cotan()
cos()
tan()
T007 9
8
HOME
asin()
acos()
E^^4
5 6
atan() acotan()
sinh()
*
1
2
3
1
cosh() tanh() cotanh()
+
-
0
END
● Degrees
Radians
BACKSPACE
DEL
CLEAR
Submit
Hint
Feedback
I give up!
2 Submission(s) Remaining
Hints: 4% deduction per hint. Hints remaining: 2
Part (b)
Feedback: 5% deduction per feedback.
If the initial voltage is 14 kV, how long does it take to decline to 6.00 x 10² V in ms?
4%
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