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Finding the capacitance charge

  1. Oct 5, 2007 #1
    1. The problem statement, all variables and given/known data

    http://books.google.com/books?id=1D...are+connected&sig=iwUiJRqZDNHDGOfQfHHvh_9aQ3Y

    problem 41 on this link. I got the equivalent capacitance for part a. for the two capacitors with 15 and 3uF- [tex]C_{eq}[/tex]=2.5uF
    for capacitor with 6uf- i got the [tex]C_{eq}[/tex]=8.5uF. final answer for a.=[tex]C_{eq}[/tex]=5.96uF. this final anwer is correct according to the answers given to me.

    2. Relevant equations



    3. The attempt at a solution
    capacitor with 15uF-[tex]Q_{1}[/tex]
    capacitor with 3uF-[tex]Q_{2}[/tex]
    capacitor with 6uF-[tex]Q_{3}[/tex]
    capacitor with 20uF-[tex]Q_{4}[/tex]
    for part b. i know that the Q for [tex]Q_{1}[/tex] and [tex]Q_{2}[/tex] are the same because they are in series. the Q i get for them is Q=[V/(1/C1+1/C2)]. If V=15V and
    C1=15 and C2=3, then the answer i get is 37.5uC. but the answer in the book says [tex]Q_{1}[/tex] and [tex]Q_{2}[/tex] =26.3uC

    for [tex]Q_{3}[/tex]- Q=[tex]C_{3}[/tex]V If [tex]C_{3}[/tex]=6uF and V=15V, then Q should be 90uC. OR if Q=[tex]C_{eq}[/tex]V, then Q=8.5uF(15)=127.5uC. but the book says it should be 63.2uC

    for [tex]Q_{4}[/tex]- Q=[V/(1/[tex]C_{eq}[/tex])] If [tex]C_{eq}[/tex]=5.96uF and V=15V, then Q should be 89.5uC- this one i got correct.

    I dont understand how i can get certain parts right, but others wrong, unless i am using the wrong equation. Please help, i need to know this for a test.
     
    Last edited: Oct 5, 2007
  2. jcsd
  3. Oct 5, 2007 #2

    learningphysics

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    Homework Helper

    Start with C4... you did that right... the charge on C4 is just the charge on the Ceq...

    What is the voltage across C4? Use V = Q/C...

    So what is the voltage across C1 and C2? this is where you made your mistake... it's not 15V...
     
  4. Oct 5, 2007 #3
    ok well, i figured that if Q4 is 89.5uC, then Q4=Q, then V3=Q/C=89.5/8.5=10.53V. if
    V3C=Q, then 10.5(6uF)=63.2uC=Q3. this appears to be correct, even though i dont understand why. as for Q1 and Q2, i am still confused, i asked my professor to explain it, but he says i should know how to solve this so to figure it out on my own. I keep thinking that it should be Q=[V/(1/C1+1/C2)]. i can see that V=V1+V2, but beyond that....i just dont know
     
  5. Oct 6, 2007 #4

    learningphysics

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    Homework Helper

    No. what I meant what you know the charge on Q4 is 89.5uC. You know the capacitance of Q4 is 20uF. So the voltage across Q4 is 89.5uC/20uF = 4.475V.

    15V - 4.475 = 10.525V

    So the voltage across Q3 is 10.525V... so the charge across Q3 is 10.525*6uF = 63.15uC. etc...
     
  6. Oct 6, 2007 #5
    i got it earlier from your earlier post, thanks so much
     
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