Find Thevenin Equiv at (a,b): Troubleshoot Currents

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Discussion Overview

The discussion revolves around finding the Thevenin equivalent circuit at terminals (a,b) of a given circuit. Participants explore various methods for circuit analysis, including mesh analysis and nodal analysis, while troubleshooting discrepancies in their current and voltage calculations.

Discussion Character

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

Main Points Raised

  • One participant expresses difficulty in obtaining consistent current values when attempting to solve for Ix, indicating discrepancies in their equations.
  • Another participant suggests that adding a test voltage changes the number of meshes in the circuit, which affects the calculations for Vth and Rth.
  • A participant proposes using a short circuit across terminals a-b to simplify the analysis, suggesting a parallel combination of resistors to find the short circuit current.
  • One participant reports high voltage and resistance values after adjusting their mesh equations, indicating potential errors in their setup.
  • Another participant points out an error in the equations, specifically regarding the units and relationships between currents, suggesting a correction to the equations used.
  • A participant mentions the challenge of expressing the voltage drop across a dependent current source and shares updated equations based on previous feedback.
  • One participant asserts that the voltage across the dependent current source corresponds to the voltage across a specific resistor in a simplified mesh case, questioning the calculated voltage value of 75.4 V.
  • A request is made for clarity on the equations used in both the 4 mesh and 5 mesh cases to identify potential issues in the calculations.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to the circuit analysis, with some proposing alternative methods and others correcting or challenging specific equations. The discussion remains unresolved regarding the accuracy of the calculations and the correct application of circuit analysis techniques.

Contextual Notes

Participants note potential limitations in their equations, such as incorrect units and relationships between currents. There is also mention of unresolved mathematical steps in the mesh analysis, which may affect the overall results.

gfd43tg
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Homework Statement


Find the Thevenin equivalent circuit at terminals (a,b) of the circuit below:


Homework Equations





The Attempt at a Solution


my currents aren't adding up right. When I try to solve for Ix using my two equations that should find it, I get different answers.
 

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When you add the 1 volt test voltage at terminals a-b, you no longer have 4 meshes; you now have 5.

Your top schematic shows 4 meshes, and this allows you to calculate Vth, because the 2Ω resistor has no effect when the output is open circuited. Vth is just the voltage across the 8Ω resistor.

But as soon as you add the 1 volt test voltage as in the bottom schematic, you have 5 meshes. You need the current in that 5th mesh to calculate Rth.

You could solve it with only 4 meshes by assuming a short across a-b; this then replaces the 8Ω resistor with the parallel combination of 2Ω and 8Ω. Solving your 4 equations would give you the current through that parallel combination. Then use the current divider rule to calculate the current through the 2Ω resistor; that would be the short circuit current through a-b.

If you solve this using the nodal method, you only have 3 equations to deal with. That's what I meant when I said "You usually can use any network analysis method. One method may be easier for a given circuit than another."
 
I'm doing the proper amount of meshes, and now my voltage is high and the resistance seems way too high. I just put those equations into a 5x5 and 4x4 solver. Here is my 2nd attempt at it
 

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Your 4th equation is incorrect. Your units are wrong. I3 is a current and 8*I4 is a voltage.

What you need is to say I4-I3 = 2*Ix, then set Ix = I1-I2

Give that a try, and make the same correction in your 5 mesh equations.

Your 5th equation for the 5 mesh system should be -8*I4 + 10*I5 = -1
 
With the dependent current source there I don't know how to express the voltage drop across it. I redid it and got 75.4 V. my first 3 equations are unchanged but the last two are now

2 i1 - 2 i2 + i3 - i4 = 0

then for the 5 mesh system similarily, I changed to the same and what you said about -8 i4 + 10i5 = -1
 
Last edited:
The voltage across the dependent current source is the same as the voltage across the 8Ω resistor in the 4 mesh case; that would be 8*I4 in the 4 mesh case. Your last two equations look good now, but I don't get 75.4 volts for that voltage.

Show the 4 equations for the 4 mesh case and the values you get for the 4 currents, and I should be able to see what the difficulty is.

Also, what do you get for the currents in the 5 mesh case?
 

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