Posted by Ayo on Wednesday, March 24, 2010 at 3:18pm.
A source of e.m.f 110v and frequency 60hz is connected to a resistor, an indicator and a capacitor in series. When the current in the capacitor is 2A.the potential difference across the resistor is 80v and that across the inductor is 40v.calculate the (1)potential difference across the capacitor (2)capacitance of the capacitor 3 inductance of the inductor

physics  Damon, Wednesday, March 24, 2010 at 4:04pm
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

physics  Damon, Wednesday, March 24, 2010 at 4:22pm
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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