Effective Resistance in a Circuit (with many paths)

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Discussion Overview

The discussion revolves around calculating the effective resistance in a circuit with multiple paths between nodes E and F. Participants explore the implications of parallel paths and how they affect the total resistance perceived by a current injected at node E and extracted at node F. The conversation includes attempts to clarify the problem and approaches to finding the effective resistance.

Discussion Character

  • Homework-related
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant expresses confusion about how to approach the calculation of effective resistance due to multiple paths in the circuit.
  • Another participant asks for clarification on what is meant by "alternative paths," suggesting that they might refer to parallel paths.
  • It is noted that if current is injected into node E, the resistance could be calculated as 3k ohms if it travels through a series of resistors, but could also be 1k ohm through a different path.
  • Participants discuss the concept that current will take all available paths, with the path of least resistance carrying the most current, in accordance with Ohm's Law.
  • One participant outlines three specific paths between nodes E and F, indicating that they are all in parallel and suggesting a formula for calculating the equivalent resistance.

Areas of Agreement / Disagreement

Participants generally agree on the concept of parallel paths affecting effective resistance, but there is no consensus on the exact calculations or methods to approach the problem. The discussion remains exploratory with varying levels of understanding among participants.

Contextual Notes

Some participants express uncertainty about the number of paths and how to calculate the effective resistance, indicating a need for further clarification on the circuit configuration and assumptions regarding the resistors.

cavalieregi
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Homework Statement


Hi I am working on a study task. This is the schematic of the circuit. (all resistors are 1 k ohms.)
ic6qvL4.jpg

I need to calculate the following values of resistance that will be seen between the pairs of nodes, indicated by the subscripts.

REF , RCD and RAB

(That is, REF refers to the total resistive effect between node E and node F. This is the resistance that would be 'felt' or 'seen' by a current injected at node E and extracted at node F. Multiple paths means multiple options, and a lowering of the effective resistance.

2. The attempt at a solution
I understand how to calculate effective resistance for series or parallel although I am not sure how I would approach this as I am confused by the alternative paths and hence how to lower effective resistance between the nodes.
 
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Not sure what you mean by "Alternative paths"? Parallel paths ?

Just have a go and show your working so we can see where/if you have gone wrong.
 
CWatters said:
Not sure what you mean by "Alternative paths"? Parallel paths ?

Just have a go and show your working so we can see where/if you have gone wrong.

Okay for example if current was injected into node E the resistance between E and F going up would be 1k ohms + 1k ohms + 1k ohms = 3k ohms but that current could also flow through the right path and resistance between E and F would be = 1k ohm. This would be the other path.

The question states 'REF refers to the total resistive effect between node E and node F. This is the resistance that would be 'felt' or 'seen' by a current injected at node E and extracted at node F. Multiple paths means multiple options, and a lowering of the effective resistance.'

I am not sure how you would get this effective resistance considering the many paths the current could take.
 
cavalieigit: 4897347 said:
I am not sure how you would get this effective resistance considering the many paths the current could take.
Many paths?? How many exactly, for E-F?

These are parallel paths here. You know how to caculate the resultant resistance for resistors in parallel?
 
RT = 1/(1/R1+1/R2+...+1/RN) right. I am just confused by the question as it says if you inject current at node E here will it travel via the right or up to get to node F. We need to find effective resistance between the nodes. The questions says you must take into account these extra paths to the node F.
 
Current will make its way taking all available paths. The easiest path will carry a heavier current, the path of highest resistance will carry the least current. These currents are always in accord with Ohm's Law.

If the paths are parallel, you can treat them as resistors in parallel and calculate a single equivalent resistance of the multiple paths.
 
Okay for example if current was injected into node E the resistance between E and F going up would be 1k ohms + 1k ohms + 1k ohms = 3k ohms but that current could also flow through the right path and resistance between E and F would be = 1k ohm. This would be the other path.

There are actually three paths between E and F...

a) R6+R5+R2
b) R3
c) R4 (assuming the one on the right is R4)

These are all in parallel with each other so the equivalent resistance is

REF = (R6+R5+R2) // R3 // R4

where // means "in parallel".
 
CWatters said:
There are actually three paths between E and F...

a) R6+R5+R2
b) R3
c) R4 (assuming the one on the right is R4)

These are all in parallel with each other so the equivalent resistance is

REF = (R6+R5+R2) // R3 // R4

where // means "in parallel".

Cheers this answers all my questions.
 

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