Finding Thevenin Equivalent Circuit for a Complex Network

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Discussion Overview

The discussion revolves around finding the Thevenin equivalent circuit for a complex network, focusing on the methods to simplify the circuit using short circuits and the implications for resistance calculations. The scope includes homework-related problem-solving and technical reasoning regarding circuit analysis.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant attempts to find the Thevenin resistance (R_th) by replacing voltages with short circuits but expresses uncertainty about the next steps.
  • Another participant suggests that combining resistances in the presence of short circuits is problematic, indicating a lack of understanding of how to handle series and parallel resistances under these conditions.
  • A participant advises that all points connected by a short circuit should be treated as a single node, recommending a redrawn circuit for simplification.
  • There is a reiteration of the previous point about condensing nodes along short-circuit wires, with a request for confirmation on whether this method is correct.
  • One participant emphasizes that only the resistance seen by the external terminals is relevant, implying that other components can be ignored.
  • A question is raised about whether the resistance could simply be 3R, which is met with a challenge regarding the need for a simplified circuit representation.
  • A later reply clarifies that a resistor with its ends shorted will not allow current to flow, suggesting it can be omitted from the circuit analysis.
  • Another participant provides a visual description of the circuit simplification process, indicating how short circuit jumpers affect the configuration and lead to an equivalent resistance of R/2 between certain points.

Areas of Agreement / Disagreement

Participants express differing views on how to approach the simplification of the circuit and the implications of short circuits on resistance calculations. There is no consensus on the correct method or final resistance value.

Contextual Notes

Participants have not fully resolved the assumptions regarding the configuration of the circuit or the specific definitions of the components involved, which may affect the analysis.

Shawkify
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Homework Statement


74df2061fc.png


Homework Equations


Thevenin Equivalents
V = IR

The Attempt at a Solution


402c55e6e5.png

To try to find R_th, I replaced all the voltages with short circuits but I do not what to do next.

Then I believe V_th would just be the same as the voltage of the rightmost resistor.
 
Last edited:
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Shawkify said:
...I do not know how to combine resistances when there are short circuits.
That implies you do not know how to combine resistances at all. I mean, if you take two series resistance, there is a short circuit (aka "a wire") between the two and it's even worse with parallel resistors since they have short circuits (aka "wires") between pairs of ends.
 
All points joined together by a short-circuit become a single node. So redraw the resistor circuit simplified by condensing nodes along each short-circuit wire into a single node.
 
NascentOxygen said:
All points joined together by a short-circuit become a single node. So redraw the resistor circuit simplified by condensing nodes along each short-circuit wire into a single node.
5faadd80df.png

This is how I ended up reducing the circuit, was this correct?
 
We are interested only in the resistance seen by the 2 external terminals. Other bits and pieces can be ignored.
 
Would the resistance just then be 3R?
 
Shawkify said:
Would the resistance just then be 3R?
No. Where is your simplified circuit showing the external terminals?

A consequence of a resistor having its ends shorted together is that no current will flow in that resistor (there being zero PD across it) so you can omit it from the circuit.
 
See if this helps.

Starting diagram:
I've replaced resistors with red lines.
The short circuit jumpers are shown in black.

upload_2017-1-27_21-37-21.png

The blue arrows show the direction the short circuit jumpers will collapse to the center.

Let's just take the upper left section and see what happens when we collapse its short circuit jumper.

upload_2017-1-27_21-39-45.png


Now I have two resistors connected between the outside and the inside of the diagram. The straight red line is the original resistor and the curved red one is the one which moves due to the short circuit jumper.

This leads to an equivalent R/2 between those points.

You need to pursue this with the rest of the circuit and see what you get.

Sorry I didn't have the energy to draw resistors but I hope you get the point.
 

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