Find the equivalent resistance

Discussion in 'Homework Help' started by mrmarshmallow, Sep 16, 2009.

  1. mrmarshmallow

    Thread Starter New Member

    Sep 16, 2009
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    I'm not sure how to find the equivalent resistance in the attached circuit diagram. I know the resistors aren't strictly in series or parallel. I don't think it can be shown as a delta-wye transformation, but I'm still a little unfamiliar with them since we haven't reviewed them in class. Our textbook doesn't appear to provide any relevant discussion and any kind of logical approach seems to be evading me at the moment. I would appreciate it if I could be pointed in the right direction. Thanks.

    P.S. Please excuse the crude drawing.
     
  2. ELECTRONERD

    Senior Member

    May 26, 2009
    1,146
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    http://forum.allaboutcircuits.com/attachment.php?attachmentid=12139&d=1253121638

    This might help you, notice that your resistors in a triangular configuration. Click on the link above for the example schematic.

    I wasn't able to place the resistors in a triangular format, but you can draw it on paper.

    Ra = (R1xR2) / (R1+R2+R3)

    Rb = (R1xR2) / (R1+R2+R3)

    Rc = (R2xR3) / (R1+R2+R3)

    Example, in the attached schematic, we want to have the resistance from point Ra to the other point (A & B). Therefore, we solve using the Ra formula. If all the resistors are 1Ω, we can equate the following:

    Ra= (1x1) / (1+1+1) = 0.33Ω

    Now, to solve for R4 and R5, we can replace our triangular resistive circuit with a simple "Ra." Notice that R4 and R5 are parallel. So 1 / (1/1 + 1/1) = 0.5Ω Thus, the total resistance is 0.33+0.5 = Approximately 0.83Ω.
     
  3. ELECTRONERD

    Senior Member

    May 26, 2009
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    Do you have an answer for your diagram that they provide?
     
  4. mrmarshmallow

    Thread Starter New Member

    Sep 16, 2009
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    I reduced the upper triangle to .33R and the lower to .67R. Adding those to the other R resistor, I get 2R.
     
  5. mrmarshmallow

    Thread Starter New Member

    Sep 16, 2009
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    Just from intuition, that answer seems a little too low. But I don't have a solution to compare it to.
     
  6. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    I get 24R/11, or 2.1818181818... ohms.

    Plainly, if the 2R resistor becomes R, then the equivalent resistance is exactly 2 ohms. But if the 2R resistor remains 2R, then the equivalent resistance must be greater than 2 ohms.
     
  7. ELECTRONERD

    Senior Member

    May 26, 2009
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    I get 0.33R for the upper triangle. Show me how you get 0.67R for the lower one. What did you do?
     
  8. ELECTRONERD

    Senior Member

    May 26, 2009
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    I get 2.83R. Because the upper triangle equals 0.33R and if you substituted that whole triangle with a simple Ra (0.33R), you would find that the other two resistors (2R and R) are in parallel. So you equate that and get 1.5R. Add them up, 1R + 0.33R + 1.5R = 2.83R. What did you do?
     
  9. AdrianN

    Active Member

    Apr 27, 2009
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    I think the easiest is to transform the star, with the center in the lower left corner, into a triangle. The star has all its resistors R. Therefore, the triangle will have all its resistors equal to 3R. The result is:

    Rtotal = R + (R||3R +2R||3R)||3R = 2.182 R,

    the same result as Electrician's.
     
  10. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    I used the admittance matrix method, which is a systematized version of the nodal method, as I explained in another thread.
     
  11. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    You're not doing it right. If you convert the upper triangle into a star, then the 2R resistor has a R/3 resistor in series with it, and the bottom R also has R/3 in series with.

    You end up with the very leftmost R in series with R/3 (from the star), and then in series with a parallel combination of (R+R/3) and (2R+R/3).

    Final result = R + R/3 + (R+R/3)||(2R+R/3) which equals (if R = 1)

    1 + .333333333 + .8484848484 = 2.1818181818
     
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