AC Steady State Analysis for Node V1 and V2 using Ohm's Law and KCL

[butleRonius]

Homework Statement

Find the steady state expressions in terms of Acos(wt+theta) for nodes v1 and v2.

Homework Equations

Ohm's law
KCL
Node Analysis

The Attempt at a Solution

Convert to phasors and find impedances-> R1= 50 Ohms R2= 30 Ohms
C1= -j2500=1/j*4e-4 Ohms C2= -j2000=1/j5e-4 Ohms
L1= j5 Ohms L2= j10 Ohms

I'm not getting an acceptable answer when I use Wolfram to solve the systems produced. I believe the error is that I need to combine the impedances of the capacitor and the inductor coming off of node 2, but I'm not sure since there is another node there. It's quite possible there are multiple errors on my part.

 

[NascentOxygen]

Hi butleRonius. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

There is no restriction on what impedance can comprise each branch; it doesn't have to be a single element. In your eqn (2) you need to replace 1/(j3E4) by 1/(z) where z is the series impedance of the capacitor and inductor.

So there's a little sub-exercise for you, determining that series impedance before using it in eqn (2).

 

[butleRonius]

so -j2000 + j10 = -j1990

Sub that in: v2/-j1990

Solve using nodal?

I actually tried combining as (Zc+ZL) || Zr but that turned into a nightmare

Thanks.

 

[butleRonius]

Solved. Thank you.

 

[NascentOxygen]

so -j2000 + j10 = -j1990
Sub that in: v2/-j1990 ✔

Solve using nodal?

Sure, go ahead.

I actually tried combining as (Zc+ZL) || Zr but that turned into a nightmare

A little bit of work involved, yes, but it simplifies the circuit to one with no nodes so the subsequent analysis is shorter. Swings and roundabouts ...