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Capacitors in Series and in Parallel

  1. Jul 9, 2011 #1
    I am trying to solve the problem another poster asked about in

    https://www.physicsforums.com/showthread.php?t=382491.

    I realize how the problem is supposed to be solved, and that you have to start with the rightmost 3 capacitors because none of the other capacitors are in series or in parallel, but I don't understand why this last fact is the case. Why can't you for instance, find the equivalent capacitor for the two leftmost triples (C1, C2, C1) individually [as these seem to me to each form a self-contained series] and the rightmost (C1, C1, C1) separately, which would leave you with three capacitors in parallel?

    Thanks!
     
    Last edited: Jul 9, 2011
  2. jcsd
  3. Jul 9, 2011 #2

    Doc Al

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    Staff: Mentor

    The leftmost triple has branches connecting to it on either side of C2, so you can't treat C1-C2-C1 as being purely in series, independent of the rest of the circuit. But the three C1s in the rightmost triple are purely in series, so you can replace them by their equivalent.
     
  4. Jul 9, 2011 #3
    Hi Doc Al,

    Thanks for your reply. I'm still not quite sure I understand -- what is it about being on the far right that makes those three capacitors independent of the rest of the circuit? I would have actually thought that that being on the leftmost end would make you the most independent, since charge flows to those capacitors directly without intermediary. You said that there are branches connecting the leftmost triple on either side of C2, but I'm not sure why this is fundamentally a different kind of connection than the branches at points c and d which connect the rightmost series with the rest of the circuit.

    Thanks!

    sweetreason
     
  5. Jul 9, 2011 #4

    Doc Al

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    Staff: Mentor

    For the rightmost series, points c and d are outside the triple (on the ends), so there's no problem replacing what's between c and d with its equivalent. But on the leftmost triple the branches are in the middle of the triple. You cannot replace those three with a single equivalent without breaking the branches somehow (and thus changing the circuit).
     
  6. Jul 10, 2011 #5
    Okay, I think I can work with that. Thanks a lot!
     
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