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Capacitor charge in series and parallel circuit

  1. Sep 23, 2013 #1
    1. The problem statement, all variables and given/known data

    Three capacitors, with capacitances C1 = 4.0 μF, C2 = 3.0 μF, and C3 = 2.0 μF, are
    connected to a 12 -V voltage source, as shown in the figure. What is the charge on
    capacitor C2 ?


    2. Relevant equations
    Q=VC
    Q = charge
    V = voltage
    C = capacitance
    Ceq = equivalent capacitance


    3. The attempt at a solution

    C1 and C2 are in series and therefore should be summed as in Ceq=1/((1/C1)+(1/C2)) because the voltage across them will be shared proportionally. If I multiply Ceq by the voltage I get a charge of Q12 which is for both capacitors. I need to find the charge on C2 so I am multiplying the charge on both, Q12, by the proportion of C2 to C1 which is 3/4. I am getting 15.4 microCoulombs but my answer should come to 16 microCoulombs. I need help on this questions. Am I doing this correctly? Is my thought process correct?
     

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    Last edited: Sep 23, 2013
  2. jcsd
  3. Sep 23, 2013 #2

    tiny-tim

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    (try using the X2 button just above the Reply bos

    Hi Joshb60796! :smile:

    (try using the X2 button just above the Reply bos :wink:)
    Nooo :redface:

    C2 and C3 are in parallel, and then C1 is in series with their resultant.

    Try again. :smile:
     
  4. Sep 23, 2013 #3
    I was aware that C2 and C3 were parallel which is why I didn't include C3. Aren't C2 and C3 seeing the same voltage so if the voltage is connected for a long time C3 doesn't factor in to the charge that is held on C2? I thought it was just Capacitance and Voltage that has to do with Charge, Q.
     
  5. Sep 23, 2013 #4

    tiny-tim

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    yes :smile:
    yes it does, C2 has to share its charge with C3 (and btw, not equally)
    yes, but you have to use the total capacitance of C2 and C3 combined to find how that combines with C1 :wink:

    find the capacitance of C2 and C3,

    then find the total capacitance …

    show us what you get :smile:
     
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