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Another one for the R/Os

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#1




Another one for the R/Os
Not sure if putting this in the quiz forum is fair because I have no idea of the solution.
For those who have difficulty with the drawing the array is a cube with voltage applied at diagonally opposite corners. Maybe the answer is very simple or maybe one should apply Maxwell's, Thevenin's or Norton's theorems either seperately or together. Maybe someone has the solution already complete for our enjoyment. I've never had the patience, since the answer didn't come easily, to try and work it out. I did, once, try to build it and cheat but ran out of resistors. By the way. Has anybody ever suggested a forum for sparkies? There are severeal threads knocking about that could do to be lumped together. 

#2




I think I know how you work this out Fubar. You use the formula for resistors in parallel which I can't write here because it needs fractional notation. Basically it's the 1/R(equivalent = 1/R1 + 1/R2 + .....
So you need to work out all the paths from + to  through the threedimensional network, work out the resistance of each path and apply the formula. So the paths are: R1+R6+R9 = 300 ohms R1+R6+R11+R8+R7+R4+12 = 500 ohms R1+R6+R11+R8+R7+R4+R10 = 700 ohms R1+R5+R10 = 300 ohms R1+R5+R4+R7+R12 = 500 ohms R1+R5+R4+R7+R12 = 700 ohms R2+R11+R9 = 300 ohms R2+R8+R12 = 300 ohms R2+R11+R6+R5+R10 = 500 ohms R2+R11+R6+R5+R4+R7+R12 = 700 ohms R2+R8+R7+R4+R10 = 500 ohms R2+R8+R7+R4+R5+R6+R9 = 700 ohms R3+R7+R12 = 300 ohms R3+R7+R8+R11+R6+R5+R10 = 700 ohms R3+R7+R8+R11+R6+R5+R10 = 700 ohms R3+R4+R10 = 300 ohms R3+R4+R5+R6+R9 = 500 ohms R3+R4+R5+R6+R11+R8+R12 = 700 ohms Blimey  my brain hurts after that. I will leave it to you to do the 1/r1 etc formula on it! Have to take the dog for a walk before it starts raining again. Brian 
#3




So your answer would be
0.0399993 ohms? Eh? But R1, R2 and R3 are already in parallel with all the other combinations so doesn't that modify their value before you start? In actual fact each path from positive to negative is made up of only three resistors in each chain so you could say that there 6 x 300 ohm paths in parallel. They're coming to take me away, haha, heehee, to the funny farm, dada, deedee....... 
#4




I must admit I have been having second thoughts about my answer while taking the dog for a walk and getting pelted with hailstones and running back home.
I think maybe the longer pathways may be invalid as the electrons would maybe have to flow "uphill". I am sure Charles Wheatstone would have had no trouble with this question! Never got anything that complicated to work out as an electrician thank goodness! Maybe there is a real genius with the answer out there somewhere? Tell you what  would you like me to buy you 12 100 ohm resistors for Christmas? Regards, Brian 
#5




I have to venture into the "storage" area this afternoon. In other words the loft 'cos I'm trying to make room for plumbers to gain access.
One of my jobs is to dispose of boxes full of 74 series ICs, capacitors, resistors, PLCs, PSUs, etc., etc., that have been there for 20 years gathering dust. I'll see if I can find enough 100 ohm resistors knocking about before I dump it all. It's not a common size unfortunately. 1k DIL resistors I can find easily. It may be the simplest solution. 
#6




There is an input node with 3 resistors connected to it.
The other end of those 3 resistors must all be at the same potential. There is an output node with 3 resistors connected to it. The other end of those 3 resistors must all be at the same potential. There are 6 other resistors that, each connect between these two internal nodes, so there must be an equal drop across each of these 6 resistors. To solve for the total resistance or the current in each of the resistors, just connect all points that must be at an equal resistance, together. This will not change any current. But with this pair of connections, the whole circuit collapses into 3 resistors in parallel, that are in series with 6 resistors in parallel, that are in in series with 3 resistors in parallel. 
#7




Hmmmm....
Not sure I agree with you there Mike. If we consider a simpler circuit with three resistors each 100 ohm. Point A is joined to point B by two pathways  the first is one 100 resistor, the second is two 100 ohm resistors in series. The potential at the midpoint of the resistors in series would be midway between the potential at point A and point B. Does this not contradict the argument in your second sentence? Fubar  you have a lot to answer for posing this damn question!!! Brian Brian 
#8




Interesting proposition.
Yes the voltages across R1, R2 & R3 must be the same as the voltages across R9, R10 & R12 but what is that voltage supposing the voltage across the network to be 100 volts for simplicity sake. The modifying factor is the centre network of resistors which cannot be said to be in parallel because each one connects to a different resistor at each end node of three. R1 connects to R5 in parallel with R6. R5 connects to R10 and R6 connects to R9 so what is in parallel with what? Attached "simplification" (?) shows the dilemma. PS curses Brian beat me to it again. Too much thunking time! 
#9




