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Find the Capacitance when the switch is closed

  1. Apr 2, 2013 #1
    1. The problem statement, all variables and given/known data

    When the switch is closed, Potential drops to 40v across C1 & C2(Parallel series). Find the Capacitance of C2

    2. Relevant equations
    Before switch is closed
    C1=300uF and holds a charge of Q1=3e4 uC V=100v


    After switch is closed
    V'=40v


    3. The attempt at a solution
    V1=V2

    So Q1/C1=Q2/C2
    Q=Q1+Q2
    How could I find the charge on C2?
     
  2. jcsd
  3. Apr 2, 2013 #2

    PeterO

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    Might be useful if there was an indication how the capacitors and the switch are connected to each other.
     
  4. Apr 2, 2013 #3
    jaIxytR.jpg
     
  5. Apr 2, 2013 #4

    PeterO

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    SO where is the switch?
     
  6. Apr 2, 2013 #5
    2ciTXkP.jpg
     
  7. Apr 3, 2013 #6

    CWatters

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    That's a two way switch so the concept of it being "open" or "closed" is meaningless. However in this case it's clear they mean it starts off to the right and moves to the left.

    V1=V2 is incorrect (they even tell you it falls to 40V). It's the charge that stays constant.

    Edit: Ah I see you what you mean about V1=V2. You are correct but that's irrelevant really. The charge is constant but the total capacitance changes when the switch is made.
     
    Last edited: Apr 3, 2013
  8. Apr 8, 2013 #7
    I've solve it

    Q=q1+q2

    if q1=12,000

    30,000-12,000=q2

    q2=18,000

    so C=q2/v

    C=450uF
     
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