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

$V_{0}$=$I_{0}$*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???

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)}$\|I_{rms \ AC\ term}\|$

$X_c=\frac{1}{\omega C}$