# How to make total resistance of 1 ohm?

1. Nov 30, 2016

### Abs321

1. The problem statement, all variables and given/known data

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.
2. Relevant equations
Parallel: 1/R1+1/R2=1/R
In series: R1+R2=R

3. 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

2. Nov 30, 2016

### BvU

Hi Abs,

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. so you must take the first next step

 With a little trial and error I stumbled on the (or: a ) solution. How far are you ?

Last edited: Nov 30, 2016
3. Nov 30, 2016

### Abs321

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.

4. Nov 30, 2016

### BvU

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 !

5. Nov 30, 2016

### Abs321

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?

6. Nov 30, 2016

### BvU

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
 turns out to be mistaken. See below .

Last edited: Nov 30, 2016
7. Nov 30, 2016

### Abs321

Oh, I've just found an answer. It's (6 ∥ 5+1∥ 4+2) ∥ 2