Very basic resistors, but cannot figure them out

Discussion in 'Homework Help' started by dapEK, Oct 7, 2015.

  1. dapEK

    Thread Starter New Member

    Oct 7, 2015
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    I'm a mechanical engineering student, so taking this course which has an electrical component is very hard to understand.

    We were given a "black box" with four terminals sticking out of the top, and told to take the voltmeter and do a set of readings for each combination of terminals. 6 combinations total, and they are listed below:

    Segment Resistance(Ω)
    AB 500
    AC 500
    AD 11.4
    BC 11.4
    BC 499
    DC 499

    Then, the instructions are to find 6 equations that will give you the internal resistors. I have found 4 of those:

    1/Rab= (1/AB ) + (1/ BC+CD+AD)
    1/Rbc= (1/BC) + (1/ CD+AD+AB)
    1/Rcd= (1/CD ) + (1/AD+AB+BC)
    1/Rad= (1/AD ) + (1/ AB+BC+CD)

    Now, I can't seem to figure out the last two? A classmate said that the last two had to do with if the arrangement inside is somehow diagonal. I don't understand how to get it from there.
     
  2. dannyf

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    Sep 13, 2015
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    Tough to get resistance readings off a voltmeter.
     
  3. dannyf

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    Between four terminals, you can have six total connections / resistors. It is not difficult to solve them.

    However, given the pattern of the readings, you know that they are only three uniquely valued resistors - or they wouldn't give symmetrical readings.

    That greatly simplifies it for you.
     
  4. dapEK

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    Oct 7, 2015
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    I used the same theory to derive these two equations in case there's a diagonal arrangement:

    1/Rac=(1/AC) +(1/AD+CD) +(1/BC+AB)

    1/Rbd= (1/BD) +(1/BC+CD) + (1/AD+AB)

    does this seem correct? it's all I can think of.
     
  5. WBahn

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    Mar 31, 2012
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    You need to learn about order of operations.

    1/AD+CD = (1/AD) + CD

    Get in the habit NOW of paying attention to these details -- it will save you MUCH grief down the road.

    You should be very suspicious of your equations because the form is different for the diagonals than for the adjacent terminals. But since every terminal is connected to every other terminal, how can that be? I could physically swap two of the terminals without disconnecting a thing inside and now the diagonal terminals are adjacent and some terminals that were adjacent are diagonal. Yet nothing has changed. For that matter, I could put the four terminals on the four corners of a tetrahedron and then one resistor along each edge. Now the absolute symmetry of the topology is obvious and you KNOW that the form of the equations for the resistance between any two terminals MUST be the same.
     
  6. dapEK

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    Oct 7, 2015
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    No, I know order of operations. That's just how the TA emailed me to explain the formulas. He sent me one, I derived the rest.

    When it says 1/BD +CD that means: 1/(BD+CD) but I figured you'd know what I meant since I knew what he meant, if that makes sense.
     
  7. dapEK

    Thread Starter New Member

    Oct 7, 2015
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    Here's the list of formulas, the format was suggested by the TA:
    1/Rab= (1/AB ) + (1/ (BC+CD+AD) )
    1/Rbc= (1/BC) + (1/ (CD+AD+AB) )
    1/Rcd= (1/CD ) + (1/(AD+AB+BC))
    1/Rad= (1/AD ) + (1/(AB+BC+CD))
    1/Rac=(1/AC) +(1/(AD+CD)) +(1/(BC+AB))
    1/Rbd= (1/BD) +(1/(BC+CD)) + (1/(AD+AB))

    I've added parenthesis in case you were confused.

    I just am having a hard time understanding the theory behind it. Can anyone explain that.
     
  8. WBahn

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    Don't make your readers guess what you mean (and your TA shouldn't have done it to you -- feel free to point out his sloppiness to him). And figuring that your readers will "know what you mean" does not in any way excuse such sloppiness in basic notation. If you don't mean (1/BD) + CD, then don't write an expression that means (1/BD) + CD.

    I can't begin to tell you how many times I have see someone type something like A/B+C when they meant A/(B+C) and, as a direct result of that, made a mess of things shortly thereafter.

    For instance, what they have is (A/(B+C)) + D but what they write is A/B+C + D and then, sometimes on the very next line, they treat it as though it were A/(B+C+D). All because they couldn't be bothered to pay attention to the most basic details.

    I also can't tell you how many times I've seen people put A/B+C into a program statement or spreadsheet formula and then not be able to figure out why their results are wrong. If you are in the habit of being sloppy when you should type A/(B+C) into a forum post, you will follow those habits when you type it into something that a computer has to act on, if for no other reason than you have trained yourself that it is an acceptable way to type that expression.
     
