# 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
73

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

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
783
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)}$\|I_{rms \ AC\ term}\|$

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