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February 1, 2015

February 1, 2015

Posted by **Ayo** on Wednesday, March 24, 2010 at 3:18pm.

- physics -
**Damon**, Wednesday, March 24, 2010 at 4:04pmThe 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

- physics -
**Damon**, Wednesday, March 24, 2010 at 4:22pmNow 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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