Resistance of cube

Discussion in 'Homework Help' started by zamansabbir, Aug 22, 2008.

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  1. zamansabbir

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    May 27, 2008
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    Think about a qube which has resistance of 2 ohms in its each arm. no resistance physically in its diagonals.Now what will be its equvalent resistance.
     
  2. Ratch

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    Mar 20, 2007
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  3. Dave

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  4. zamansabbir

    Thread Starter Member

    May 27, 2008
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    Thanks for your kind information.
     
  5. Lilienfeld

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    Nov 8, 2012
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    Dear Ratch, for me this problem is really interesting. Remember when you were a student and had to solve it. You just know Kirchhoff's laws and have no intuition whatsoever... Then, if you understand the problem, you will be able to generalize it - like putting capacitances or inductances in the cube, (see e.g. here) and be able to apply more advanced methods.

    For me, there are no easy problems, it's just a matter of perspective!
     
  6. Austin Clark

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    Dec 28, 2011
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    Has anyone found a solution assuming the resistors AREN'T of the same value? What is the equation, given all the separate resistor values? THAT would be a challenge, I think, but fun regardless.
     
  7. tshuck

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    Oct 18, 2012
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    I believe someone did toward the end of the link Dave posted...
     
  8. atferrari

    AAC Fanatic!

    Jan 6, 2004
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    Theory....

    If you proceed carefully, you will arrive to the right result.
     
  9. WBahn

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    Mar 31, 2012
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    I have to disagree. Almost all courses use a set of classic problems that have been used by similar courses for decades. Remember, others may have thoroughly plowed the ground in general, but the students taking that course are working on a whole new field (their mind).

    Problems such as this are quite rich and useful because there is nothing magical about them and, at first glance, most people think it should be a very straightforward application of the techniques they already know. But, for most of them, when they try to actually work it they quickly discover that it isn't nearly as easy as they first thought. If you step back and apply the fundamentals, you can push to a solution fairly handily. But you can also discover the power of looking at a problem and capturing behaviors due to symmetry that can lead to a solution in just a couple of lines. In doing so, you continue building up your tool chest with more problem techniques and also build upon your ability to know when, and when not, to use them.
     
  10. Lilienfeld

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    Nov 8, 2012
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    Yes, WBahn, this is exactly what I was trying to say! Once we understand something, it may become trivial to us but we should never forget how dumb we looked at things in the first place :)
     
  11. The Electrician

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    Oct 9, 2007
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    I get a different result. Your final result is 411.7 ohms, but I get 419.47342.

    It's well nigh impossible not to make an error with so many calculations involved, and, sure enough, I note a substantial error in this calculation:

    1 / S = 1 / MO + 1 / PN, then, S = 645.71

    This result should be 657.86; this one error probably accounts for the error in your final result.

    Rather than using all these Δ-Y and Y-Δ transformations that you have to keep track of and perform in the right order, just write down the admittance matrix, invert it and there's your result.
     
  12. atferrari

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    Checked by my grandaughter yesterday.

    She confirmed it is 411.7 Ohms.

    She cheated using LTSpice. Not allowed in the kindergarden.

    Quoting myself:

    If you proceed carefully, you will arrive to the right result.
     
  13. The Electrician

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    Oct 9, 2007
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    I made a single digit error when I copied your resistor values into my matrix, so, yes, I didn't proceed with absolute care. The correct final result is 411.733653166

    But, you didn't either. Your calculation for S gave a value of 645.71 when in fact the calculation 1 / S = 1 / MO + 1 / PN can be easily verified to give 657.86. You gave a 5 digit result but only 1 digit was correct.

    I saw a couple more errors. For example, you have:

    HB = H x B / (H + B + EA) = 396,000 / 3193.9 = 24

    but it should be:

    HB = H x B / (H + B + EA) = 396,000 / 3193.9 = 124

    but you used 124 rather than 24 in subsequent calculations.

    You have:

    SQ = S x R / (S + R + Q) = 559.966.2 / 2177.65 = 257.14

    but it should be:

    SQ = S x Q / (S + R + Q) = 559.966.2 / 2177.65 = 257.14

    I carried out your calculations with infinite precision arithmetic (this is done by using rational arithmetic) and rounded the final values for all your intermediate variables to 12 digits. Comparing them to the values you have on your web page, we see that your variables become increasingly less accurate as all the calculations proceed. Near the end your values are only accurate to 1 or 2 digits.

    [​IMG]

    So how did you manage to get 4 correct digits in your final result? You lucked out. :)

    It just so happens that the errors in your values for the variables QR and V cancel, and by this fortuitous circumstance, give a result accurate to 4 digits.

    And, a sensitivity analysis shows that the final result is very insensitive to the value used for the intermediate variable, S.

    For example, if S is taken to be 600, Rtot = 411.448. If S is taken to be 700, Rtot =411.918. When I first posted in this thread, I guessed that your error in calculating S might account for what I thought was your error in the final result; I hadn't done a sensitivity analysis. But I see now that your calculations will give a good result in spite of errors in S (assuming no other errors creep in).

    Also, if you really want to proceed carefully, you should use more than 4 digits in all your calculations.

