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Combining Resistors

  1. Apr 25, 2015 #1
    1. The problem statement, all variables and given/known data

    P26_36.jpg Assuming R = R' in the network shown above, determine the net resistance between the points a and c. Express your answer as a number times R.

    b. Assuming R = R', determine the net resistance between the points a and b.

    c. Now suppose R' = 2.5R, and find the net resistance between points a and b.

    2. Relevant equations

    Equivalent Resistance for Resistors in Series = R1+R2......etc
    Resistors in Parallel = 1/Equivalent Resistance = 1/R1+1/R2...etc

    3. The attempt at a solution
    I am comfortable with knowing how to combine the resistors but I am just unsure about what it means to "determine the net resistance between the points a and c" does it mean all the resistors (all the resistors within the big triangle)? And for the net resistance between points a and b does it mean just the right small triangle of the bigger triangle?
     
  2. jcsd
  3. Apr 25, 2015 #2

    gneill

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    The "net resistance" between two points takes into account all the resistances of the circuit, replacing them all with one equivalent resistor between the two points. Think of it as the resistance that you would measure if you were to connect an ohmmeter between the points.
     
  4. Apr 25, 2015 #3
    So there's 3 resistors between a and b and 6 between a and c?
     
  5. Apr 25, 2015 #4

    berkeman

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    It means if you took your DVM and put your probes at points a and c, what would the DVM readout be?

    There may be some tricks you can use here, but for complicated arrangements of resistors like this, I just write KCL equations for the setup, and solve that way. Like, put a 1 volt source across the bottom resistor R from a to c, and solve for the resulting node voltages and currents. You define the other 2 nodes in the middle at b and at the top of the triangle. Can you set up those 4 KCL node equations and see if the solution isn't too bad?


    EDIT -- Beat to the punch by gneill again! :smile:
     
  6. Apr 25, 2015 #5
    Its been a long time since I have done ohms law...I will consider this problem... Nice puzzle.. Your math look correct.. Net resistance it the total resistance between A-C points. It would be a lot easier if all the resistor were numbered.. then we could refer to them in an math example. I normally convert this in crude numbers, for example add "aR and cR"... 5 ohms, while aRc is just 2.5 ohms... Link to:http://www.sengpielaudio.com/calculator-paralresist.htm This program computed a solution of 0.533 ohms... -e
     
  7. Apr 25, 2015 #6

    gneill

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    Let's put it this way, ALL of the resistors will contribute to the net resistance between any two points (unless you can show that no current would flow through a given resistor if a potential were placed across the points in question. Then that particular resistor could be ignored).
     
  8. Apr 25, 2015 #7

    berkeman

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    :smile:
     
  9. Apr 25, 2015 #8
    So I'm just going to explain how I labeled them for the sake of being on the same page when talking about what's in parallel and what's in series.

    Going clockwise around the big triangle, left side is R1, right side is R2, bottom side is R3. Then on the inside, left side is R4, resistor going up is R5 and the one on the inside right is R6.

    R1 R2 and R3 are in series
    R1 R4 and R5 are in parallel
    R2 R5 and R6 are in parallel
    and R1 R4 R5 R2 are all in parallel with R3

    ?
     
  10. Apr 25, 2015 #9

    berkeman

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    Use gneill's hint that I highlighted in my previous post...
     
  11. Apr 25, 2015 #10
    From point a to c I got the total resistance of 4/R+3R How is that not right?
     
  12. Apr 25, 2015 #11

    berkeman

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    I got something a lot simpler using gneill's hint. What is his hint? How can it be applied to this circuit. Look for symmetries that lead to components with no voltage drop across them....
     
  13. Apr 25, 2015 #12

    gneill

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    The units don't work for one. You can not add 1/Ω to Ω.

    Can you show your work?
     
  14. Apr 25, 2015 #13
    I added R1 and R4 first (1/R+1/R=2/R) and equivalent resistor for those two is R/2
    Then I added R5 and R2 (1/R+1/R=2/R) and equivalent resistor for those two is R/2
    So not it's a single triangle with another resistor on the right side (in parallel)
    I added those two (1/(R/2)+1/R=3/R) and equivalent resistor is R/3
    then I just have one triangle with three equivalent resistors: R/2 R/3 and R
    Adding those in series, I get 11/6 R
     
  15. Apr 25, 2015 #14

    gneill

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    How did you determine that you could treat R1 and R4 as being in parallel? (I'm not saying that it is wrong! You just need to justify the operation)
    Again, can you justify this? (The answer should be no this time! Why? What assumptions are you making?)

    Some of the combinations you've chosen to treat as being in parallel can only be justified if you make explicit assumptions and then stick to them. This didn't happen :oops: As a result you've arrived at an incorrect result.

    Did you spot any resistors that can be ignored? Think about what potentials might appear at the various nodes if you were to place a voltage source across a and b.
     
  16. Apr 25, 2015 #15
    So the one on the inside going up is the one to ignore. Because all the rest are connected to the nodes from a to c
     
  17. Apr 25, 2015 #16

    gneill

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    Correct conclusion, but I'm not comfortable with your argument the way you state it. To clarify, Suppose you imagine a voltage source placed across a-c:

    Fig1.gif
    What can you say about the potentials at nodes b and d (I added d to your diagram)? Note that symmetry is a factor here since all the resistors have the same value (R1 and R2, R4 and R6 in particular). R1 and R2 form a potential divider, as do R4 and R6.
     
  18. Apr 26, 2015 #17
    So adding R1 and R4 in parallel I get an equivalent resistor of 2/R and using symmetry, R2 and R3 have an equivalent resistor of R/2
    now it's just one triangle with a resistor on each side (2/R, 2/R and R) adding those in series, I get 2R. Why is that not right?
     
  19. Apr 26, 2015 #18

    gneill

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    2/R is not a resistance. It has units of Ω-1, or S (Siemens), which is the unit of conductance. When you combine resistors in parallel, don't forget final inversion: 1/R = 1/R1 + 1/R2, and you're looking for R.

    I think you meant R2 and R6 as the symmetric pair, right?
    Fix your parallel resistor calculations.
     
  20. Apr 26, 2015 #19
    1/R1+1/R4=1/R1/R=2/R then equivalent R is R/2 and same for 1/R2+1/R6 it also has equivalent R of R/2
    So adding those two in series with the last one, I got 2R
     
  21. Apr 26, 2015 #20

    gneill

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    Are all three really in series from a to c?
     
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