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Finding the equivalent resistance

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the equivalent resistance of the following circuit as seen from terminals a and b.
    BTyU8.jpg

    2. Relevant equations

    For resistors in series we can sum their values;
    For resistors in parallel we use [tex]R_{eq} = 1 / (1/{R_1} + 1/{R_2} + ... 1/{R_n )}[/tex]

    3. The attempt at a solution

    I don't recall we talking about it in class but from reading a book i know another possible configuration apart from parallel and series is in Δ or π(pi), and so that we can know the equivalent we should transform those to wye (Y) or T respectively. Since we didn't talk about that in class i suppose there must be another way to solve this. But i'm at a lost since i can't identify any parallel or series resistors besides the ones with 40 and 20 ohms which are in series.
     
    Last edited: Mar 27, 2012
  2. jcsd
  3. Mar 27, 2012 #2

    NascentOxygen

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    Staff: Mentor

    It looks formidable at first glance, but once you get into it you'll see there are plenty of places where it can be simplified by combining two (or more) resistors into just one resistor. https://www.physicsforums.com/images/icons/icon3.gif [Broken]

    See that 25Ω and the 15Ω resistor below it in the figure? These are in series because every electron that passes through one of them must also pass through the other. There is simply no other alternative path.

    See the "vertical" 10Ω resistor and the 15Ω it is joined to. They are in parallel because they are joined together at one end, and at their other end they are similarly joined together---by a piece of wire. Any electron that passes through one of them will not pass through the other.

    See how you go now. (I hope delta or pi don't come into the picture, otherwise I won't be able to do it. :wink: )
     
    Last edited by a moderator: May 5, 2017
  4. Mar 27, 2012 #3
    For those, you can always apply a test voltage and compute the resistance as

    [tex] R = \frac{V}{I}[/tex]

    I believe it is doable.
     
    Last edited by a moderator: May 5, 2017
  5. Mar 28, 2012 #4
    Hello, I can confirm that you can do this without Δ-to-Y/Y-to-Δ transformations!

    Redrawing the circuit after making simplifications can really help make connections more clear (with a few explicit nodes). I've done the simplifications NascentOxygen mentioned above and have drawn how I moved resistors after simplifying. There's still one big step left but hopefully this gets you on the right path:

    Req.jpg
     
  6. Mar 28, 2012 #5
    So in trying to understand what you did ( btw nice job :) ) i made it a bit more colorful . Same color = same voltage (node), except red where it means they're in series. (i switched green and yellow by mistake from [2] to [3])

    BbemJ.jpg

    Now i know 15+25 are in parallel with 40+20. But the other 4 resistors i'm not sure what i can do. I would be tempted to say that 10||15 is in series with 30, but if i put those two as one, there's no space for the wire a to be.

    Also, a and b aren't in parallel as well since a has blue and yellow voltage while b has yellow and green (so to say).
    hm...
     
  7. Mar 28, 2012 #6

    NascentOxygen

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    Staff: Mentor

    That's looking good. :cool:
     
  8. Mar 28, 2012 #7
    In trying to solve that i came across another circuit whose equivalent resistance i can't solve. Perhaps knowing to do this one would me do the one of the topic. ( this one is related to Thevenin's and Norton's theorems )

    Wanting to know the equivalent resistance as seen from a and b i opened and removed the current source. Then i'm left with a combination of resistors that don't seem to be neither in series or parallel. Yet my teacher solved this as

    (6+6)||4, and i'm not really sure why. For example, if we remove wires a and b (and it should be no problem since there's no current from a to b anyway, and the voltage is equal) it's symmetrical, so why can't it be (4+6)||6 ? Is it because if we send a current through a, then after the first node it only has one path, which is 6+6 ohms, so they are in series? That means that if we put a and b on the other side the equivalent resistance would be (4+6)||6..

    bLRrQ.jpg
     
    Last edited: Mar 28, 2012
  9. Mar 28, 2012 #8
    With this new logic (i hope it's correct) i think i was able to finish what was asked:

    7s9Dk.jpg

    Finally the last 3 resistors are in series.

    Please correct me if i'm wrong (and the post before too :) )
     
  10. Mar 28, 2012 #9
    Correct, the placement of a and b mean everything (can't move/remove them). After all, if we placed a and b on the same node, we could get zero equivalent resistance.

    Your solution to the problem also looks good, well done! I really like your use of color for the different voltage levels. I hope you don't mind if I use that technique in the future. :)
     
  11. Mar 28, 2012 #10
    Awesome. :) Of course not, go ahead. I'm going to have test on 16th april so any help is appreciated, and yours was good. NascentOxygen's aswell. Thanks :)
     
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