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Capacitors Problem: find potential differences

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data
    The 2290.-nF capacitor is then disconnected from the 19.2-V battery and used to charge three uncharged capacitors, a 175.-nF capacitor, a 280.-nF capacitor, and a 375.-nF capacitor, connected in series.
    After charging, what is the potential difference across each of the four capacitors in Volts?


    2. Relevant equations
    q=cv


    3. The attempt at a solution
    Its been said in my homework forum that C1 and C2,3,4 are in parallel. This means that potential difference across them is equal. Thusly, the potential difference of C1 is 19.2V. To find q for C2,3,4, I divided V or 19.2 by (1/C2 + 1/C3 + 1/C4) because V should be v2 + v3 + v4. Then I divided q by each capacitor value for 2,3,4. They added up to 19.2, which I would expect. My answers were wrong. I also tried a method where C1 and C2,3,4 were all in series, and it was still wrong.
    I think my problem is that my q values are wrong?
    If q, which i have calculated using 19.2*C1, is supposed to be q1+q2, where q1=C1V and q2=C2,3,4V, then it would seem that the V's between the two different equations are different, but that doesnt make sense as they are (supposedly) in parallel.
    Help :(
     
  2. jcsd
  3. Feb 4, 2012 #2

    vela

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    C1 is indeed parallel to the C2-C3-C4 combo. The capacitors are also all in series.

    Your error is assuming that the voltage across C1 remains 19.2 V. When it's connected to the uncharged capacitors, a current will flow, so the charge on C1 decreases and hence its voltage.

    Say an amount of charge q leaves C1. Can you figure out how much charge ends up on C2, C3, and C4?
     
  4. Feb 4, 2012 #3

    SammyS

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    Hello, hofferry. Welcome to PF !

    Adding to what vela said:

    The charge lost by C1 is gained by the series combination of the others. The resulting voltage across C1 is the same as the resulting voltage across the series combination of the others.
     
  5. Feb 5, 2012 #4
    Up to this point i thought that I had a pretty firm understanding of the material, as ive been able to answer all the other questions, but I guess not. What I'm getting from what youre saying is that V is not 19.2 on C1, but is spread across the other capacitors, which makes sense. How much it is spread across, ive no idea. I assumed that V for C1 = ƩV for C2,C3,C4, which would mean V1 for C1 = 9.6. Therefore 9.6/Ceq = charge on C2,3,4 and their respective potential differences would be that charge(q) value divided by their individual capacitances. My answer was wrong, so I'm assuming V is not halved. How else then is the charge distributed?
     
  6. Feb 5, 2012 #5

    SammyS

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    Do you have the equivalent capacitance for the C2, C3, and C4 capacitors in series?

    Treat them as one capacitor with capacitance C2,3,4, or whatever you like to call it.

    Suppose C1 has a charge Q0 when charged to 19.2 Volts. After C1 is connected to C2,3,4, the amount of charge C1 loses is equal to the amount of charge C2,3,4 gains. Let Q be the amount of charge on C1 and q be the amount of charge on C2,3,4[/SUB after they're connected together. Then Q + q = Q0.

    How does the potential difference across C1 compare to the potential difference across C2,3,4 when they're connected?
     
  7. Feb 5, 2012 #6
    systems of equations for the win. thanks guys
     
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