Simple Resistor Puzzle

2Ω.

I count 3Ω. No?

Don't spill the beans until SSCpal states it isn't homework!

OOPS!

Gotta get better at watching for 'Homework Trollers'.

I'm pretty sure it's 8.67 Ω, but I might have rounded off wrong.

Standby, I am modeling it on a computer (what's the resistance of the interconnections between the resistors?).

Fellows, I'm 65, retired and haven't done homework since 1967. I saw it on a bathroom wall at Purdue back then. This puzzle does not need a computer or math beyond algebra, just some logical thought. And the answer is not 2 or 3. I will wait a day and give hints if no one has got it.

Well, alright then. Lessee, You take 1, then half the other side, that's..um.. 2.5. Then add 1 in parallel with 167/549 (no, crap!) ok, in parallel with n/n²+1. Now, I take the log. Um, can't find Smolley's Tables, so flip lucky coin. Hah! 2.6666... Ω

But the real question is why you were in the Ladies Bathroom at Pudue in 1967? Wasn't that the Summer of Love? Mmmm. Durn hippy Boilermakers had all the fun.

8.143729²x pi I think. What type of solder is being used and what gauge wire? Do we account for humidity or is it in a nitrogen enviroment?

2+ 1/∞

Nice problem, SSCpal, but you should have left out the resistors along one side. Had you done so, the problem would have had a much, much more mathematically-interesting solution: the net resistance would have been phi ohms, and the set of partial solutions along the way would have formed a perfect Fibonnaci sequence.

No donut.

2 ohms

Carl try this

http://www.1728.com/quadratc.htm

It's interesting to see people so confidently give the wrong answer.

And interesting to see people state that it's a waste of time doing these problems.

The skills you develop doing these problems, help you solve real world problems.

I've managed all these years...
I could solve it, if necessary...but I've yet to meet a situation where such a calculation was necessary.
I shall put on my tin helmet in case of any more sniping
Del

I get 1+(3)^0.5 ohms or

2.732051... ohms

But my proof is weak and I'd be interested in seeing a proper one.

8/3 Ω

2.7320508075689............ Ohm

i am a mechanical engineer, so plz go easy on me, but here's my approach

i divided the circuit like the above image, each star to triangle, if each resistance = R, then each star will have 2R/3 resistance as seen from any two points, then we have n stars in parallel, hence the total resistance of the stars would be 2R/3n, which are in parallel to 1 and 2, thus now we have 2R/(3n+1), which are in series with 3, thus the total resistance of the circuit is R + 2R/(3n+1), when n goes higher, the second factor goes towards zero, so for n -> infinity, the total resistance approaches R, and R=1, so the complete circuit has a resistance of 1 Ohm.

0.5 Ohm

Rtotal=2,73 R

As an explanation for the above equation:

In the front end of the network we have 3 resistors in series: R + R₁ + R

The R₁ consists of the (middle) resistor R in parallel with the rest of the network. But the "rest of the network" is exactly the same as the total (initial) network (and it is so exactly because the network is extended to infinity). So, the R1 is [the R in parallel with Rtotal]

You are probably right, i used your approach, made a small excel iteration, and i had 2.732051... with R=1

Sorry... When I wrote: ..... R_total² - 2*R*R_total - 2*R²

I (obviously) meant: ..... R_total² - 2*R*R_total - 2*R² = 0

(I forgot the "= 0" ...These happen when you are in a hurry... ...)

1+sqrt(3) ohms

Well, I don't get to use any of my witty clues, as several of you got it. Since the ladder extends to infinity, we can add another section on the front end and it still reads R. So R must equal 2+(1 parallel R). Rewrite it as a quadratic equation, solve for R and we get 1+sqrt 3. It is an irrational (mathematically), but not as irrational (logically) as some of the answers were. There, I managed to get a little wit in.

As an interesting asside (or maybe not) how many sections would you need to consider if you were working it out longhand to get an answer within the (say) 5% tolerance of the resistors...?
My guess is prob just one or two sections, but being a cat I can't be bothered.
I expect one of you guys who's put it into Excel could give an answer.
I'll wub you if you do...prrrr prrrr.
Del

You are very close, actually, you start with 3 ohms, which is already at an error of less than 10%, then you go to 2.75 and that's less than 1%, then to 2.733333 and that's almost like 0.0469%. I hope that's what you were asking for.

Ta,
Yes, that confirms what I suspected about the daftness of the Q in real terms
Del

Del Quote "As an interesting asside (or maybe not) how many sections would you need to consider if you were working it out longhand to get an answer within the (say) 5% tolerance of the resistors...?"

One added section to the front resistors get the answer to 2.75.

Well D'Oh, that's the hard way. Do it like I do. Start out at ∞ and work your way back. Piece of cake. Say, can you spare a fellow a pencil? Mine's down to a nub.

It appears that the infinite resistor seem to equate to 0.70 ohms

Ok, the result is 1±√3.

What (if anything) is the physical significance of the 1-√3 (that's -0.732Ω) solution?

Ah, grasshopper, - 0.732 is the sound of one Ohm clapping. Please light one candle and put 50 pee in the collection. Move along quickly.

Pretty sure its 1+√3.