# Finding the resistor values in a mesh from the voltages at the nodes

Discussion in 'General Electronics Discussion' started by striplar, Dec 13, 2012.

1. ### davennModerator

13,866
1,957
Sep 5, 2009
how can you assume that... as Bob said even for just a 2 resistor network there are many many variations with the size of network you are looking at the possibilities are almost endless for resistor value variables

you seem to be unwilling to accept that ... why ?

Dave

2. ### striplar

14
0
Dec 13, 2012
I can assume that all of the resistors will be very similar because the method of their construction is designed to make sure that they are.
I have no problem with the special case that Bob rightly points out as having many solutions. Let's take the simple case I posted where there are effectively two parallel arms with a resistor across the middle. Removing that middle resistor by making it infinite gives multiple solutions, so mathematically you are right to argue that the circuit as drawn has multiple solutions.
What I'm saying is that the bridging resistor will have a value, in fact all of the resistors will have quite similar values in practice, so there is an additional constraint that forbids any resistor taking on an infinite value. Once you take that constraint into account, I'm suggesting that there is only one unique solution.

Last edited: Dec 15, 2012
3. ### striplar

14
0
Dec 13, 2012
Just a quick update on this. A member of another forum has shown numerically that within a range of values, there are in an infinite number of solutions. Bob had already shown that the case of the bridging resistors being infinite gave this too, and he correctly concluded that this was probably the case with non infinite values too.
What I needed was proof of this and now I have it.
Many thanks to all those who have contributed, your efforts and comments have been most welcome and useful. I now have a much better understanding of the problem, so I need to look at this in a different way.

4. ### BobK

7,682
1,688
Jan 5, 2010
I am glad you are finally convinced. I started out to do a proof on the 2x2 network, but I got distracted. Glad to know someone did it for me! In that network (two parallel legs with 2 resistors in series and one bridge resistor), you get 4 equation for currents, in 5 unknowns. Set the value of any resistor and you can solve it for the others.

Bob

5. ### striplar

14
0
Dec 13, 2012
There's no substitute for proof in my book. I was taught not to take someone's word or opinion for something just because they are knowledgeable and probably right, that's a dangerous path if you need certainty. I hope you're not offended, it was not intended as a sleight.
It's proven to be a useful exercise though, and I've learned an awful lot of things that are new to me. This is the joy of being able to dabble outside of my primary discipline.
The numerical method for finding the values if I define one or more resistors may well prove to be useful.
Once again, thanks for your input and patience.

6. ### BobK

7,682
1,688
Jan 5, 2010
No problem. It was proof enough to me that there was 1 less equation than needed to constrain the problem to a signle solution, but with less experience, one might not see that.

Bob