Equivalent resistance of a purely resistive network

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Discussion Overview

The discussion revolves around calculating the equivalent resistance of a purely resistive network between two points, X1 and X2. Participants explore different methods and reasoning for determining the equivalent resistance, including the application of circuit laws and visual aids.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses uncertainty in solving the problem despite having done similar calculations in the past.
  • Another participant proposes that the equivalent resistance could be 2 ohms based on their analysis of the circuit with three 6 ohm resistors in parallel.
  • A different participant agrees with the 2 ohm conclusion but questions the reasoning, suggesting the use of Ohm's law and Kirchhoff's Current Law for a more thorough analysis.
  • One participant suggests improving the clarity of the explanation by using color-coded diagrams to represent voltage levels in the circuit.
  • Another participant shares their own calculations, stating that with 1 ohm resistors, the effective resistance would be 1/3 ohm, and confirms that with 6 ohm resistors, it would be 2 ohms.
  • One participant elaborates on the importance of visualizing how the resistors are connected to clarify their arrangement in the circuit.

Areas of Agreement / Disagreement

There is some agreement on the equivalent resistance being 2 ohms when using 6 ohm resistors, but participants express differing views on the reasoning and methods used to arrive at that conclusion. The discussion remains unresolved regarding the best approach to calculate the equivalent resistance.

Contextual Notes

Participants rely on various assumptions about resistor values and circuit configurations, and there are references to visual aids that may not be fully described in the text.

asp_physics
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Homework Statement


Calculate the equivalent resistance between the points X1 and X2.


Homework Equations


R series = sum of resistances
R parallel = reciprocal of resistances


The Attempt at a Solution


I'm sorry I do not get any idea on solving the question. I have done it in the past and remember that it's a one step answer, but can't recollect now. Please help. Thanks.
 

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Hi, Could it be 2 ohm?

Here is my attempt.
If we connect a 1V source, the voltage at various nodes are as shown. In effect, it's the three resistances in parallel.

So three 6 ohms in parallel = 2ohm.
Please correct if I'm wrong.
 

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I believe you are correct with 2 ohms, but I'm not so sure about your reasoning.

Assuming a 1V voltage source is a good thought - then the voltages at each point are determined. Instead of thinking about resistors in series or parallel, apply basic circuit laws - specifically Ohm's law will give you the current through each resistor, and Kirchoff's Current Law (sum of currents into a node = 0) let's you see the total current flow from X1 to X2 as a result.

Equivalent resistance can then be worked out by applying ohm's law again.

Does that make sense? I'll work it through if not. :)
 
While your diagram is correct, your explanation seems lacking. Trying using different colors to highlight each voltage level on the diagram and then redraw it (see attached picture).
 

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@ Jmf
Thanks for the response.
Is this what you were referring to? [Please refer to the image :) I've assumed the resistances to be 1 ohm each for simplicity]

Source voltage = 1V, Total current = 3A. Therefore effective resistance = 1/3 ohm
If the resistances were 6 ohms each, then effective resistance = 2 ohms.

@Zryn
Thanks for your response too; though I could not exactly figure out what you were referring to with the colour highlighting remark.
 

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asp_physics said:
Source voltage = 1V, Total current = 3A. Therefore effective resistance = 1/3 ohm
If the resistances were 6 ohms each, then effective resistance = 2 ohms.
Yeah, that's exactly what I meant. :)
 
When you highlight the different area's you can see how the resistors are connected together in a way that can usually undo the deliberate attempt to confuse you by creator of the diagram.

By looking at the red section you can see that all three resistors are connected on one side to A, and by looking at the blue section you can see that all three resistors are connected on the other side to B.

Redrawing the diagram then shows them in parallel.
 
Oh. Now I get what you meant. Thanks.
 

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