Use the information given in the figure for the series RCL circuit to determine the phase angle between the current and the voltage.
(a) zero degrees (b) +5.3° (c) –9.6°(d) –84°(e) +90°
L= .035 H
V = 60 Hz
R = 150 ohms
C = 1.75 x 10^-6 F
Ive been trying to figure it out for an hour, the answer is -84 degrees but i dont understand how to get that.
2007-12-01 10:28:25 · 3 answers · asked by Abc 3 in Science & Mathematics ➔ Physics
urgh I got it, i was being stupid... i dont know why but i was taking the frequency as voltage. UFF!
2007-12-01 11:02:22 · update #1
angular frequency:
ω = 2πf = 2π ( 60 Hz ) = 377
capacitive reactance:
XC = 1 / ωC = 1 / ( 377 )(1.7 x 10^-6 F ) = 1560
inductive reactance:
XL = ωL = ( 377 ) ( .035 ) = 13.2
phase angle:
tan(-1) ( ( XL - XC / R ) )
tan(-1) ( -1547 / 150 )
-84 degrees
2007-12-01 10:51:23 · answer #1 · answered by jgoulden 7 · 1⤊ 0⤋
The complex impedance of the series RCL circuit is :
Z = R + j(wL - 1/wC) ( j^2 = - 1) , w = 2 pi V = 120 pi.
The phase angle is : tan(phi) = (wL-1/wC)/R
phi = tan^-1[(120 pi*0,035 - 10^6/120pi*1.75)/150] = - 84°
2007-12-01 10:59:23 · answer #2 · answered by Luigi 74 7 · 1⤊ 0⤋
so ap physics is merely relatively not elementary, and as constantly relies upon on the instructor as properly (is he/she a stable instructor? plenty or little artwork, etc...) additionally relies upon on what else you're taking once you're taking ap physics, if the different instructions are not not elementary and you will concentration on the class, then it relies upon and it relies upon on you, and how the difficulty is for you, and how not elementary you prefer to artwork at it sorry if this would not help, yet there are diverse factors...
2016-11-13 04:37:31 · answer #3 · answered by ? 4 · 0⤊ 0⤋