a capacitor of unknown capacitance is wired to an alternating voltage source with a frequency of 60 hertz. If the resulting capacitive reactance is 663 (3.14), find the capacitance of the capacitor.
2006-07-02 11:34:17 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Given that there are only two values given in the problem, it's obviously not that hard of a problem. (It's been a while since I've taken circuits, but I've taken other higher level classes of varying natures and this *tends* to be true.)
Thumb through the book and find the section on capacitive resistance and find the proper equation in there. Maybe you could even use Google.
Good luck.
2006-07-02 11:42:57 · answer #1 · answered by scuazmooq 3 · 0⤊ 0⤋
Capacitive reactance is the "effective reasistance" of a capacitor in an AC circuit. An uncharged capacitor offers little resistance. To charge, a capacitor must have current flow in one direction for a period of time. (The time to fully charge decreases with a larger current and decreases with a larger capacitance.) Since AC current constantly changes directions, the capacitor will probably not fully charge. The larger the capacitance, the less fully charged the capacitor will be. Also, the higher the frequency, the less time the current spends flowing in one direction before reversing, so the less charged the capacitor becomes. Therefore, capacitive reactance is inversely related to both capacitance and to frequency. This at least tells you that both f and C will be in the denominator. To get the constants that go in the equation, you'll need to look it up. It should be in your textbook, or you can check it out at the reference below.
2006-07-02 11:51:05 · answer #2 · answered by not_2_worried 2 · 0⤊ 0⤋
Capacitive reactance = 1/(@ x pi x f x c), where f is the frequency in Hertz and c is the capacitance.
You have some of these values, which you can fill in to give you:
663(3.14)= 1/(2 x 3.14 x 60 x c)
so we know that
2081.82= 1/(388.8 x C)
Solve for C!
2006-07-02 11:52:14 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Xc= 1/(2*pi*fC)
=> C = 1/(2*pi*f*Xc)
= 1/(6.28*60*663)
The math is left for the student
PS capactive reactance is expressed in OHMS
2006-07-02 11:48:38 · answer #4 · answered by revicamc 4 · 0⤊ 0⤋
Xc = 1 / (2π. f. c)
c = 1/ (2π .f .Xc)
c = 1/ (2* 60 * 663 *π² )
c = 1.2735 * 10^-6 farad
Capacitance (c) = 1.2735 µf
2006-07-02 14:45:48 · answer #5 · answered by M. Abuhelwa 5 · 0⤊ 0⤋
1.2569130216189039718451483157366e-5
= 1.257 MF
2006-07-02 15:12:57 · answer #6 · answered by mohammadsaleh a 1 · 0⤊ 0⤋
what if its a flux capacitor using 1.21 gigawatts?
2006-07-02 16:36:28 · answer #7 · answered by winstonsmithratm 2 · 0⤊ 0⤋