# arbitrary resistor network analysis

Discussion in 'Electronic Basics' started by Alex Hunsley, Apr 8, 2004.

1. ### Alex HunsleyGuest

I'm aware of the basic rules for calculation resistance networks where you have
resistors in series, and in parallel combinations, but is this knowledge enough
to calcuklate the resistance between two points in an arbitrary network (or
think connected graph!) of resistors?

For example, take the following graph where every edge represents a resistor:
(needs to be viewed in a propertional font like courier)

A
/|
/ |
/ |
R0 / | R1
/ |
/ |
/ R2 |
---------
\ |
\ |
\ |
R3 \ | R4
\ |
\ |
\|
B

What is the effective resistance betwen A and B in terms of R0, R1 etc?
Or, to put another way, can I analysis this circuit just thinking in terms of
parallel and series resistors?

thanks
alex

2. ### CBarn24050Guest

It's a basic problem covered in any text book. Basicly you mark in the current
flow around each loop (it doesn't matter that there is no actual flow) and
produce a voltage equasion for each loop. in this example there are 3 loops
giving you 3 equasions. The currents are also related ie. ir0=ir1, ir3=ir4,
ir2=(ir0-ir3). A bit of algebra and your there.

3. ### Alex HunsleyGuest

This makes some sense to me, but I don't understand completely. Do you have any
resources on the web you could point me at to explain this?
Or could you explain a little more?
How do I start marking current in? Do I just assume there is a current, call it
I, flowing from A to either connections there, and find out the corresponding
voltage (knowing that V = IR)?
Suppose I do this for the three loops that exist, and come up with three
voltages across them, V0, V1, V2, what do I do then with these three voltages?

thanks
alex

4. ### John PopelishGuest

Not for the general case, you can't. When there is symmetry that
forces zero drop across the center resistor, you can. Do a search for
"loop equations" and "node equations". They are the general methods.

5. ### Alex HunsleyGuest

Oh yes, I found this one out by putting a few resistors together - when you
have symettry like that, it makes sense that the resistor crossing the middle
might as well not be there as far as the resistance calculation goes.

Aha, thanks, these were the terms I was looking for!

alex

6. ### John LarkinGuest

The general solution, as has been noted, is to write the loop
equations and solve them simultaneously, or use equivalent matrix
methods.

You can also do this by simple breakdown:

Apply 1 volt at A and ground B.

Ignoring R2 for a bit, compute the Thevenin equivalent voltage and
resistance at the R0:R3 junction (call that Vx, Rx), and repeat for
the R1:R4 junction (Vy, Ry.) Now add in R2. The two Thevenins and R2
make a simple voltage divider string, and you then get "loaded" values
for Vy and Vx. Knowing these, the supply current is apparent and the
effective net resistance is just its reciprocal.

More fun than solving simultaneous equations.

Interestingly, it's often easier to design networks than to analyze
them, especially since you have the option of picking ones that are
easier to think about. I once designed a duty-cycle DAC that had two
trimpots to set gain and offset. The interactions were so bizarre that
*nobody* could set the pots.

John

7. ### cirip cipciripGuest

Another approach, more related to parralel/series calculation is to do a
delta-Y conversion. Google for delta-Y, then apply the formula. I guess the
derivation of the formula is beyond the scope of the answer here, but it can
be done.

Cirip