\(V_{0}\)=\(I_{0}\)*R
\(V_{0}\)=3*100=300v

\(V_{1}\)=(\(I_{1}\))(R+\(\frac{1}{jwC}\))
=(4)(100+\(\frac{1}{j(2pi60)(.00005)}\)
=(4)(100+\(\frac{1}{j0.0188}\)
=(4)(100-j53.19)
=(4)(113.26@-28deg)
=453.04@-28deg

\(V_{2}\)=(\(I_{2}\))(R+\(\frac{1}{jwC}\))
=(6)(100+\(\frac{1}{j(4pi60)(.00005)}\)
=(6)(100+\(\frac{1}{j0.0377}\)
=(6)(100-j26.52)
=(6)(103.46@-14.86deg)
=620.76@-14.86deg

V(T)=300+453.04cos(2pi60t-28deg)+620.76cos(4pi60t-14.86deg)

\(P_{0}\)=300*3 = 900w
\(P_{1}\)=906.08cos28deg = 800
\(P_{2}\)=1862.28cos14.86 = 1800
Ptotal=900+800+1800=3500w

formulas for P are based off of this formula

also
P=\(I^{2}_{rms}\)*R
=((3^2)+(4/1.414)^2+(6/1.414)^2)*100
=3500w

did the work two ways and got 3500w both times. 3500w is not the correct answer.

I have a feeling I went wrong in the formulas for V1 and V2...i.e.
\(V_{1}\)=(\(I_{1}\))(R+\(\frac{1}{jwC}\))

where did I go wrong???

 

It's a parallel load.

The DC component only goes through the resistor.
The two AC components will divide between resistor and capacitor.

Your AC calculation assumes the R and C are in series. This is inconsistent with your DC calculation - a series RC has 0 DC current.

 

so instead of (R+(1/jwC))

what should it be???

((1/R)+(1/jwC))

how do I split up the capacitor and resistor?

 

For each AC term you could apply the current divider rule to find the current contribution to R from that term using ..

\(I_R=\[\frac{-jX_C}{R-jX_C}\]I_{AC\ term}\)

Since you only need the rms values for power calculation it could be 'simplified' by conversion to the magnitude form ...

\(\|I_R\|=\[\frac{X_C}{\sqrt(R^2+X_C^2)}\]\|I_{rms \ AC\ term}\|\)

\(X_c=\frac{1}{\omega C}\)