Mastering Thevenin/Norton Problem Analysis: Tips and Tricks | Homework Help

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Discussion Overview

The discussion revolves around solving Thevenin and Norton circuit analysis problems, focusing on techniques for circuit analysis, particularly nodal analysis. Participants share their struggles with specific problems and seek guidance on their approaches and calculations.

Discussion Character

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

Main Points Raised

  • One participant expresses difficulty in performing circuit analysis for Thevenin/Norton problems and describes their attempt using KCL, noting confusion in relating vx and vAB.
  • Another participant suggests using nodal analysis to find the potential at a controlled current source when the output is open-circuited to determine Thevenin voltage.
  • A participant shares their mathematical manipulations but expresses uncertainty about the correctness of their results, indicating that things do not add up when checked.
  • Another participant advises against treating Vx as a known value and encourages writing node equations without assuming values for variables, suggesting a focus on deriving expressions based on circuit parameters.
  • There is a discussion about the correct expression for Vx in relation to vAB and vs, with participants exploring different formulations.
  • One participant acknowledges a mistake in their calculations and confirms the open-circuit voltage as 18 V, seeking clarification on the next steps for finding Rth or Isc.
  • Another participant confirms that the nodal analysis can be repeated with the output shorted to determine the Norton current, advising to leave k as a variable in the Thevenin voltage solution.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to the problem, with multiple viewpoints on how to handle variables and node equations. Uncertainty remains regarding the correctness of calculations and the interpretation of circuit parameters.

Contextual Notes

Participants express limitations in their understanding of circuit analysis techniques and the relationships between variables, indicating a need for further clarification on specific mathematical steps and assumptions in their approaches.

Who May Find This Useful

Students and individuals seeking assistance with Thevenin and Norton circuit analysis, particularly those struggling with nodal analysis and circuit relationships.

RoKr93
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Homework Statement


2013-06-25215856_zps5b4eb0e5.jpg



Homework Equations


V = IR
I1 + I2 ... + In = 0 for a node (KCL)
V1 + V2 ... + Vn = 0 for a loop (KVL)

The Attempt at a Solution


I've really been struggling with this problem and Thevenin/Norton problems in general. I just can't seem to perform the proper circuit analysis. I know that I can set IA = 0 and VAB = VVoc in order to find Thevenin values, but I just keep getting stuck. I tried to do KCL on the top node, which got me -kvx - (vx/50Ω) + ((vs - vx)/200Ω), but that appears to be getting me nowhere. I've been trying to find a relation between vx and vAB so I can substitute in, but to no avail. I'd really appreciate a nudge in the right direction here as well as general advice, as I'm pretty lost.
 
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Are you familiar with nodal analysis? If so, use it to find the potential at the top of the controlled current source when the output is open-circuited. That'll be your Thevenin voltage. Show your work so we can see how you're doing.
 
-vx/50Ω + vAB/200Ω -kvx = 0

vAB/200Ω = vx/50Ω + kvx

vAB = 4*vx + 4*kvx

additionally,

vAB - vs + vx = 0

vx = vs - vAB

so

vAB = 4*(vs - vAB) + 4*k*(vs - vAB)

vAB = 4*vs - 4*vAB + 4*0.025*vs - 4*0.025*vAB

5.1*vAB = 4.1*vs

vAB = 16.078... V

I don't think that's right. Things aren't adding up when I plug it back in. Where'd I go wrong?
 
Okay, for starters don't treat Vx as "known" value for your node equation, except where it is referenced on controlled sources. Vx is a variable that you'll find an expression for based upon fixed circuit parameters. Write your node equation for the 50 Ω branch as though Vx wasn't labeled.

What's the expression for the current in the 50 Ω branch supposing that the node voltage is VAB? Given that current flowing through the 50Ω resistor, what's an expression to replace Vx?
 
Isn't that what I did? Can't vx be written as -(vAB - vs) or vs - vAB if we're treating it as the voltage difference between the nodal voltage and vs?
 
RoKr93 said:
Isn't that what I did? Can't vx be written as -(vAB - vs) or vs - vAB if we're treating it as the voltage difference between the nodal voltage and vs?

Okay, you're right, you can do it that way. My own preference is to write the node equations ignoring the "sampled" values except for their use in the controlled sources, then later replace those instances with expressions derived from the circuit.

Looking at your node equation manipulations:

-vx/50Ω + vAB/200Ω -kvx = 0

vAB/200Ω = vx/50Ω + kvx

vAB = 4*vx + 4*kvx​

check the coefficient for the last term on the right hand side.
 
Well that was stupid of me. Figures that I thought to check my circuit before I thought to check my actual math, lol. VAB = Voc = 18 V.

So now I need either Rth or Isc. If I do the same nodal analysis on the top node but this time include a -IA, does that make my VAB = 0? It would be a short circuit, correct?
 
RoKr93 said:
Well that was stupid of me. Figures that I thought to check my circuit before I thought to check my actual math, lol. VAB = Voc = 18 V.

So now I need either Rth or Isc. If I do the same nodal analysis on the top node but this time include a -IA, does that make my VAB = 0? It would be a short circuit, correct?

Your open-circuit voltage looks fine.

Yes, you can repeat the analysis with the output shorted to determine the Norton current. Just find the node voltage and divide by the 40Ω "load" to find the current in that branch.

Note that for part (b) you're going to need a version of the solution for the Thevenin voltage that leaves k as a variable...
 
Got it. Thanks so much for your help- it's much appreciated.
 

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