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Reducing Equivalent Resistors and Capacitors

  1. Jan 16, 2008 #1
    1. The problem statement, all variables and given/known data

    Reduce the circuit as much as possible. All resistors have R and all capacitors have same C.

    2. Relevant equations

    3. The attempt at a solution


    And this is my reduced one (I only reduced two capacitors in series and two parallel resistors)


    The reduced capacitors (the ones in the right corner of that little square) =

    1/C_eq = 1/C + 1/C = 2/C
    C_eq = .5C

    1/R_eq = 1/R + 1/R = 2/R
    R_eq = .5R

    (Ignore the random squiggly at the bottom)

    Anyway - that's as far as I got. I'm not sure if the two capacitors in the center are considered parallel? And if so, is that other capacitor on the far left considered in parallel with them too? (even though there are resistors between them?)
    Last edited: Jan 16, 2008
  2. jcsd
  3. Jan 16, 2008 #2

    Tom Mattson

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    Can you attach the images to your post? I really don't feel like joining facebook just to look at your problem. :smile:
  4. Jan 16, 2008 #3
    oh, sorry - they were supposed to show up!
  5. Jan 17, 2008 #4
    Yeah, the two caps in the middle are in parallel, but the other one you are asking about would not be. Think of parallel as whether or not the wires are connected at both ends. Think of series as whether or not two elements share a common connection. (A fancy name for connection is a node.)
  6. Jan 18, 2008 #5
    Is this as far as it can be reduced?


    Hmm...is this it? It seems like something simpler could be achieved somehow...*sigh*
  7. Oct 22, 2010 #6
    Can anyone give me a hint how to calculate the impedance of the following circle ?
    Thank you

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