How to make total resistance of 1 ohm?

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Homework Help Overview

The discussion revolves around creating a circuit with a total resistance of 1 ohm using a set of resistors: 1Ω, 2Ω, 2Ω, 4Ω, 5Ω, and 6Ω. Participants explore various configurations and the principles of series and parallel circuits to achieve the desired resistance.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss different combinations of resistors, considering series and parallel arrangements. There is an exploration of how to achieve a total resistance of 1 ohm, with some questioning the feasibility of certain configurations and the necessity of multiple branches.

Discussion Status

The conversation includes attempts to find a solution, with some participants offering insights into the requirements for parallel and series circuits. There is recognition of the complexity involved in achieving the target resistance, and while some potential configurations are suggested, no consensus on a definitive solution has been reached.

Contextual Notes

Participants note that the total resistance cannot be achieved with a single parallel branch and discuss the implications of using multiple branches. There is also mention of trial and error in finding a suitable arrangement, indicating that the problem may require further exploration.

Abs321
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Homework Statement



We have 1Ω, 2Ω, 2Ω, 4Ω, 5Ω and 6Ω resistors. We have to make a circuit with all of them and total resistance must be 1 ohm.

Homework Equations


Parallel: 1/R1+1/R2=1/R
In series: R1+R2=R

The Attempt at a Solution


I have only made a circuit from 4 resistors:
1 ohm and 2 ohms in series, 6 ohms parallel to them and 2 ohms parallel to them all
 
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Hi Abs, :welcome:

Unfortunately, that doesn't count as an attempt (more like an arbitrary guess).
You need a bit more than that; considerations, for example. Like:
Your last step will be a parallel circuit (because a series will exceed 1 ##\Omega##).
That parallel circuit will have ##\ge## two sub-circuits, etc. etc.
Fortunately, I don't have the answer either. :smile: so you must take the first next step

[edit] With a little trial and error I stumbled on the (or: a ) solution. How far are you ?
 
Last edited:
Oh, I suppose there could be two parallel circuit from 2 ohms and another sub-circuit totally of 2 ohms and so on, I understand how these laws work, but I can't construct such circuit. Just playing on with resistors.
 
Yes, one of the considerations is: the 1 ##\Omega## can't be a single parallel branch (total would go under 1 ##\Omega##).

So the first candidate as a parallel branch would be the 2 ##\Omega##. That reduces the problem to making another 2 ##\Omega## with the remaining 5 resistors (1,1,4,5,6) (since 2 ##\Omega\, \parallel \, 2 \Omega = 1 \Omega ## )

And so on. You're almost there :rolleyes: !
 
And this is where I stop. Is it possible to combine 2 ohms in 2 parallel branches from remaining 5 resistors? Or do we need 3 branches?
 
Yes. So in total you end up with three branches: one of 2 ohm.
The remainder needs to be 2 ohm. Can't be one branch (I think -- why not ?) but two branches is worth investigating
[edit] turns out to be mistaken. See below .
 
Last edited:
Oh, I've just found an answer. It's (6 ∥ 5+1∥ 4+2) ∥ 2
Was your solution simpler?
 
Brilliant ! Well done.

My solution was for the wrong problem (1,1,2,4,5,6) o:) instead of (1,2,2,4,5,6). Want to try that too ?
 
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