# physics

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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 - ,

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 - ,

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