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Capacitance in series and parallel circuit

  1. Feb 15, 2008 #1
    [SOLVED] Capacitance in series and parallel circuit

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

    Determine the equivalent capacitance between A and B for the group of capacitors in the drawing.

    2. Relevant equations

    Cseries = 1/Cs = 1/c1 + 1/c2 + ...
    Cparallel = Cp = C1 + C2 + ...

    3. The attempt at a solution

    The 4 and 12 micro farad capacitors are in parallel, so Cp = 16mf.
    The other four are in series, so 1/5+1/24 = 1/C = 4.138mf (top)
    1/6+1/8 = 1/C = 3.429mf (bottom)
    All of these are in series, so 1/4.138+1/3.429+1/16 = 1/c; c = 1.678mf

    This is not right. Where is my logic faulty?

    Attached Files:

  2. jcsd
  3. Feb 15, 2008 #2


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    Staff: Mentor

    The 24 uF, 12 uF and 8 uF capacitors are in series - since they are end to end between the two nodes in common with the 4 uF capacitor.

    So find the capacitance of the series of 24 uF, 12 uF and 8 uF, and that combination is parallel with the 4 uF capacitor. That group is in series with the 5 uF and 6 uF capacitors between nodes A and B.
  4. Feb 15, 2008 #3

    Doc Al

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    Staff: Mentor

    Capacitors are in parallel only if their ends are directly connected by nothing but a wire. That's not the case here, since the 24 and 8 micro farad capacitors are in the way.

    Similarly, capacitors are in series only if there's nothing else in the path.

    Do it step by step. Find the obvious combinations and replace them with their equivalents. Then redraw it, looking for more combinations.

    The first combination that I see are several capacitors in series. Start there.
  5. Feb 15, 2008 #4
    Thanks guys; apparently I didn't quite understand how to combine initially. I get it now!
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