Resistance and Capacitance in a Circuit (From Sullivan, Precalculus, ch. 5, p. 2
ID: 3630898 • Letter: R
Question
Resistance and Capacitance in a Circuit
(From Sullivan, Precalculus, ch. 5, p. 276)
The equation governing the amount of current I (in amperes) after time t (in microseconds) in a single circuit consisting of a resistance R (in ohms), a capacitance C (in volts), and E is an electromotive force (in microfarads).
For each question below, fill in the blank with a number with decimal point.
(a)
If E=120, R=2,000, and C=1, how much current is flowing initially (t=0)? Enter decimal number as the answer.
Answer: amperes.
How much current is flowing after 1,000 microseconds? Enter decimal number as the answer.
Answer: amperes.
How much current is flowing after 3,000 microseconds? Enter decimal number as the answer.
Answer: amperes.
(b)
If E=103, R=1015, and C=(16)/10.0, how much current is flowing initially (t=0)? Enter decimal number as the answer.
Answer: amperes.
How much current is flowing after 1,000 microseconds? Enter decimal number as the answer.
Answer: amperes.
(c)
If E=136, R=1299, and C=(29)/10.0, how much current is flowing initially (t=0)? Enter decimal number as the answer.
Answer: amperes.
How much current is flowing after 3,000 microseconds? Enter decimal number as the answer.
Answer: amperes.
Explanation / Answer
(a) Part I: I = (120/2000)*e^(-0/((2000)(1))) I = .06 amperes Part II: I = (120/2000)*e^(-1000/((2000)(1))) I = .0363918396 amperes Part III: I = (120/2000)*e^(-3000/((2000)(1))) I = .0133878096 amperes (b) Part I: I = (103/1015)*e^(-0/((1015)(16/10.0))) I = .1014778325 amperes Part II: I = (103/1015)*e^(-1000/((1015)(16/10.0))) I = .0548211917 amperes (c) Part I: I = (136/1299)*e^(-0/((1299)(29/10.0))) I = .1046959199 amperes Part II: I = (136/1299)*e^(-3000/((1299)(29/10.0))) I = .0472140533 amperes
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