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Sum of capacitors connected in Series and in Parallel

  1. Sep 19, 2012 #1
    1. The problem statement, all variables and given/known data
    2z3xi04.jpg

    Find the sum value of capacitors.

    2. Relevant equations



    3. The attempt at a solution

    I tried to redraw the scheme like in the picture below, but my teacher told me it's not correctly redrawn. I hope You'll prove to me that my redrawn scheme is correct... If not I'm more than happy to know the both: correctly redrawn scheme and the right answer.

    dwyl1g.jpg
     
  2. jcsd
  3. Sep 19, 2012 #2

    vela

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    Looks fine to me the way you redrew it.
     
  4. Sep 19, 2012 #3
    That looks correct to me.
    Anyways, why did you spend time redrawing this when its clear which one are in series and parallel? :)
     
  5. Sep 19, 2012 #4

    CWatters

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    Hint. Start at the right hand side of the original drawing. Simplify the three caps in series then see what you have.
     
  6. Sep 20, 2012 #5
    looks correct.
     
  7. Sep 20, 2012 #6
    Yep it's correct.
    I like moving from the far end of the circuit to the source (from left to right now). It's easier for me.
     
  8. Sep 20, 2012 #7
    Because I didn't know how those 4 parallel capacitors' sum will look like in the drawing...

    Say I use the old drawing:

    2z3xi04.jpg

    Could Anyone tell me How will it look like after summing those 4 parallel capacitors?

    This?:

    zkpxd1.jpg


    OR maybe This?:

    2isbeoo.jpg

    oh, and Please explain Your choice.
     
  9. Sep 20, 2012 #8

    vela

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    Neither is correct as the circuit doesn't have four capacitors in parallel.
     
  10. Sep 20, 2012 #9

    vela

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    If two elements are in parallel, they have to be connected to the same two nodes in the circuit. In the original circuit, there are six nodes: the four dots and the two corners on the right. You can see that no two elements are connected to the same two nodes. None of the elements is in parallel with another.

    If two elements are in series, all of the current through one element has to flow through the other element. For this to happen, the two elements have to be connected end to end, and no other element can be connected to the node that connects the two elements because it will siphon off some of the current. In the original circuit, only the three capacitors on the right are in series. (Can you see that?) Using the formulas you have for combining capacitors, you'll find you can replace those three with an equivalent capacitance of 1/3 μF.

    Now redraw the circuit with those three capacitors replaced by a single 1/3 μF capacitor. Are any capacitors in the new circuit in parallel or in series?
     
  11. Sep 20, 2012 #10

    PeterO

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    Actually C1 = 2 μF so that total should be 2/3 μF
     
  12. Sep 20, 2012 #11

    vela

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    Yes, of course. I mixed up the two values. Thanks for catching that.
     
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