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Calculating Capacitance in a Mixed Circuit?

  1. Feb 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Please calculate the equivalent capacitance if C1 = 10.3 microfarads, C2 = 41.1 microfarads, and C3 = 95.5 microfarads in the diagram.

    2. Relevant equations
    1 / Ctotal for Parallel = (1/C1) + (1/C2) + ...
    Ctotal for Series = C1 + C2 + ...

    3. The attempt at a solution
    So I tried to split it up. The total for the parallel part was 28.73389458 microfarads. Then the series was 10.3 microfarads, so when I add it up, I get 39.03389458 microfarads. Is this right?
     

    Attached Files:

  2. jcsd
  3. Feb 9, 2016 #2

    cnh1995

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    Are you sure?
     
  4. Feb 9, 2016 #3
    Oops, sorry! It's the opposite, my bad >.<

    Now I'm getting 146.9 microfarads?
     
  5. Feb 9, 2016 #4

    cnh1995

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    Are you sure you should add the 10.3 microfarad directly to the parallel combination?
     
  6. Feb 9, 2016 #5

    gneill

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    Staff: Mentor

    A quick reality check is to realize that any time you have capacitors in series the result must be less than the smallest in the string. In your figure capacitor C1 is definitely in series with whatever net capacitance follows it...
     
  7. Feb 9, 2016 #6
    Well, being that I have absolutely no idea how to face this situation, I'm as sure as can be with a wild guess :/ I'm guessing I did it wrong, though, right?

    So should I be adding its inverse?
     
  8. Feb 9, 2016 #7

    cnh1995

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    Yes. Series capacitors add like parallel resistors.
     
  9. Feb 9, 2016 #8
    Ah okay, I think I got it: 9.55 microfarads

    Thank you!
     
  10. Feb 9, 2016 #9

    cnh1995

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    Good! In such addition, the result is always less than the least of the component values, as gneill mentioned in #5.
     
  11. Feb 10, 2016 #10
    thank you for the help!
     
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