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Equivalent Resistance in partial parallel/series cicuit

  1. Feb 15, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the equivalent resistance between points A and B in the drawing.

    2. Relevant equations

    Rseries = R1+R2+R3...
    Rparallel: 1/Rp = 1/R1+1/R2+1/R3...


    3. The attempt at a solution

    Reducing it down:
    1/8+1/9 = 1/R1 = 4.235
    That's in parallel with 20, so:
    1/20+1/4.235 = 3.495 ohms

    3, 4, and 6 are in series:
    3+4+6 = 13 ohms
    Add that to the previous and you have 16.495 ohms, which is incorrect.

    Is one parallel and I'm not adding it as such? Or should I combine differently?

    Thanks so much!
     

    Attached Files:

  2. jcsd
  3. Feb 15, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Good. R1 = 4.235 (not 1/R1).
    Nope. The 3 ohm resistor is in the way.
     
  4. Feb 15, 2008 #3
    So then 20 and 3 are in series, right?
    20+3=23 ohms

    So now my drawing has:

    23 ohms . 4 ohms

    4.235 ohms

    . 6 ohms

    Would the 4 and 6 be considered series?
     
  5. Feb 15, 2008 #4
    4+6 are in series
    9+8 are in parallel
    3 and 20 are not in series!!!

    You need to do this step by step...

    6+4 = 10
    1/8 + 1/9 = 17/72
    now you have

    ------3 -----------
    --|--------|------|
    -20------17/72--10
    --|--------|------|
    -------------------

    Next you have 17/72 and 10 in parallel.

    1/10+72/17 = 737/170

    -- -- 3 -------
    --|---------|
    -20------737/170
    --|---------|
    --------------

    Now the 3 and 737/170 are in series so just add them to get

    -------------
    --|---------|
    -20-------1147/170
    --|---------|
    -------------

    Now these are in parallel you can solve!
     
  6. Feb 15, 2008 #5
    When you added for 17/72 and 10 in parallel, shouldn't it have been 1/4.235+1/10 = 1/R = 2.975?

    so then with
    ------3--------
    | |
    20 2.975
    | |
    ------------

    So now which should I combine first?
     
  7. Feb 16, 2008 #6
    first take this and then you shall have the feel of the question much better..""

    General rules for doing the reduction process include:

    1. Two (or more) resistors with their heads directly connected together and their tails directly connected together are in parallel, and they can be reduced to one resistor using the equivalent resistance equation for resistors in parallel.

    2. Two resistors connected together so that the tail of one is connected to the head of the next, with no other path for the current to take along the line connecting them, are in series and can be reduced to one equivalent resistor.

    Finally, remember that for resistors in series, the current is the same for each resistor, and for resistors in parallel, the voltage is the same for each one
     
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