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3 capacitors, a switch, and battery:simple configuration!

  1. May 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Initially switch is at A, and C1 is fully charged (batter=12volts). C2,C3 are initially uncharged.
    Then switch is moved to B and the charge is redistributed.

    What is the voltage across C1? [Answer: 8.00volts]

    What is the voltage across C3? [Answer: 2.67volts]


    2. Relevant equations
    C=Q/V
    Series:
    1/Ceq=1/C1+1/C2
    Q=Q1=Q2
    V=V1+V2


    3. The attempt at a solution
    The question does not state a time after the switch is closed. Does the time not matter or should I assume it means A LONG TIME AFTER?
    If it meant a long time after, I assumed that the original voltage of 12 volts is shared and the charge is equivalent for all capacitors. Thus C123=40/3 pF and Q123= C123*Vb=(40/3)*12=160... Then V1=Q123/C1=160/40=4pF but this is incorrect
     

    Attached Files:

  2. jcsd
  3. May 1, 2013 #2

    ehild

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    It means enough time after switching over. No resistors are shown, the resistance of the wires is very small. The time constant is proportional to the resistance, you can take it very small.
    No, the charge is not equally shared by the capacitors. Check the sign of the charges on the individual plates. If case of series capacitors, one of two connected plates has positive charge, the other has negative one.

    ehild
     
  4. May 1, 2013 #3
    Right, one side of a capacitor is positive and the other side is negative adding to a total of zero. So the capacitors in series have equivalent charges. Since all three capacitors are in series then they have equivalent charges. Is this what you are saying?
     
  5. May 1, 2013 #4
    So initially the capacitor becomes charged with 12 volts across it. The current stops. Then the switch is in place and the charges redistribute. With the battery disconnected,the capacitor has 12 volts that gets redistributed through the 3 capacitors. Initially C1 has 12v(40pF)=480pC of charge to redistribute. Since they are in series 480pC/3=160pC per capacitor. So V1=160pC/40pF=4volts.
     
  6. May 1, 2013 #5

    ehild

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    The charge is not equally distributed on series connected capacitors: They posses equal charges.

    The capacitors in the problem are not in series. Series and parallel has only sense with respect to a battery, or with respect to an other element of the circuit.
    See picture: What are the signs of charge on the individual plates? C1 with its stored charge and voltage of 12 V serves as a source for the other capacitors, which are really connected in series with respect to C1. It shares the charge, some of that 480 pC goes over to the empty plate of C3...

    ehild
     

    Attached Files:

  7. May 1, 2013 #6
    So how do you find the voltage on c1 if you don't have a way to relate it to the other capacitors? I'm really trying here but it isn't clicking.

    I'm having a tough time taking this concept and applying the equations for capacitors to this problem...

    Also, if the other capacitors only take some of the charge, how do you know how much charge is redistributed to them?
     
    Last edited: May 1, 2013
  8. May 1, 2013 #7
    If C1 isn't in series, or in parallel, how can you apply any equations to relate how the charge is distributed to the other capacitors?
     
  9. May 1, 2013 #8

    ehild

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    The voltage across the chain of C2 and C3 becomes the same as the voltage across C1.
    The sum of the charge that stays on C1 and the charge that goes over to the other capacitors is equal to 480 pC. C1 can be considered parallel to the series resultant of C2 and C3.

    ehild
     
  10. May 1, 2013 #9
    Wohoo! 3hrs later! Now I can continue on and do a MILLION other physics problems! YES!!!!

    Grand finale of solutions:
    C1 is in parallel with C23:

    1) 480pC/(C1+C23)=480pC/(60pF)=8V across C1, C23

    2) V3=Q23/C3=8/3V
     
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