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Charging capacitors in series

  1. Feb 24, 2013 #1
    1. The problem statement, all variables and given/known data

    A 2600-nF capacitor is disconnected from the 15-V battery and used to charge three uncharged capacitors, a 100-nF capacitor, a 200-nF capacitor, and a 300-nF capacitor, connected in series.

    After charging, what is the potential difference across each of the four capacitors?


    2. Relevant equations

    Qo = CV = 3.9e-5 C

    Qo = Q234C1[(1/100)+(1/200)+(1/300)] + Q234

    3. The attempt at a solution

    Qo = 5.33Q234

    Qo = 3.9e-5 C

    Q234 = 7.32e-6 C

    Q1/C1 = (Qo-Q234)/C1 = [(3.9e-5 C) - (7.32e-6 C)]/2600 nF = 12.2 V

    V2 = (7.32e-6 C)/100 nF = 7.32 V

    V3 = (7.32e-6 C)/200 nF = 3.65 V

    V4 = (7.32e-6 C)/300 nF = 2.43 V

    This is incorrect, and I haven't been able to figure out what's wrong.
     
  2. jcsd
  3. Feb 24, 2013 #2

    NascentOxygen

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    Staff: Mentor

    In charging the others, the big capacitor loses a charge Q and drops its voltage from 15V to V.
     
  4. Feb 24, 2013 #3
    I express that in this equation,

    Qo = Q234C1[(1/100)+(1/200)+(1/300)] + Q234

    which is derived thusly:

    (Qo-Q234)/C1 = V1

    Q234[(1/100)+(1/200)+(1/300)] = V234

    V1 = V234

    (Qo-Q234)/C1 = Q234[(1/100)+(1/200)+(1/300)]

    Qo = Q234C1[(1/100)+(1/200)+(1/300)] + Q234
     
  5. Feb 24, 2013 #4

    NascentOxygen

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    So, you can finish this now?
     
  6. Feb 24, 2013 #5
    No, I cannot.

    Correct answers are:
    V1 = 14.7 V
    V2 = 8.01 V
    V3 = 4.01 V
    V4 = 2.67 V

    These differ from my answers in the op.
     
  7. Feb 24, 2013 #6

    NascentOxygen

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    Your method looks right. When I finish it I get 14.692V, 8.0137V, etc.

    So it looks like you may be making an arithmetic error.
     
  8. Feb 24, 2013 #7
    Yes, I figured out what I was doing wrong. instead of adding [(1/100) + (1/200) + (1/300)]
    I was doing 1/600. lol. Gotta be careful. Thanks for your help.
     
  9. Feb 24, 2013 #8
    So is it like this:
    the previous charge of the energized capacitor is used to charge other capacitors. The previous energized charge is distributed to new charge of C1 and the equal charge of the series capacitors.
    [tex]Q_{\mbox{energized C1}}=Q_{\mbox{new charge of C1}}+Q_{234}[/tex]
    Hence the new charge is
    [tex]Q_{\mbox{energized C1}}-Q_{234}=Q_{\mbox{new charge of C1}}=V_{1}C_{1}[/tex]
    is my analogy correct?
    forgive me for hijacking but this also came up in the exam but the difference is a capacitor is charging two parallel capacitors.

    I already also solved the problem in the op
    Q_234 is 8.01x10^-7Coloumb btw
     
  10. Feb 24, 2013 #9

    NascentOxygen

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    That looks right.
     
  11. Feb 24, 2013 #10

    NascentOxygen

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    Staff: Mentor

    As a check, you could consider charging one capacitor of 600/11 nF, since that's the equivalent capacitance of those 3 capacitors in series.
     
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