Solve Thevenin Resistance at Node 2: 10k || 10k

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Discussion Overview

The discussion revolves around calculating the Thevenin resistance (Rth) at node 2 in a circuit involving two 10kΩ resistors. Participants explore the application of Thevenin's theorem and the treatment of voltage and current sources in circuit analysis.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions the solution guide's assertion that Rth = (10k || 10k), suggesting that the resistors should be in series based on resistance equivalent rules.
  • Another participant clarifies that the presence of a fixed 10V ideal source implies that the two resistors are in parallel due to the source having zero resistance.
  • A third participant explains that during Thevenin resistance calculations, voltage sources are treated as short circuits, which results in the two 10kΩ resistors being in parallel.
  • A repeated point from the first participant emphasizes the need for clarification on the parallel configuration, reiterating the reasoning behind the paths to ground through the resistors.
  • The explanation includes that for a path to ground, current can flow through either resistor independently, which characterizes them as parallel rather than series components.

Areas of Agreement / Disagreement

Participants express disagreement regarding the configuration of the resistors, with some asserting they are in parallel due to the circuit conditions, while others maintain that they should be considered in series based on different interpretations of the rules.

Contextual Notes

Participants rely on specific assumptions about the treatment of voltage and current sources in circuit analysis, which may not be universally agreed upon. The discussion reflects differing interpretations of Thevenin's theorem application.

Will15
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For the following circuit:
http://imagizer.imageshack.us/v2/640x480q90/674/aBkkRU.png

I'm solving for the Rth at node 2. In the solution guide, it shows that the Rth = (10k || 10k). I don't understand how this is so, by following the resistance equivalent rules, these two resistance should be in series.

I'd appreciation for the clarification.
 
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The fixed 10V ideal source has zero dv/di = zero resistance, so the two resistors are actually in parallel.
If it was a fixed current source, it would be treated as an infinite resistance element.
 
When computing the Thevenin resistance, voltage sources become short circuits and current sources become open circuits. In your example, when the voltage source is shorted, the 10k resistors are in parallel
 
Will15 said:
I'm solving for the Rth at node 2. In the solution guide, it shows that the Rth = (10k || 10k). I don't understand how this is so, by following the resistance equivalent rules, these two resistance should be in series.

I'd appreciation for the clarification.
For Thévenin analysis, we first replace the voltage source with a short circuit. Then examine what we are left with.

To get from node 2 to ground, an electron can go via a path through one of the 10k resistors to ground, OR through the other 10k resistor to ground, but no path to ground takes it through both resistors. That's the characteristic of parallel paths: you can take one path OR the other, and don't go through both in sequence.

If one path leads inevitably through the other, then and only then, are they are in series.
 

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