What's the total resistance of this network?
Dave Farrance
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?
--
Dave Farrance
John F
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 [b] :-)
--
Johannes
You can have it:
Quick, Accurate, Inexpensive.
Pick two.
Dave Farrance
>Easy :-) Rearrange it a bit:
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
>Now calculate :-)
>
>Hint: use Thevenin's theoreme to get the voltage across [b] :-)
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
--
Dave Farrance
Genome
"Dave Farrance" <DaveFa...@OMiTTHiSyahooANDTHiS.co.uk> wrote in message
news:jrldu2hq6dev76ls2...@4ax.com...
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
Dave Farrance
>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.
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.
--
Dave Farrance
Genome
There's always some damn excuse isn't there :-)
DNA
Jon
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
Jim Thompson
It's called loop and nodal analysis.
...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
John F
Similar... Transform into
> v1 o---- ---o v1
> | |
> | |
> / /
> \ \
> /e /a
> \ \
> | b |
> v2 +--/\/\/\/--+ v3
> | |
> | |
> / /
> \ \
> /c /d
> \ \
> | |
> 0v o---- ---o 0v
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.
krw
DaveFa...@OMiTTHiSyahooANDTHiS.co.uk says...
> "John F" <call_08...@gmx.at> wrote:
>
> >Easy :-) Rearrange it a bit:
>
> 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.
Thevanize e, c, and v1, and a, d, v1. Then use superposition to
combine those with b. Stir and serve cold.
>
> v1 o-----------o v1
> | |
> | |
> / /
> \ \
> /e /a
> \ \
> | b |
> v2 +--/\/\/\/--+ v3
> | |
> | |
> / /
> \ \
> /c /d
> \ \
> | |
> 0v o-----------o 0v
--
Keith
Dave Farrance
>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
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
answer after a non-trivial effort.
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.
--
Dave Farrance
John Larkin
<DaveFa...@OMiTTHiSyahooANDTHiS.co.uk> wrote:
Hell, Spice it!
John
Greg Neill
news:015eu25jld76nspt3...@4ax.com...
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.
Dave Farrance
>"John Larkin" <jjla...@highNOTlandTHIStechnologyPART.com> wrote in message
>> <DaveFa...@OMiTTHiSyahooANDTHiS.co.uk> wrote:
>> >
>> > ---------------/\/\/\/-----------
>> > | d |
>> > a | b c |
>> > o----+----/\/\/\/----+----/\/\/\/----+----/\/\/\/----+----o
>> > | |
>> > | e |
>> > ----------/\/\/\/----------------
>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.
Aha. Thanks. Problem solved.
-----------------/\/\/\/----------
| d |
| ab c |
o----- \ ----- -----/\/\/\/----+----o
| / a+b+e |
| \ |
| ae / be |
| ----- \ ----- |
| a+b+e | a+b+e |
----/\/\/\/----+-----/\/\/\/-----
--
Dave Farrance
John Popelish
resistance of this arrangement, but I can't find it through
Google.
The Thevenin's equivalent solution is shown here:
http://www.broadcast.net/hallikainen/theory6.html
The Phantom
<DaveFa...@OMiTTHiSyahooANDTHiS.co.uk> wrote:
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.
Robert Baer
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.
Robert Baer
I prefer funny resistor networks, as the silly ones keeps the clowns
agitated.
Fred Bloggs
Dave Farrance wrote:
> 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:
>
>
>
> 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?
>
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.
.
.
.
.
. ---------------/\/\/\/----------
. | d |
. a | b e |
. o----+--/\/\/\/----+----/\/\/\/----+----/\/\/\/---+----o
. | |
. | c |
. --------/\/\/\/----------------
.
.
.
.
. R - R
. sc oc
. R= R + ----------
. oc b
. 1 + --
. R
. m
.
.
.
.
. R = bridge resistance
.
. R = R when b is open circuit
. oc
.
. R = R when b is short circuit
. sc
. R + R
. oc sc
. R = value of b to make R= ---------
. m 2
.
.
.
.
. By inspection:
.
. R = (a+d)||(c+e)
. oc
.
. R = (a||c)+(d||e)
. sc
.
. and
.
. R = (a+c)||(d+e)
. m
.
Rich Grise
If Mr. Thevenin fails you, you might check with Mr. Kirkhoff. :-)
Good Luck!
Rich
John F
> If Mr. Thevenin fails you, you might check with Mr. Kirkhoff. :-)
>
> Good Luck!
> Rich
No... Too much StarTrek... Thats "Chekoff" and Kirk... not a bastard
named Kirkoff... ;-)
BTW.: It's only named after Thevenin ... it's not his
idea/discovery... Got me beat up at an oral exam at university...
"Don't change history, my colleague...", the prof said ... Got a B
anyways. Long time ago...
David L. Jones
If you are lucky enough to have A=E and C=D then B=0 ohms, in which
case it's super easy ;-)
Dave :)
John F
> On Mar 2, 1:07 am, Dave Farrance
> <DaveFarra...@OMiTTHiSyahooANDTHiS.co.uk> wrote:
[snip]
>> (fixed pitch needed to display this, of course)
>>
>> ---------------/\/\/\/-----------
>> | d |
>> a | b c |
>> o----+----/\/\/\/----+----/\/\/\/----+----/\/\/\/----+----o
>> | |
>> | e |
>> ----------/\/\/\/----------------
>>
[snip]
> If you are lucky enough to have A=E and C=D then B=0 ohms, in which
> case it's super easy ;-)
More general but harder to see: e*d=a*c
kell
Here's a closely related problem:
http://groups.google.com/group/sci.electronics.basics/browse_frm/thread/fb68740a97c6c22f/1ae15894955a1365?lnk=st&q=&rnum=2#1ae15894955a1365
Robert Baer
Equality is not a necessary condition for a balanced bridge.
Fred Bloggs
And how some moron can conclude that B must be zero is beyond me. The
David L. Jones sounds a lot like that fag Grise in many ways with the
trivial non-informational content of his posts and the inclusion of the
smiley, which is quite sickening when you think of the marijuana stained
brown teeth....
David L. Jones
You are all class "Fred"
In case you can't figure it out on your own and you really want to
know... a common (and often taught) way to solve these resistor
problems is by finding equipotential nodes, and once found, the
technique is to short those nodes together, thus reducing the
complexity of the circuit and making it easier to analyse.
So in this case, if it happened that A=E and C=D, then the two inner
nodes are obviously equipotential so you could short them out, thus
giving B=0 using this technique.
I was not implying that B *must* be 0, in fact, if A=E and C=D then B
can be infinite as well.
Dave :)