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Potential Difference across capacitors in parallel.

  1. Jul 31, 2011 #1
    1. The problem statement, all variables and given/known data

    Two capacitors, C1 = 20 micro Farads and C2 = 5 Micro Farads are connected in parallel and the resulting combination is connected to a 12-V battery. Find (a) the Equivalent capacitance of the combination, (b) the electric charge on each capacitor, (c) the potential difference across each capacitor, and (d) the energy stored in each capacitor.


    2. Relevant equations

    Ceq = C1 + C2

    Q = CV

    Ec = QV/2 = (C(V^2))/2 = (Q^2)/2C


    3. The attempt at a solution

    (a) 2.0x10^-6 F + 5x10^-6 F = 2.5x10^-5 F


    (b) Q = CV,

    Q1 = (20x10^-6 F)(12V) = 2.4x10^-4 C

    Q2 = (5x10^-6 F)(12V) = 6x10^-5 C


    (c) Not quite sure what this is asking for?


    (d) Ec = CV^2/2

    Ec1 = c1V^2 / 2 = 0.00144 F*V^2

    Ec2 = c2V^2 / 2 = 3.6x10^-4 F*V^2




    As you can see I've worked most of this out, seems simple enough, but I dont know what to do for (c)?
     
  2. jcsd
  3. Jul 31, 2011 #2

    kuruman

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    It's simpler than you think. Draw yourself the circuit and look at it.
     
  4. Jul 31, 2011 #3
    errrr charge over capacitance?
     
    Last edited: Jul 31, 2011
  5. Jul 31, 2011 #4

    kuruman

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    How did you do part (b)? What voltage did you assume across each capacitor? :rolleyes:
     
  6. Jul 31, 2011 #5

    kuruman

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    Please don't edit previous posts. Just post a new response.

    Your "attempt at a solution for (b) says


    (b) Q = CV,

    Q1 = (20x10^-6 F)(12V) = 2.4x10^-4 C

    Q2 = (5x10^-6 F)(12V) = 6x10^-5 C

    Tell me in English what do the numbers highlighted in red represent?
     
  7. Jul 31, 2011 #6
    Battery Voltage/Potential Difference?
     
  8. Jul 31, 2011 #7

    kuruman

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    It is not a ratio. In plain English the equation Q = CV means

    The charge on a capacitor is the same as the capacitance multiplied by the potential difference (or voltage) across the terminals of the capacitor.

    Can you answer (b) now?
     
  9. Jul 31, 2011 #8
    So the potential difference is the same as charge on a capacitor divided by the capacitance.

    V1= Q/C1
    V2= Q/C2

    Thanks!
     
  10. Jul 31, 2011 #9

    kuruman

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    Actually, you are doing it backwards. Memorize this definition

    "Two circuit elements in parallel share the same potential difference (or voltage) across their terminals"

    and use it constructively.

    Cheers.
     
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