Nodal Analysis and Power Balance in AC Circuits

[sandy.bridge]

I'm working on an ac circuit, and we were told to use nodal analysis. Furthermore, we are asked to show a power balance between the sources and the power dissipated in the resistors, along with a reactive power balance between the sources and the inductors and capacitors.

I have this formula for the sources:
P=I_{rms}V_{rms}cos(\theta{_v}-\theta{_i})

However, when I am using this forumla, some of power in regards to the sources is negative, and some is positive. Therefore, upon summing together each of their contributions, the power is really low in comparison to the power dissipated in the resistors. Am I supposed to be taking the absolute value of these powers or what?

 

[sandy.bridge]

I decided to post an example:

At nodes 1, 2 and 3 respectively I have
(0.6-j0.2)V_1+0.2V_2-0.1V_3=0, 0.2V_1+(0.2+j0.05)V_2-0.2V_3=8.75+j2.1651, -0.1V_1-0.2V_2+(0.3+j0.25)V_3=-12.5
Upon solving the system of equations, I find that
(V_1, V_2, V_3)=(-18.2797-9.09145j, 61.7064+8.49923j, -2.44807+5.00906j)=(20e^{-j156}, 62.3e^{j7.84}, 5.575e^{j116})
The current through the j10V source is
(V_3-j10)/(-j4)=1.38975e^{-j26.1}
and the power generated by this source is
P=(1.38976)(10)cos(90-(-26.1))=-6.1W
Current through the 50V source is
(V_3+50-V_2)/5=2.916e^{-j166}
and the power is
P=(2.916)(50)cos(0-(-166))=-141.5
The current in the final voltage source is
(V_2-10cos(30)-j10sin(30))/(-j4)=13.29e^{j94}
and the power is
P=(13.29)(10)cos(30-94)=58
Total power supplied is 205W.
The power dissipated in the 10-ohm resistor is
|V_3-V_1|^2/10=42W
In the 5-ohm resistor,
|V_3+50-V_2|^2/5=42W
and in the 2-ohm we have
|V_1|^2/2=200W
Combined we have total power dissipated as 284W.
Can anyone see where I have gone wrong here? Obviously this is not adding up, but I've been staring at this problem for a very long time and cannot seem to pin point it.

 

[gneill]

I believe that the second term of your first node equation (for node 1) should be imaginary.
That is, j0.2V₂.