What is the combined resistance here?

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Homework Help Overview

The discussion revolves around determining the combined resistance in a circuit between points A and B, specifically focusing on series and parallel resistor configurations.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore the concept of series-parallel reductions and discuss the configuration of resistors in the circuit. Questions are raised about the connections between resistors and how to simplify the circuit step by step.

Discussion Status

Participants are actively engaging in the problem, with some providing guidance on how to approach the simplification of the circuit. There is an ongoing exploration of the connections and configurations of the resistors, but no consensus has been reached on the final equivalent resistance.

Contextual Notes

The original poster expresses uncertainty about how to begin solving the problem, indicating a potential lack of familiarity with the relevant concepts.

Eitan Levy
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Homework Statement


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What is the combined resistance in this circle between A and B.

Homework Equations

The Attempt at a Solution


Honestly I have no idea what to here, any explanation will be great.
 

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Have you studied series-parallel reductions?
Start from the extreme right. What can you say about the two Rs in the rightmost loop?
 
cnh1995 said:
Have you studied series-parallel reductions?
Start from the extreme right. What can you say about the two Rs in the rightmost loop?
This is the only thing I knew how to do here. They both combine to one resistor with a resistance of 2R.
 
Eitan Levy said:
This is the only thing I knew how to do here. They both combine to one resistor with a resistance of 2R.
Right. Then what happens to the two 2Rs? How are they connected?
 
cnh1995 said:
Right. Then what happens to the two 2Rs? How are they connected?
Their connection is parallel.
 
Eitan Levy said:
Their connection is parallel.
Yes. So now you can go on simplifying the network until you find the equivalent resistance.
 
cnh1995 said:
Yes. So now you can go on simplifying the network until you find the equivalent resistance.
Alright, thank you! I think I got it.
 

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