Help solving a puzzle:

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These puzzles are fun somtimes even I did not get the correct answer.

Starting from the left and working my way to the right first time i got 333.33hm: second time different approach yielded 454.54hm:

Then get the soldering iron warm and solder 9 of these 1 khm: resistors together 617hm:

( I checked the R values and most were around 993-995 hm: )

Good exercise, I have forgotten the tricks how to do these

Regards, Raymond
 
The answer....

It seems the answer is indeed 333

Can you explain how you arrived at that? (Just for my own understanding)

How come th eactual measured difference is so different? Is this due to a change in th ecircuit by adding the ohm meter itself?

Hmmmm.....

-Michael
 

Then get the soldering iron warm and solder 9 of these 1 k resistors together 617
Click to expand...
Using a typical bench DVM, with an input resistance of say 1Mhm: the resistance would still be a nominal 618.2hm:

The fact that someone has taken the trouble to assemble and measure with a meter, which gave 618hm: should convince you its 618hm:, UNLESS you havn't given us the FULL puzzle...
 
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I suspect you are joking with us, but nevertheless here is step by step result.
 

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Alternately, you can do a Δ-Y translation on the center two triangles. But after trying that, I wouldn't recommend it over sniper007's method.
 
The key to this puzzle is where you start from. I found my notes again and decided to have an other go at it. I hadn't look at this thread so all the newer solutions were unknown to me.

Firstly the connecting leads are at the RHS of the puzzle so the minimum value of the RHS string can never be less than 500hm: It will be the parrallel value of 1khm: in parrallel with (1khm: + Xhm

Then I started working the way through from the LHS of the puzzle by series parralleling the sections of three resistors.

2khm://1khm: yields 666.6hm:
1.666khm://1khm: yields 625hm:
1.625khm://1khm: yields 619hm: = X
1.619khm://1khm: yields 618hm:

I didn't put all the decimals in the example but you get the idea.

Regards, Raymond
 
Hi,

I did the same thing, collapsing the network from left to right using
series and parallel formulas. The result i got was:

6800/11 ohms exactly,

or

618.1818181818181 ohms approximately.
 
I solved from the left also and got 618.18 ohms. I can't see how you could do it from the right when your parallel values are unknown
 
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