Quote:
To solve for the total resistance or the current in each of the resistors, just connect all points that must be at an equal *potential*, together. PS I also got a bit of a head working this out so I went to another forum for a second opinion that about agrees with my thoughts on this. Mike 
#10




Another reply say's why not simply first remove R5,R8.
This, with symmetry considerations, will reduce the network to 2 dividers to which you can now reconnect R5//R8. What think ye? Mike This is a bit like that tune that gets into your head and repeats with variations 
#11




Mike
If you do that shouldn't you remove 3 resistors from the centre array? One from each inner end of R1, R2 & R3? I have a sneaky feeling I know the answer to this without working it out or maybe I've just lost my nerve. 1100100  Binary is so much easier. 
#12




Fubar.........Keep remembering Napoleon XV and those nice young men in their clean white coats............sigh, days of my youth...........................pete

#13




The definitive answer my friend is blowing in the wind:
http://www.radioelectronicschool.net...be_problem.pdf Have fun. Mike 
#14




Bugger me!
What was the question?
__________________
Ray . . . . A closed mouth gathers no feet! 
#15




I've been sat on the sidelines watching this one. every so often breaking off with what I have been doing during the day the grab my calculator.
Thank goodness someone has worked it out  The second solution appeals most. Just a thought (tongue in cheek)  What difference if any would it make if one constucted the cube with 10% tolerance resistors? I'll go back into my corner now.
__________________
Malcolm. 
#16




Well, well.
As simple as that. For voltage read current. It's the approach that counts. That original diagram of mine has been knocking around in my college notes since 1974 when I took my electronics ticket. I never got around to solving it and now I don't have to. Great stuff. I'm naturally drawn to the simpler solution. Equations have never been one of my strong points. In this array that would make the answer 83.3333(recurring) ohms. Don't be naughty Clockman! 
#17




Thank God someone provided a definitive answer.
I don't think we will take you up on your offer Ray  kind as it was! Now does anyone have a fivedimensional cube version of this puzzle?  that one was a bit too easy really. Brian 
#18




Never mind Ohms Law! During the late 50's and early 60's, which was the best receiver out of the following?
CR300 Mercury & Electra Atalanta R50M (Redifon) Without doubt, the worst one of all was the standby 500Kc/s Rx, "Alert". It had no BFO, and if some idiot was transmitting in CW the whole thing was blocked out! 
#19




IMHO definitely Atalanta, but I'm prejudiced.
Wasn't the "Alert" a TRF and excellent homing device for Uboats? The "Monitor" was a great improvement. The only good thing about the Mercury and Elettra was the ability to monitor 500 with a decent receiver while listening to GRL TFC LST. Never seen a Redifon R50M. 
#20




Watt a shocking puzzle, I've been sat at ohm trying to un earth the current solution before some other bright spark got switched on to volt it was.
__________________
[SIGPIC][/SIGPIC]Oul hand Money can't buy happiness but it's more comfortable to cry in a Porsche than a Skoda 
#21




Redifon Receiver
Now weren't you the lucky one, Fubar?
__________________
Ron _____________________________________________ Never regret growing older. It is a privilege denied to many. Don't worry about old age  it doesn't last. 
#22




Trevorw
Sailed with the first 4 (Marconi) receivers and had the illluck to have to service the R50M and its equallyrubbish successor the R408. What a mess, don't think they had ever heard of selectivity or protection against IF breakthrough.
__________________
Ron _____________________________________________ Never regret growing older. It is a privilege denied to many. Don't worry about old age  it doesn't last. 
#23




R408 rings a bell but I don't know where from. It may have been at college when doing electronics. Martec had all sorts of scrap cr*p for us to play with during practical sessions.
Does anyone remember what a "Seacall" was? Was it any relation to a "Pennant"? I feel a faint stirring of the grey cells but I don't think I had occasion to actually use either of 'em. I did a coastal on the "City of Ripon" in '75 after getting an electronics ticket. That was a blast from the past. 1955 kit with the only concession to the "modern" age being a Raymarc 16 radar. Does anyone remember what a "Nebula" looked like? 
#24




Quote:
__________________
Ron _____________________________________________ Never regret growing older. It is a privilege denied to many. Don't worry about old age  it doesn't last. 
#25




Thanks Ron
I recognise the description of the R408 but never had the "pleasure" of sailing with one. Oh yes I did! You can just see one of the handles on my profile pic which was taken on Big Geordie. I don't remember having any problems with it. I had problems with just about everything else on that ship. The Pennant I sailed with but it seems to have been paired with an Atalanta as a SSB receiver. I usually used to "phone home" quite a lot but it appears I never used it all trip. The Seacall was fitted between trips on one ship but I can't remember it receiving a call. The same ship had the Nebula but I don't remember a thing about it. I don't think I've got a radio room pic so I shall have to do some digging. 

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