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  9. WBahn

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    Again, does it make sense that the form of the first four equations is different than the last two? Doesn't symmetry dictate that all six equations must have the same form?
     
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  10. WBahn

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    Also, if all you give us are your final results, we aren't in a position to look over your work and point out where you went wrong. All we can do is give you a Yes/No on whether your results are correct. That's best case. But to do that, you are requiring that we work the problem ourselves from scratch so that we can compare our results to yours. Most of us aren't willing to do that.

    Instead, SHOW YOUR WORK. That way we can follow along through your work and point out the first spot where you've gone astray. You can then fix that, update your work, and we can make another pass.
     
  11. dapEK

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    Oct 7, 2015
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    I used the same principle as he seemed to point us towards, but I figured because they were, as he said "In case there is a diagonal arrangement" they would be different. I treated them as though there was an extra resistor going diagonal for each of those, and then do three different segments in parallel.

    The diagram I'm about to insert is the one he drew on the board for this lab:[​IMG]

    For the first four equations, I think what he did was segment it into two parts, and treat it as parallel. So for the diagonal, I did three sections parallel assuming there's now one in the middle.
     
  12. dapEK

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    Oct 7, 2015
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    http:// imgur.com/rUSjRZn

    the image wont show, take out the space in the link.

    There is no work, I used logic. He wrote 1/Rab= (1/AB ) + (1/ (BC+CD+AD) ) to find the resistance of Rab

    This I broke down in more steps to use for the other equations:
    1. section AB
    2. sections BC CD and AD in series

    so, that would make a funny looking parallel circuit, right? That's why the parallel formula is used
     
  13. dapEK

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    Oct 7, 2015
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    I received an email from the TA just now, and he said, "The equation u sent is wrong but you are very close. Just remember there can be only a total of 2 connections in a parallel" so I figure I can remove one of the segments in each of the diagonals, and it will be correct? So, for example, if you look at the diagram and see Rac, I can choose to eliminate the series AD+CD, or AB+BC
     
  14. WBahn

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    This is correct for the case of FOUR resistors connected in a ring with no diagonal resistors. You have two current paths that are in parallel. One through the direct connecting resistor and the other through the series combination of the other three.

    But once you add in the diagonal resistors you know longer have a series connection of resistors for the second path.

    These are the parallel sections for the four resistor case. No problem there.

    I don't follow this at all.
     
  15. WBahn

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    This sounds like you are looking for "a happening", which means that you make a change to what you have and hope that it just happens to work. Not the way to do things.
     
  16. WBahn

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    But they aren't in parallel. Parallel requires that two paths have the same voltage across them and that the current entering each path is the same as the current leaving that path. You have a resistor that interconnects the two paths and allows current to shift from one path to the other.
     
  17. dapEK

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    Oct 7, 2015
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    I don't understand the theory, so that's making things very difficult. I'm trying to understand. So what's the best way to approach a "possible diagonal arrangement"? It would still be a parallel, but now I realize you can't parallel with three lines, only two.
     
  18. dannyf

    Well-Known Member

    Sep 13, 2015
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    Think about a square,

    A <-> B
    | |
    | |
    D <-> C

    Two additional resistors on the diagnose, AC and BD.

    Given the symmetry of readings presented earlier, It must be true that

    AB = AD=R1, CD = CB=R2, AC = DB = R3

    Take AC for example, there must be three legs:

    1. A-B-C,
    2. A-C,
    3. A-D-C

    You can work out the math there easily -> alternatively you can imagine A-C being split in half and calculate over two legs.

    Hope it helps.
     
  19. dapEK

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    Oct 7, 2015
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    Actually, that does. Thanks, dannyf. Is it possible to use pythagorean and treat this as a right triangle?
     
  20. WBahn

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    Sure you can parallel with three lines. I don't know what point he was trying to make with that comment.

    Perhaps if we redraw the circuit so that things are nice and rectangular:

    upload_2015-10-7_19-26-25.png

    First, convince yourself that this circuit is identical to a black box with four terminals having a resistor connected between each of the terminals.

    Now, if we want the find the total resistance we would measure between points A and B, we have Rab in parallel with everything else. But everything else is neither in parallel nor in series. If Rcd weren't there, we would have a total of three parallel paths, namely Rab, (Rac+Rbc), and (Rad+Rbd). But Rcd IS there, and so the right two vertical paths are NOT in parallel.
     
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