Thevenin Voltage & Impedance with Dependent Source: Homework & Solution"

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Discussion Overview

The discussion revolves around calculating Thevenin voltage and impedance in a circuit that includes a dependent source. Participants explore various methods for solving the problem, including nodal and mesh analysis, while addressing specific equations and values related to the circuit components.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks guidance on how to determine Thevenin voltage and impedance with a dependent source.
  • Another suggests solving for the voltage across a specific resistor to find Vth, but questions arise about the correct nodal equation.
  • There is mention of a supernode in the context of nodal analysis, with a suggestion to consider mesh analysis instead.
  • Participants discuss the correct formulation of equations, with some expressing confusion over the voltage values and the impact of a capacitor in the circuit.
  • One participant calculates a voltage value and discusses the relationship between the voltage across the capacitor and the dependent source.
  • Another participant proposes a method for finding Zth by using the short circuit current and the ratio of Thevenin voltage to Norton current.
  • There are multiple equations presented, with participants questioning the correctness of their calculations and results.
  • Discrepancies arise regarding the values obtained from different equations, leading to further inquiries about the methods used for solving them.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to solving the problem, with no consensus reached on the final values or methods. Some agree on the need to revise equations, while others challenge the results obtained by their peers.

Contextual Notes

Participants reference specific equations and calculations, but there are indications of confusion and potential errors in the application of methods. The discussion highlights the complexity of dealing with dependent sources and the nuances of circuit analysis.

eehelp150
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Homework Statement


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Homework Equations

The Attempt at a Solution



How do I do Thevenin voltage and impedance with a dependent source? Any hints would be greatly appreciated.
 
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You just did a network with a dependent source using nodal analysis. Solve for the voltage across the 300 ohm resistor in this network; that will be Vth.
 
The Electrician said:
You just did a network with a dependent source using nodal analysis. Solve for the voltage across the 300 ohm resistor in this network; that will be Vth.
At the "right side" of node V, is it (V+2V)/300 or (V-2V)/300?
Would the nodal equation be:
(V-9)/600 + V/(-j300) + (V+2V)/300 = 0?
 
The thing is, you have a supernode here if you're going to do nodal analysis.

See how they have two mesh currents, I1 and I2 shown? I would take that as an invitation to use mesh analysis.
 
I see some of your posts changing, but there is no indication of editing. I think you may be deleting a post and then reposting. Don't do that; use the edit function.

When you delete and then repost it can be very confusing.

Do you know how to deal with a supernode? If you don't, perhaps you should do a mesh analysis first.
 
eehelp150 said:
At the "right side" of node V, is it (V+2V)/300 or (V-2V)/300?
Would the nodal equation be:
(V-9)/600 + V/(-j300) + (V+2V)/300 = 0?

That equation is correct, but you need the voltage on the right side of the dependent source--that will be Vth.
 
The Electrician said:
That equation is correct, but you need the voltage on the right side of the dependent source--that will be Vth.
No worries. That would just be 3V. Solve for V and multiply by 3 :smile:
 
gneill said:
No worries. That would just be 3V. Solve for V and multiply by 3 :smile:
Solving for V in that equation, I get V = 9/5 Volts. Is that correct? What would V300ohm be?
 
eehelp150 said:
Solving for V in that equation, I get V = 9/5 Volts. Is that correct? What would V300ohm be?

No, the voltage should be complex thanks to the capacitor in the circuit.
 
  • #10
gneill said:
No, the voltage should be complex thanks to the capacitor in the circuit.
Interesting. I "re-pressed" the solve button on Wolfram alpha and got:
V = (63-18i)/53.
 
  • #11
eehelp150 said:
Interesting. I "re-pressed" the solve button on Wolfram alpha and got:
V = (63-18i)/53.
And that would be a much better result!
 
  • #12
gneill said:
And that would be a much better result!
So now I have the voltage across the capacitor and the value of the dependent source (which is simply 2*Vcapacitor). The voltage across the 300ohm resistor is Voc which would be: (Vcapacitor + 2Vcapacitor) for a total of 3 * Vcapacitor right?
 
  • #13
eehelp150 said:
So now I have the voltage across the capacitor and the value of the dependent source (which is simply 2*Vcapacitor). The voltage across the 300ohm resistor is Voc which would be: (Vcapacitor + 2Vcapacitor) for a total of 3 * Vcapacitor right?
Right.
 
  • #14
gneill said:
Right.
How would I find Zth?
 
  • #15
eehelp150 said:
How would I find Zth?
One method is to find the short circuit current across the output (i.e. the Norton equivalent current). The impedance is the ratio of the Thevenin voltage to the Norton current.
 
  • #16
Replace the 9 volt with a short and add a 1 amp current source at the output. Solve the network for V again. The voltage at the A-B terminals will be 3*V. The value of that voltage will be equal to the resistance at the A-B terminals which will be the desired Rth.

Since you already have an equation for V due to the 9 volt source, it will be a small change to add the 1 amp current source.
 
  • #17
The Electrician said:
Replace the 9 volt with a short and add a 1 amp current source at the output. Solve the network for V again. The voltage at the A-B terminals will be 3*V. The value of that voltage will be equal to the resistance at the A-B terminals which will be the desired Rth.

Since you already have an equation for V due to the 9 volt source, it will be a small change to add the 1 amp current source.
I ended up getting I1 = -0.10975609756098 +0.01219512195122i
and I2 = -0.13414634146341 -0.20731707317073i

Did I do it right?
3*V = 197.56 -21.95i

V300ohm = (I2+1)*300 = 259.756-62.19i

Which value is right?
 
  • #18
Neither is right. Let's see the equation you used.

Your original equations was (V-9)/600 + V/(-j300) + (V+2V)/300 = 0

What would it be if you short the 9 volt source and apply 1 amp to the A-B terminals?
 
  • #19
The Electrician said:
Neither is right. Let's see the equation you used.

Your original equations was (V-9)/600 + V/(-j300) + (V+2V)/300 = 0

What would it be if you short the 9 volt source and apply 1 amp to the A-B terminals?
V/600 + V/(-j300) + 3V/300 - 1 =0
 
  • #20
So is this the equation that gave you the values in post #17? If so, something's wrong with the solution because that's the right equation. Did you use MyAlgebra to solve it?

Here's what I get:

Solu.png
 
  • #21
The Electrician said:
So is this the equation that gave you the values in post #17? If so, something's wrong with the solution because that's the right equation. Did you use MyAlgebra to solve it?

Here's what I get:

View attachment 107413
redid with wolfram and got it. Thanks!
 

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