# What's the total resistance of this network?

Discussion in 'Electronic Design' started by Dave Farrance, Mar 1, 2007.

1. ### Dave FarranceGuest

I'm designing a circuit and found myself unable to figure out the
resistor values. How embarrassing. After trying to break down the
problem using equivalent circuits, I found that the problem in its most
basic form is that I can't figure out the total resistance of this:

(fixed pitch needed to display this, of course)

---------------/\/\/\/-----------
| d |
a | b c |
o----+----/\/\/\/----+----/\/\/\/----+----/\/\/\/----+----o
| |
| e |
----------/\/\/\/----------------

Anybody know? Or maybe somebody could point me to a web resource with
tips on how to get to grips with a tangled network like this?

2. ### John FGuest

Easy Rearrange it a bit:

o-----------o
| |
| |
/ /
\ \
/e /a
\ \
| b |
+--/\/\/\/--+

| |
| |
/ /
\ \
/c /d
\ \
| |
o-----------o

Now calculate Hint: use Thevenin's theoreme to get the voltage across 3. ### Dave FarranceGuest

Thanks for the suggestion, but I dunno. It seems to me that Thevenin's
theorem only works where you can avoid that construct. The interaction
between the resistances is still overtaxing my brain.

v1 o-----------o v1
| |
| |
/ /
\ \
/e /a
\ \
| b |
v2 +--/\/\/\/--+ v3
| |
| |
/ /
\ \
/c /d
\ \
| |
0v o-----------o 0v

I think that you're suggesting that I add voltages as above, work out
the equations at each node, and substitute for the voltages until
they're eliminated. It doesn't look solvable to me I'm afraid:

(v1-v2)/e = (v2-v3)/b + v2/c

(v1-v3)/a + (v2-v3)/b = v3/d

4. ### GenomeGuest

I usually find that not designing in silly resistor networks helps a lot in
that you don't have to solve them.

I suppose it's a matter of preference but, if you try my method, I can
assure you that it's much easier.

DNA

DNA

5. ### Dave FarranceGuest

Yes, yes, surely I could rearrange the design so that it didn't have
tangles like that. I know. Trouble is, I've got to fit in with a
previous design and with cost constraints, and it's hard to avoid.

6. ### GenomeGuest

There's always some damn excuse isn't there DNA

7. ### JonGuest

Thevenin eqiuvalent resistance = (open ckt voltage)/Short Circuit
Current.

---------------/\/\/\/-----------
| d |
a | b c |
o----+----/\/\/\/----+----/\/\/\/----+----/\/\/\/----+----o
| | |
| e | |
Vin ----------/\/\/\/------------- |
|
o
-----------------------------------------------------
|
|
---
-
Write 3 voltage loops, e.g.,
a, b, e (I1)
b, d, c (I2)
e, c, Ground (I3)
Solve for I3

Open circuit voltage = Vin
Req = Vin/I3

8. ### Jim ThompsonGuest

It's called loop and nodal analysis.

...Jim Thompson

9. ### John FGuest

Similar... Transform into
and now you are able to do thevenin twice:

e||c v2 b v3 a||d
v1*c/(e+c)----/\/\/\/----/\/\/\/-----/\/\/\/----v1*d/(a+d)

now you can easily calculate v2 and v3 as a function of v1.
then the current across b ...

Another idea... use superposition of
1) left v1 and
2) right v1

SO many possibilities.

10. ### krwGuest

Thevanize e, c, and v1, and a, d, v1. Then use superposition to
combine those with b. Stir and serve cold.

11. ### Dave FarranceGuest

Thanks Jon and Jim Thompson for pointing out loop analysis.

This web page gives a technique which I finds works OK if I put in
actual numbers for the resistor values and I can then get a numerical

http://mathonweb.com/help/backgd4.htm

If I try to retain the algebraic form while using this technique,
though, it quickly turns into a monster. Hmm. I'll have to think if I
can simplify the original form of the puzzle a bit.

12. ### John LarkinGuest

Hell, Spice it!

John

13. ### Greg NeillGuest

Well, that's certainly one way to do it.

If I had to analyze this circuit, I think I'd start
by performing a Delta-Y transformation for resistors
a,b,e. Should be clear sailing from there.

14. ### Dave FarranceGuest

Aha. Thanks. Problem solved.

-----------------/\/\/\/----------
| d |
| ab c |
o----- \ ----- -----/\/\/\/----+----o
| / a+b+e |
| \ |
| ae / be |
| ----- \ ----- |
| a+b+e | a+b+e |
----/\/\/\/----+-----/\/\/\/-----

15. ### John PopelishGuest

A few months ago, someone posted the formula for total
resistance of this arrangement, but I can't find it through

The Thevenin's equivalent solution is shown here:

16. ### The PhantomGuest

Just say "Om mane padme hum", and stare at it for a while. Say to
yourself, "if it weren't for that pesky b resistor, the total resistance
would be the parallel combination of the two arms; something like:"

(a+d)(c+e)
-------------
a+c+d+e

Just a plain product over the sum.

Ah, yes. Now I see it. It's:

a(de + b(c+e) + c(d+e)) + d(ce + b(c+e)
---------------------------------------
a(b+c+d) + b(c+d+e) + e(c+d)

I know you're looking for a solution method which I haven't given. But I see in another post that you have been tipped off to the delta-Y
transformation. This result provided for checking purposes only.

17. ### Robert BaerGuest

Here is a rare case when delta-to-wye conversion is useful; take the
3 points described by b, c, and d (note the delta configuration, and
convert it to a wye resistor configuration B, C, and D (same points; now
have an extra point in "middle".
The derivation of that conversion is fairly simple but i have never
seen it mentioned in any university of note (over the past 50 years).
Once you have the wye values, the reduction is simple.
BTW, this is the classical bridge, so if there is any symmetry, even
on a ratio basis, take advantage.

There are at least 4 different ways to solve the "resistor cube"
problem (what is the resistance across the farmost corners, given all
sides are of one ohm resistors); many "cheat" based on the symmetry.

18. ### Robert BaerGuest

I prefer funny resistor networks, as the silly ones keeps the clowns
agitated.

19. ### Fred BloggsGuest

It's one thing to come up with a tangled formula and entirely something
else to make sense of it:
View in a fixed-width font such as Courier.

20. ### kellGuest

Here's a closely related problem:  