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Thevenin Equivalent Resistance for a Black Box

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data
    I am trying to find the theoretical Thevenin equivalent resistance for a black box that I used in an experiment. The measured Thevenin equivalent resistance was roughly 4786[tex]\Omega[/tex]. However, I have tried several times to solve the problem, but keep arriving at a theoretical value of around 3511[tex]\Omega[/tex]. That just seems to be a little too low...

    I have attached the circuit schematic, with resistor values included.
    Could someone please look at the problem and see if I am on the right track. If so, any pointers on how I should finish it?

    2. Relevant equations
    R[tex]_{Th}[/tex]= V[tex]_{Th}[/tex]/I[tex]_{sc}[/tex]
    Resistors in Parallel: R[tex]_{eq}[/tex]= (1/R[tex]_{1}[/tex]+...+1/R[tex]_{n}[/tex])[tex]^{-1}[/tex]
    Resistors in Series: R[tex]_{eq}[/tex]=R[tex]_{1}[/tex]+...+R[tex]_{n}[/tex]
    Y to Delta Transformation:
    3. The attempt at a solution
    STEP ONE: R2, R3 in parallel
    R23 = [(1/R2)+(1/R3)]^-1 = [(1/9.97k[tex]\Omega[/tex])+(1/1.001k[tex]\Omega[/tex])]^-1 = .9096k[tex]\Omega[/tex]

    R235 = R23 + R5 = .9096k[tex]\Omega[/tex] + 10.0k[tex]\Omega[/tex] = 10.9096k[tex]\Omega[/tex]

    STEP THREE: Y TO [tex]\Delta[/tex] TRANSFORM R4, R6, R7
    Rc = [(R4*R6)+(R4*R7)+(R6*R7)]/R4 = 24.95k[tex]\Omega[/tex]
    Ra = [(R4*R6)+(R4*R7)+(R6*R7)]/R6 = 24.85k[tex]\Omega[/tex]
    Rb = [(R4*R6)+(R4*R7)+(R6*R7)]/R7 = 12.39k[tex]\Omega[/tex]

    [(1/24.95k[tex]\Omega[/tex] )+(1/10.9096[tex]\Omega[/tex] )]^-1 = 7.59k[tex]\Omega[/tex]

    From here, I have tried different combinations of source transformations, converting the delta back to a Y, etc. to try and get an equivalent resistance.

    Attached Files:

  2. jcsd
  3. Mar 21, 2009 #2

    The Electrician

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    If you post your image on a free image hosting site, and put the link here, we won't have to wait for the attachment you already posted to be approved.
  4. Mar 21, 2009 #3
    Thanks for the tip. I've uploaded it:

    http://www.mountainescapesproperties.com/theveninpf.jpg" [Broken]
    Last edited by a moderator: May 4, 2017
  5. Mar 21, 2009 #4

    The Electrician

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    Have you studied any of the general network solution techniques, such as the nodal method or the loop method?

    If you were to use the nodal method, you only have 3 nodes in this circuit, and the solution would be fairly straightforward.

    You are assuming the DC voltage source has zero internal resistance, right?
  6. Mar 21, 2009 #5

    The Electrician

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    By the way, notice that R2, R3 and R4 are all three in parallel. If you replace then by their equivalent, and replace R5 and R6 by an equivalent, you will have a simple ladder network. You can then work that out very easily, without any need for delta-y or similar transformations.
  7. Mar 21, 2009 #6

    The Electrician

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    I get 3784.959 for the theoretical value.

    I assume you got your measured value in the lab with a simple ohmmeter measurement. Did you remember to replace the DC voltage with a short circuit before you made your measurement?
  8. Mar 21, 2009 #7

    Thanks for all of your advice.

    I came to my measured value of R_th by measuring the open circuit DC voltage between the terminals of the black box (V_th) and then measuring short circuit DC current between the terminals of the black box (I_sc)
    then calculating R_th = (V_th)/(I_sc)...
  9. Mar 21, 2009 #8

    The Electrician

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    Well, that should work.

    I think if I were you, I would get an ohmmeter and verify that the resistance of each of the resistors in your circuit is what you think it is. Also verify the connections.
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