AC nodal anaylsis to find V thevenin

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The discussion revolves around solving for V Thevenin (Vth) in an AC circuit using nodal analysis. The user is struggling to reconcile their calculated Vth of -0.4 - 0.8i with the TA's posted solution of 0.4 - 0.8i, believing there may be a sign error in the TA's work. The user sets up equations based on node voltages and attempts to substitute values to solve the system but questions the accuracy of the TA's equation, VL - Vth = 0.25Ic. The conversation highlights the importance of careful attention to signs and the correct application of nodal analysis in circuit calculations. Ultimately, the user seeks clarification on the discrepancies in their results.
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Homework Statement


Here is the circuit. I have to solve for V Thevenin

I am having trouble figuring out what I am doing wrong here. I need to solve for Vth and I do, but it does not match with what my TA posted as the solution. [/B]
upload_2015-4-25_14-55-39.png


Homework Equations


The equations I set up are the same as the above. VL is the node voltage on the jΩ inductor and Vth is the voltage on the right.

The Attempt at a Solution


1. -1∠0° + (VL)/j + (Vth)/(-.5j) = 0

2. Vth - VL = .25Ic

3. Ic = Vth/(-.5j)Substitute 3 into equation 2.

4. Vth - VL = .5iVth

Now solve the system of equations for 1. and 5.
1. -1∠0° + (VL)/j + (Vth)/(-.5j) = 0
5. Vth(1-.5i) -Vth = 0

Vth = -0.4-0.8i

VL = -0.8-0.6i

Am I doing this correctly? The solutions my TA posted does not get this answer for Vth. She got Vth = .4 -.8i.
 

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Your TA has forgotten the "-" in front of ".4".
 
Hesch said:
Your TA has forgotten the "-" in front of ".4".
He used VL-Vth = .25Ic. That is wrong correct?
 
Swagbinger said:
2. Vth - VL = .25Ic

3. Ic = Vth/(-.5j)Substitute 3 into equation 2.

4. Vth - Vth = .5iVth
What happened to VL?
 
gneill said:
What happened to VL?
Sorry, a typo, the second Vth should be VL
 
Swagbinger said:
He used VL-Vth = .25Ic. That is wrong correct?

According to arrow and signs, it's wrong.
 

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