Finding equivalent resistance in complicated circuit

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SUMMARY

The discussion focuses on calculating the equivalent resistance of a circuit with resistors R1 = 4Ω, R2 = 5Ω, R3 = 14Ω, and R4 = 9Ω. The correct approach involves recognizing that R2 and R4 are in parallel, forming R5, which is then in series with R3 (R6). Finally, R6 is in parallel with R1. The key takeaway is to analyze connections rather than the physical layout of the resistors.

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Homework Statement



Problem: What single, equivalent resistor could replace all of the resistors in this circuit?
Given R1 = 4, R2 = 5, R3 = 14, R4 = 9

Homework Equations



In series, R1 + R2... = Req
Parallel, 1/R1 + 1/R2... = 1/Req

The Attempt at a Solution



I have tried multiple different possibilities of series/parallel. I thought for sure that R2/R4 are in parallel, that R1 was in series with 2/4 and that R3 was in parallel with 1/2/4. When I tried this answer, I got feedback that I was wrong.
 

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I would say that R2 and R4 are in parallel, call it R5
R5 is in series with R3, call it R6
R6 is in parallel with R1
 
Your first conclusion was correct, But R1 is not in series with R2/R4.

Two resistances are in series:
- if they are connected to each other with a wire.
- if nothing else connects to this wire.

The second condition fails because the battery is also connected to the connection between
R1 and R2/R4

try to look only at what is connected to what, and not where the resistances are drawn. direction.
 

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