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Parallel and series capacitors question

  1. Feb 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Three capacitors are connected to a battery as shown. Their capacitances are
    c1=3C
    c2=C
    c3=5C

    -------c1----------------------
    |........................|................|
    |........................|................|
    V.......................c2-----------c3
    |_______________|__________|

    2. Relevant equations
    state the ranking of the capacitors accoriding to the charge they store?
    How do you find out the individual charges?



    3. The attempt at a solution
    for the
    (1)parallel capacitor:
    c2+c3=
    (C+5C)=6C
    (2)for the series: 1/c1= 1/(3C)=1/3C
    dividing (1) by (2) to get Ceq which is 2C
    I know that Q=C*delta V
    Qeq=Ceq*V
    Qeq=(2C)(V)
    Qeq= Q1+Q2+Q3 etc
    to find
    Q1 then Qeq-Q2-Q3
    Q2 then Qeq-Q1-Q3
    Q3 then Qeq-Q2-Q1
    ????
    thank you in advance:confused:
     
  2. jcsd
  3. Feb 21, 2009 #2

    Delphi51

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    I am thinking of the series circuit of C1 and C4(C2 and C3 combined).
    Since V = Q/C for each part, we have V = Q1/3 + Q4/6 = Q1/3 + (Q2+Q3)/6

    I'm having trouble finding another equation and I'm not convinced that your Qeq is equal to Q1 + Q2 + Q3. Something is bothering me about that.
    I'm wondering if Q1 = Q2+Q3 because current can't flow through a capacitor so the only charge that can be on the top plates of C2 and C3 must be charge that came away from the right plate of C1.

    Oh, the voltage on c2 equals the voltage on c3, so we have Q2 = Q3/5
     
    Last edited: Feb 21, 2009
  4. Feb 21, 2009 #3

    Yes, I think Im wrong there too. I see how c4 is the combination opf c2 and c3.
     
  5. Feb 21, 2009 #4

    Delphi51

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    The voltage on c2 equals the voltage on c3, so we have Q2 = Q3/5
     
  6. Feb 21, 2009 #5
    sorry...I guess I'm a little slow..
    so.. do you mean the charge (Q2 is equal to (Q3/5) or that the voltage on C2=Q3/5?
     
  7. Feb 21, 2009 #6

    Delphi51

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    The voltages are equal so the Q/C's must be equal so
    Q2/1 = Q3/5.

    Or we can just sum the voltages as V = V1 + V2
    to get V = Q1/3 + Q2.
    Down to 2 variables, one equation!
    Do you think my thought that Q1 = Q2 + Q3 could be right?
     
  8. Feb 21, 2009 #7
    I think you're right...
    I just don't know how to set up the solution to finding the individual charges:
    Q1=C (Q1/3c +Q2)? I can't solve for Q1


    the answer to thi problem is: Q1>Q3>Q2
     
  9. Feb 21, 2009 #8

    Delphi51

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    V = Q1/3 + Q2 (1)
    Q2 = Q3/5 (2)
    Q1 = Q2 + Q3 (3)
    Use (2) to eliminate Q3 from (3).
    Sub the result into (1) so you only have Q2 and it is found (in terms of V).
    Then use (2) to find Q3. Finally, use (3) to find Q1.

    Yes, it works out so Q1 is biggest, Q2 smallest.
     
  10. Feb 21, 2009 #9
    Can you please verify my work:
    (3) (Q3/5)+Q3 [tex]\rightarrow[/tex] 6Q3/5
    (1) (6Q3/5) +Q2
    v=(Q1/3) +Q2
    Q2=(V)(3/Q1)
    Q2=(Q1/3 + Q2 )(3/Q1)[tex]\rightarrow[/tex]3Q1+3Q2/(3Q1)[tex]\rightarrow[/tex]{3Q1+3(Q3/5)} /(3Q1) [tex]\rightarrow[/tex] Q2=(7/6)
    *v=(Q1/3) +Q2
    v=(6Q3)/5 +(7/6)[tex]\rightarrow[/tex] Q3=35/6
    so then Q1= 6Q3/5 [tex]\rightarrow[/tex] {(6(35/6) / 5}= 7

    I'm so sorry...I feel like an idiot!
     
  11. Feb 21, 2009 #10

    Delphi51

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    You can't find the 3 charges numerically - all depends on V!
    should be Q1 = Q3/5+Q3 = 6Q3/5
    should be V = 6Q3/5 + Q2
    is correct.
    should read Q2 = V - Q1/3
    doesn't make sense - all the V's have been lost!

    Use (2) to eliminate Q3 from (3).
    Q3 = 5Q2 so Q1 = Q2 + 5Q2 = 6Q2
    Sub the result into (1) so you only have Q2 and it is found (in terms of V).
    V = 6Q2/3 + Q2 so V = 3Q2 and Q2 = V/3
    Then use (2) to find Q3.
    Q3 = 5Q2 = 5V/3
     
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