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Capacitance 1/4 correct need b,c,d

  1. Oct 15, 2009 #1
    1. The problem statement, all variables and given/known data
    25-25alt.gif V=8.8V
    (a) Find the equivalent capacitance between the terminals
    Answer= .242uF

    b.) Find the charge stored on the positively charged plate of each capacitor?
    C1=.3uF , C2=1uF , C3= .25uF
    So I have to find the charge on each plate of the three

    c.)Find the voltage across each capacitor?
    Same deal but with voltage

    d.) Find the total stored energy?
    U=1/2*Cep*V
    U=1/2 (.242)* 8.8^2
    U=9.37uJ
    2. Relevant equations
    Cep= Q/V=(Q1+Q2+Q3)/V= C1+C2+C3
    CV=Q

    3. The attempt at a solution
    I tried CV=Q with V=8.8V and my answers for .3uF=2.64uC, 1uF=8.8uC, .25uF=2.2uC
    all wrong... I tried to think of C2 and C3 in a parallel Combo where there net capacitance is 1.25uF=Q/8.8V wrong for that too. I've only tried b because I thought V=8.8V was the potential and it was going to be constant but I guess not.
     
    Last edited: Oct 15, 2009
  2. jcsd
  3. Oct 15, 2009 #2
    Maybe you could use one of Kirchhoff's laws?
     
  4. Oct 15, 2009 #3
    The method you need to use is as follows:
    1) Calculate total capacitance . Ct = 1/((1/c1)+(1/(c2+c3))
    2) Find charge on Ct at given voltage. q = C*V
    3) Find potential V1 on C1 at charge: V = C/q
    4) Subtract V1 from V to find voltage on C2 and C3: V2 = V3 = (V-V1)
    5) Now find charge on C2 and C3 by q = q=C*V
     
  5. Oct 15, 2009 #4

    The method you need to use is as follows:
    1) Calculate total capacitance . Ct = 1/((1/c1)+(1/(c2+c3))
    2) Find charge on Ct at given voltage. q = C*V
    3) Find potential V1 on C1 at charge: V = C/q
    4) Subtract V1 from V to find voltage on C2 and C3: V2 = V3 = (V-V1)
    5) Now find charge on C2 and C3 by q = q=C*V
     
  6. Oct 15, 2009 #5
    The thing to realise is that an electric current represents an amount of charge moving through a circuit. For every electron which goes into a series circuit, there is an electron which comes out the other end. So in a series circuit like the series of 2 capacitors in your problem (the parallel caps are 1 cap for the purposes of this remark) the same charge is displaced on both cap.
    Then you can deal with the parallel circuit of C2 and C3. The potential must always be the same on each capacitor, if potential were different you could expect current to flow until the potentials were equalised.
    Don't worry it all becomes second nature once you've done all the homework and sat an exam or 2.
     
  7. Oct 15, 2009 #6
    nice that worked and made sense thanks man
     
  8. Oct 15, 2009 #7
    you're welcome
     
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