AC nodal anaylsis to find V thevenin

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Discussion Overview

The discussion revolves around the application of AC nodal analysis to determine the Thevenin voltage (Vth) in a given circuit. Participants are attempting to reconcile their calculations with a solution provided by a teaching assistant (TA), highlighting discrepancies in their results.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • The initial poster presents their equations and calculations for Vth and VL, noting a discrepancy with the TA's solution.
  • Some participants suggest that the TA may have made an error by omitting a negative sign in the solution for Vth.
  • There is confusion regarding the correct formulation of the equations, particularly the treatment of the relationship between VL and Vth.
  • One participant questions the validity of the equation used by the TA, suggesting it may be incorrect based on the signs and arrows in the circuit analysis.
  • Clarifications are made regarding typographical errors in the equations presented, specifically the mislabeling of variables.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correctness of the TA's solution and the equations used. There is no consensus on the proper formulation or the accuracy of the calculations presented.

Contextual Notes

Participants note potential typographical errors and misunderstandings in the equations, which may affect the analysis. The discussion does not resolve these issues, leaving the correctness of the equations and solutions open to interpretation.

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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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