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Finding Equivalent Resistance

  1. Sep 14, 2003 #1
    I'm stumped by this problem.

    There's no symmetry in the circuit, so I can't ignore the 1 ohm resistor that bisects the triangle. Is there something else I'm missing? I'm supposed to show the Rab = 27/17 ohms.

    Please see the attached diagram.

    Thanks.
     

    Attached Files:

  2. jcsd
  3. Sep 14, 2003 #2

    dduardo

    User Avatar
    Staff Emeritus

    You have to apply a delta-wye transformation. See attached picture to see what I mean.

    a = A*B / ( A + B + C )
    b = A*C / (A + B + C )
    c = B*C / (A + B + C)

    Once you get a, b and c, then...

    b is in series with the 1ohm resister
    c is in series with the 5ohm resister

    These two series are parallel with each other and in series with a
     

    Attached Files:

  4. Sep 15, 2003 #3
    Hmmm..... This technique isn't mentioned in my Physics book.

    a, b, and c are in ohms? a = 3/5, b = 1/5, c = 3/5.

    so b + 1 ohms = 6/5 ohms
    c + 5 ohms = 28/5 ohms
    The equivalent of these parallel resistors is 85/84 ohms + 3/5 ohm
    = 677/420 ohms.

    This doesn't agree with the answer in the book which is 27/17 ohms.

    The book has been known to be wrong.

    Can anyone else verify?

    Thanks.
     
  5. Sep 15, 2003 #4

    dduardo

    User Avatar
    Staff Emeritus

    the answer does come out to be 27/17 ohms
    .check 85/84 ohms + 3/5 ohm

    Here are the steps for people interesting in knowing how to get the answer.:

    a = A*B / ( A + B + C )
    a = 1*3 / ( 1 + 3 + 1)
    a = 3/5 ohms

    b = A*C / ( A + B + C )
    b = 1*1 / ( 1 + 3 + 1)
    b = 1/5 ohms

    c = B*C / ( A + B + C )
    c = 3*1 / ( 1 + 3 + 1)
    c = 3/5 ohms

    The circuit should look like this after the transformation
    .............................b=1/5ohm.....1ohm
    ...a=3/5ohm...---XXXX-----XXXX---
    ---XXXX----|.................................... |----
    ........................ ---XXXX-----XXXX---
    .............................c=3/5ohm.......5ohm

    please ignore the periods ( . ). They are there for spacing.

    The two resistors on top are in series and combined equal 6/5 ohms

    The two resistors on the bottom are in series and combined equal 28/5 ohms

    so now you have a circuit that looks like this

    ........................6/5ohms
    3/5ohms....---XXXX----
    --XXXX--|.......................|-----
    ...................----XXXX----
    ..........................28/5 ohms

    The top and bottom resistors are in parallel. The equation is R1*R2/(R1+R2)

    So the combined parallel resistors equals 84/85 ohms

    finally to have a circuit that looks like this


    ....3/5ohms...............84/85ohms
    ----XXXX-----------XXXX-----

    The final two resistors are in series so just add them up

    3/5ohms + 84/85ohms = 27/17ohms
     
  6. Sep 16, 2003 #5
    Thanks. You're right. I goofed with the math.

    I was able to figure it out the hard way going through the three loops to get three equations using Kirchhoff's rules.

    I = I1 + I2 => for the currents entering the the various junctions in the network.
    E = electromotive force attached to the network.

    5I1 - I2 = 3I
    7I2 - I1 = 5I
    E - 1.0ohm(I1) - 1.0ohm(I2) = 0

    Since E = I(R equivalent) if you do the algebra with the equations above you get E = (27/17ohm)I

    Thanks for the help and the shortcut. They're much appreciated.
     
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