nonsinusoidal source and linear RC load

Discussion in 'Homework Help' started by notoriusjt2, Sep 23, 2010.

  1. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
    209
    0
    [​IMG]

    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
    [​IMG]
    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???
     
  2. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    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.
     
  3. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
    209
    0
    so instead of (R+(1/jwC))

    what should it be???

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

    how do I split up the capacitor and resistor?
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    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}
     
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