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Help distinguishing between series and parallel capacitors

  1. Mar 2, 2008 #1
    Edit: I found out how to solve this but I'd still appreciate some tips (see bottom).

    1. The problem statement, all variables and given/known data


    2. Relevant equations

    Capacitors in series:
    Q = Q1 = Q2 = ...
    V = v1 + v2 + ...
    1/Ceq = 1/C1 + 1/C2 + ...

    Capacitors in parallel:
    Q = Q1 + Q2 + ...
    V = V1 = V2 = ...
    Ceq = C1 + C2 + ...

    3. The attempt at a solution

    I tried combining the capacitors C1 and C2 on the left and right sides together in parallel, but apparently that is not correct. I have trouble understanding when to combine in series and when to combine in parallel. It says to combine in parallel if the capacitors appear to be connected like the rungs of a ladder. In this problem, they appear to be connected like the rungs of a ladder, yet they are in series.

    Is there some "trick" to understanding when to combine in parallel and when to combine in series? Thanks for your help.

    Edit: Upon searching more in this forum, I found a post where someone said capacitors are only in parallel if there is nothing between the capacitors except wire. So in this case, C3 is between C1 and C2, thus C1 and C2 cannot be parallel. Is this correct? If so, I would solve by combining C1 and C2 in series on each side, then combining those with C3 in parallel. Also on the bottom, combine C2 and C2 in parallel, then combine that result with the previous result in series. I did that, and got the right answer. However, if there's any kind of tricks you use to understand how to combine capacitors, I would appreciate it if you'd still share them.
    Last edited: Mar 2, 2008
  2. jcsd
  3. Mar 2, 2008 #2


    User Avatar

    The trick is this:

    If you want to see if two circuit elements are in series (this applies to capacitors, resistors, anything): try the following:

    If you can get from one terminal (one side) of the first element to one terminal of the second element without ever crossing a junction in the wire (a point where two or more wires connect), then those two elements are in series. It does not matter if along the way you go through another circuit element (a battery, a resistor, a capacitor, etc), as long as you don't cross a junction. You don't have to be able to do that for both terminals of the two elements. As long as you can get from one of the terminals fo the first element to one of the terminals of the second element without crossing ajunction, they are in series.
  4. Mar 2, 2008 #3
    Thank you, that really clears a lot up for me. But just to clarify: is the reverse of that true as well? - If you cross a junction no matter which side you choose, the capacitors/resistors/etc. must be in parallel.
  5. Mar 2, 2008 #4


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    No, it's a bitmore involved for parallel.

    You have to satisfy two conditions. You must be able to go from one terminal of the first element to a terminal of the second element without crossing any other circuit element but while crossing one junction AND you must be able to do the same for the second terminal (you must be able to get from the second terminal of teh first element to the second terminal fo the second element without crossing any other circuit element but while crossing a junction).

    For the series, there is only one condition. But for parallel, the two terminals must satisfy the rule.
  6. Jun 4, 2009 #5
    I have a question based on this. Can there be a circuit that is neither series or parallel? Such as if there conditions for parallel held for one terminal of the first element, but for the other terminal, there are 2 junctions and another circuit element in the path to the second terminal of the second element?
    Last edited: Jun 4, 2009
  7. Jun 4, 2009 #6


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    Homework Helper

    It's a difficult topology to imagine as a circuit element will necessarily connect between 2 nodes of a circuit. If you have a single loop then the elements will necessarily be connected in a string or in series. Once you have multiple loops then you must give consideration to the Kirchhoff and Ohms law considerations. Of course you can always simply circuits with equivalent substitutions, but ultimately these are all derived from the solutions to the circuit equations aren't they?
  8. Jun 4, 2009 #7
    Attached is a sample circuit, also, note that all the capacitors have equal capacitance except fot the one on the top left, which is less than the others.

    Attached Files:

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