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Diamond-shaped Circuits

  1. Oct 16, 2011 #1
    How would I find the equivalent resistance of a diamond-shaped circuit? Is the middle resistor simply omitted?

    By diamond-shaped, I mean it looks like a Wheatstone bridge, but it has a resistor in the middle instead of a current detector.
     
  2. jcsd
  3. Oct 16, 2011 #2
    Say R1 to R5 form the diamond shape, where there are three chain in parallel.

    1) R1 in series with R2.
    2) R3.
    3) R4 in series with R5.

    So the equivalent R is very simple.......All 3 chain parallel, or if you like, find;

    1) (R1+R2)//R3.

    2) [(R1+R2)//R3]//(R4+R5).

    Simple!!!

    Am I getting it right? You testing us?!!!
     
  4. Oct 16, 2011 #3
    I thought you couldn't use series-parallel relationships to solve a Wheatstone bridge, and had to use mesh analysis or something.
     
  5. Oct 16, 2011 #4
    yungman, the OP might want the resistance between the other two points on the diamond circuit, which would be the resistance that would be the load on the power supply if the diamond circuit was powered like a Wheatstone bridge.


    Code (Text):


            A
            |
           /_\
           \ /
            |
            B

     
     
    Last edited: Oct 16, 2011
  6. Oct 16, 2011 #5
    No, your circuit looks like this:

    |
    /|\
    \|/
    |

    My circuit looks like what the person above just posted.

    I created the circuit using a circuit simulator, and it showed that there would be no current flowing through the middle resistor. I have no idea whether or not this is correct.
     
  7. Oct 17, 2011 #6
    It's only correct for balanced bridge. If R1/R4 = R2/R3, then no current would flow through the "middle resistor," but otherwise there would be current there (when VAB ≠ 0).

    Code (Text):

           A
           |
     R2   /_\  R1
     R3   \ /  R4
           |
           B

     
     
    Last edited: Oct 17, 2011
  8. Oct 17, 2011 #7

    jim hardy

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    simplify: think of a bridge as two voltage dividers on same supply

    IF they're both set to the same fraction of that supply
    THEN they're the same voltage so no difference and no current flows
    whether it's a resistor, current or voltage detector between the dividers
     
    Last edited: Oct 17, 2011
  9. Oct 17, 2011 #8
    Hmm, so there's no way to simplify the circuit into one resistor unless R1/R4 = R2/R3?
     
  10. Oct 17, 2011 #9

    jim hardy

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    well there's the old standby of write equations from Kirchoff's laws

    5 resistors, five unknown currents, five equations

    as we said in the 60's - should be plug&chug.
     
  11. Oct 24, 2011 #10
    Try Googling "wye-delta conversion"

    You should find all you need.
     
  12. Oct 24, 2011 #11

    vk6kro

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