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Circuits Wired Partially in Series and Partially in Parallel

  1. Jul 18, 2006 #1
    [​IMG]

    R1 = 3 ohms; R2 = 6 ohms

    What is the equivalent resistance between A & B?

    I just can't find the right answer with this one.

    I don't know if I'm placing the wrong ones in parallel. I've tried two different set ups. How exactly does this type of wiring set up work?




    This one is giving me the same problem:
    R1 = 3 ohms; R2 = 9 ohms ; R3 = 12 ohms
    [​IMG]
     
    Last edited: Jul 18, 2006
  2. jcsd
  3. Jul 18, 2006 #2

    chroot

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    First, collapse the three series resistors on the right into a single resistor, (3 + 6 + R2) ohms. Next, combine that resistor in parallel with the 8 ohm resistor. Next, combine that in series with the 4 ohm resistor... etc.

    - Warren
     
  4. Jul 18, 2006 #3

    Office_Shredder

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    Break it up into a linear circuit:

    10----------------------------B
    A-R1-- | |
    | 8-----| |
    4------- |
    6---R2----3---|

    I apologize for it being so bootleg, but does this help?

    EDIT: Apparently it collapsed all the empty space, so quote my post to see what it's really supposed to look like
     
  5. Jul 18, 2006 #4
    Warren,
    Thanks for the help on the first circuit. I understand the technique now.

    Not sure if I understand how to apply it to the 2nd but I'm trying right now.
     
  6. Jul 18, 2006 #5
    OS- thanks for the help but I got it.
    The spacing was still messed up in the qoute though :rofl:

    Is it the same approach for the 2nd circuit?
     
  7. Jul 18, 2006 #6

    Office_Shredder

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    For the second:

    Combine R3 and the 9 ohm. Then that's in parallel with the 6 ohm. That, R1, and R2 are in series, and that's in paralell with the 20 ohm
     
  8. Jul 18, 2006 #7
    OK, I succeeded at # 2 aswell
    thanks for allt he help
     
  9. Jul 18, 2006 #8

    Office_Shredder

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    FOr future reference, even though you can't see my diagram very well, it's useful to take point b, and "rotate" it directly horizontal with point a. Then redraw the circuit, and what's in series and what's parallel should be more obvious
     
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