What's the total resistance of this network?

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] :-)

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

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

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.

There's always some damn excuse isn't there :-)

DNA

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

It's called loop and nodal analysis.

...Jim Thompson

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

SO many possibilities.

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

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.

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.

Hell, Spice it!

John

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.

Aha. Thanks. Problem solved.

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

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

The Thevenin's equivalent solution is shown here:

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.

Here is a rare case when delta-to-wye conversion is useful; take the

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.

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

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 .

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...