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Resistors in series/parallel

Discussion in 'Electronic Basics' started by Joachim, Nov 14, 2005.

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  1. Joachim

    Joachim Guest

  2. Tim Williams

    Tim Williams Guest

  3. John - kd5yi

    John - kd5yi Guest

    (R1 || R2) + (R3 || R4).

    Or not.
     
  4. Noway2

    Noway2 Guest

    This is a classic problem for circuit analysis classis as the diamond
    shape configuration throws alot of people off. In order to determine an
    equivalent value, it is necessary to pick the problem apart in smaller
    step.

    First, redraw the circuit in a more familiar form with the resistors
    drawn horizontally and vertically.

    Once that is done, you should be able to see that the circuit is a
    simple series - parallel combination. Proceed through the circuit,
    simplifying the parallel combinations to get the single value
    equivalence.
     
  5. Joachim

    Joachim Guest

    Thanks for your help, I would be happy if you could now help me confirm
    this:

    R1 = 1/2
    R2 = 3/2
    R3 = 1/2
    R4 = 3/2
    R5 = 1/2

    Does Rtot = 3/4 ?
     
  6. Fester

    Fester Guest

    Please,

    Do your own homework... We already graduated but on our own.

    Fester
     
  7. Bob Monsen

    Bob Monsen Guest

    Use the PI -> WYE conversion formula for R1,R2,and R5. That simplifies
    it...

    ra=r1*r2/(r1+r2+r3)
    rb=r1*r5/(r1+r2+r3)
    rc=r2*r5/(r1+r2+r3)

    Then, r = ra + parallel(rb+r3,rc+r4)

    so

    r = (r1 r2 r3 + r1 r2 r4 + r1 r2 r5 + r1 r3 r4 + r2 r3 r4 + r1 r4 r5 +

    r2 r3 r5 + r3 r4 r5) / (r1 r3 + r1 r4 + r2 r3 + r1 r5 + r2 r4 + r2 r5 +

    r3 r5 + r4 r5)

    ---
    Regards,
    Bob Monsen

    One cannot inquire into the foundations and nature of mathematics without
    delving into the question of the operations by which the mathematical
    activity of the mind is conducted. If one failed to take that into account,
    then one would be left studying only the language in which mathematics is
    represented rather than the essence of mathematics.
    - Luitzen Brouwer
     
  8. Joachim,

    That's hard to say without knowing the values of the resistors involved.
    Besides, aren't you supposed to find the solution yourself?

    petrus bitbyter
     
  9. John - KD5YI

    John - KD5YI Guest

    Yes.
     
  10. PeteS

    PeteS Guest

  11. Bob Monsen

    Bob Monsen Guest

    Cute. Now, try R1/R3 != R2/R4.

    The OPs homework was probably presented with these values, and he tried to
    derive an equation for the resistance, without realizing that R5 doesn't
    pass any current... The problem is a bit harder if you don't notice that
    fact.
     
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