Posted by Ayo on Wednesday, March 24, 2010 at 3:18pm.
The current is the same in all three since they are in series, 2 amps.
Say i = 2 cos w t
where w = 2 pi f = 120 pi
then
volts across resistor = iR = 2R cos wt
We measure 80 volts, no phase given so assume 2R=80 and R = 40
volts across inductor = Ldi/dt = -240piL sin wt
again no phase given so assume 240piL = 40 and L = .053 H
volts across capacitor = q/C = integral i dt /C = [1/(60piC)] sin wt
Now we measure 110 volts across the whole circuit and that must be the sum of the three voltages.
Therefore
110 sin(wt+phi) = 80 cos wt - 40 sin wt + [1/(60 pi C)] sin wt
110 sin(wt+phi)= 80 cos wt + [-40+ 1/(60piC)]sin wt
110[sinwt cosphi + coswt sinphi] =80 cos wt + [-40+ 1/(60piC)]sin wt
sin wt terms:
110 cos phi = [-40+ 1/(60piC)]
cos wt terms:
110 sin phi = 80 solver for phi, then use in above to solve for C
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