Identifying parallel elements in a circuit

  1. In the following circuit, a) identify all combinations of 2 or more circuit elements that are connected in series and b) Identify pairs of circuit elements that are connected in parallel.

    a) Seems easy enough, the 4 and 6 ohm resistors are in series, and so are the 5 ohm resistor with the 20 V volate source. Did I miss any?

    b) Here the question says "Identify pairs" not "all pairs" so I'm not sure if I have to find all of them...aren't there a lot? I think I'm having trouble finding every last parallel pair. This is what I have so far.

    1 ohm is in parallel with 2 ohm
    4 ohm "" 5 ohm
    4 ohm "" 20 V
    6 ohm "" 5 ohm
    6 ohm "" 20 V
    3 ohm "" 5 ohm
    3 ohm "" 20 V
    1 ohm "" 3 ohm (?)
    2 ohm "" 3 ohm (?)

    I feel like I have too many or too little pairs...

    In a circular circuit where all the elements are connected in a single loop, are the elements in series or in parallel?
  2. jcsd
  3. gneill

    Staff: Mentor

    Parallel components have leads that share exactly two nodes.

    Serial components have leads that exclusively share exactly one node (no other connections by other components to that node).

    Not every pair that you've identified are in parallel.
  4. So say we have a square circuit, with one resistor on the top and 3 resistors in series on the bottom (nothing on left and right)
    Only the outer 2 of the 3 bottom resistors would be parallel with the top? Not the middle one?

    also, going by what you said my answer for series seems to be correct, but only the 1 and 2 ohm resistors are in parallel?
  5. gneill

    Staff: Mentor

    You mean like this:


    There are NO resistors in parallel in this circuit; no two resistors share exactly two nodes. That is, you must have one lead from each component connected together at one node, while the other leads of the two components connect together at another (separate) node.
    Yes, that;s good.

    Attached Files:

  6. Okay I see, thank you very much for clarifying that for me.
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