How to make total resistance of 1 ohm?

AI Thread Summary
To achieve a total resistance of 1 ohm using 1Ω, 2Ω, 2Ω, 4Ω, 5Ω, and 6Ω resistors, a combination of series and parallel configurations is necessary. Initial attempts involved using a 1Ω and a 2Ω resistor in series, followed by a parallel arrangement with a 6Ω and another 2Ω, but this did not yield the correct total resistance. The discussion highlights the importance of using parallel circuits to avoid exceeding 1Ω, suggesting that two 2Ω resistors can be combined in parallel to simplify the problem. Ultimately, a successful configuration was found to be (6Ω ∥ (5Ω + 1Ω) ∥ 4Ω) ∥ 2Ω. The conversation emphasizes the trial-and-error nature of circuit design and the need for strategic resistor combinations.
Abs321
Messages
4
Reaction score
0

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
 
Physics news on Phys.org
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 ?
 
  • Like
Likes Greg Bernhardt

Similar threads

Back
Top