    I must say, I'm impressed that you did all those Y-Δ and Δ-Y transformations, got them in the right order and made as few errors as you did. You obviously exercised a lot of care. :)

    However, I think it's better to avoid methods that are so cumbersome; they invite error. You can write the admittance matrix for the resistor cube by inspection and then let the computer (I used an HP-50 calculator) do all the number crunching involved in inverting the matrix.

    And, about your granddaughter, is she really in kindergarten? She can use spice? She's got an obvious career path.
     
  14. The Electrician

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    Oct 9, 2007
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    To show the power of nodal analysis, consider the cube on atferrari's website: http://cablemodem.fibertel.com.ar/atferrari/Solving the resistor cube.htm

    Number the nodes like this:

    [​IMG]

    Once you have entered the admittance matrix for the cube, you can calculate anything you want to know quickly.

    For example, let's calculate the resistances from not just node 1, but from all the nodes to ground:

    Node Resistance to ground
    1 411.734Ω
    2 367.824Ω
    3 154.843Ω
    4 226.732Ω
    5 131.244Ω
    6 289.328Ω
    7 261.563Ω

    It took about 1.5 seconds to calculate all those.

    Now, suppose we want to know the resistance from node 1 to node 2; in other words, connect the ohmmeter across resistor A. The result is 212.785Ω. From node to 1 to all the other nodes the result is:

    Code ( (Unknown Language)):
    1. From node        To node       Resistance
    2.      1               2          212.785Ω
    3.      1               3          402.671Ω
    4.      1               4          440.994Ω
    5.      1               5          421.633Ω
    6.      1               6          354.618Ω
    7.      1               7          400.303Ω
    Suppose we return to the original problem, which was to calculate the resistance from node 1 to ground. If all the resistors were equal, nodes 2, 4 and 6 would be equipotential and connecting them together wouldn't change the resistance from node 1 to ground. But for atferrari's cube, those nodes are not equipotential and connecting them together will change the resistance measured from node 1 to ground. Let's calculate the resistance from node 1 to ground with nodes 2, 4 and 6 connected together. The result is 327.182Ω. What is the resistance from the now connected nodes 2, 4 and 6 to ground? It's 152.657Ω

    How about if we connect nodes 3, 5 and 7 together; then what will be the resistance from node 1 to ground? It will be 392.537Ω

    Suppose we short node 2 to ground? Then what will be the resistance from node 1 to ground? It will be 193.401Ω.

    Let's connect nodes 2, 4 and 6 together and then calculate the resistance from node 1 to the 3 connected nodes 2, 4 and 6. The value is 174.525Ω.

    It has taken me about 12 minutes to write up this post. The various calculations each took no more than 2 or 3 seconds for the calculator to do.
     
  15. atferrari

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    Jan 6, 2004
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    As a Chief Officer I was used, in the pre-PC times, to run routinely the moment calculations requiered to know the final trim and mean draft of a vessel. With just a small calculator.

    Doing it with lot of care rechecking every step prior moving to the next was normal for me.

    And my grandaughter, she is the teacher. She is not allowed to run alien software in their PC.

    If I find my calculations (they should be somewhere) I could post them here. Be well.
     
  16. The Electrician

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  17. atferrari

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    Yes, they are but I will look for the paper sheets where I did it so long ago.

    I do not discard eventual errors while typing. It would be interesting to check the whole thing against yours.
     
    Last edited: Nov 15, 2012
  18. atferrari

    AAC Fanatic!

    Jan 6, 2004
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    Hola TE,

    Here you have them. Exactly what I did.

    There is at least one figure wrong in my site. It is 124 and not 24.

    I recall also, implementing this in a breadboard and measuring resistance which gave a quite close 411 ohms.

    Infinite precision, sensitivity analysis? :eek: Oh no, not at that time. :)

    It all started by a comment in the EPE forum. I think it was a Swedish contributor who said that it was possible to solve it using delta-wye transforms. I got astonished to see that once you focus yourself on how to, it is quite simple to advance as I did. More or less as when designing a new algorithm when programming a micro.

    Be well.
     
  19. The Electrician

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    Oct 9, 2007
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    I wanted to see if I could track down the error you made in the calculation of the variable "S".

    You have 1/S = 1/MO + 1/PN

    Using your numbers, you should have had 1/S = 1/985.2 + 1/1980; this should give 657.86, but you got 645.71, which has only 1 digit accuracy.

    I wondered what error you made here and with a little exploration I discovered that if I calculate 1/S = 1/958.2 + 1/1980, i get S = 645.71, the very result you got. It seems clear that you transposed two digits in the value for MO when you calculated S.

    Even though you exercised extreme care, an error crept in, but luckily the error propagated to QR and V mostly cancelled in the final calculation.

    An ordinary person, not having your history of doing moment calculations, would be unlikely to get a correct result after so many Δ-Y and Y-Δ transformations.

    So, even though it can be done, I think it's not a good way to go for a complicated network.
     
  20. Wendy

    Moderator

    Mar 24, 2008
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    Welcome to AAC.

    Congratulations, you have practiced the arcane art of necromancy, the revival of a long dead thread. Likely the OP (Original Poster) has solved his problem in the years that has passed, or thrown it away, or something.

    If you want you continue this conversation please start a new thread instead.
     
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