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Capacitator Circuit and determining unknown charges voltages

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

    Suppose in the figure (Figure 1), that

    C1 = C2 = C3 = 18.0μF and C4 = 28.2μF.

    If the charge on C2 is Q2 = 37.4μC, determine the charge on each of the other capacitors
    Determine the voltage across each capacitor.
    Determine the Voltage Vab across the entire combination.

    https://dl.dropbox.com/u/47465778/MP24.24.PNG [Broken]


    2. Relevant equations

    [tex]Q=CV[/tex]

    Rules for Series:
    Q is the same across a series
    [tex] \frac{1}{Ceq} = \frac{1}{C1} + \frac{1}{C2} + . . . [/tex]
    [tex] Vtotal = V1 + V2 . . . [/tex]

    Rules for Parallel:
    [tex] Qtotal = Q1 + Q2 . . .[/tex]
    [tex] C_(eq) = C1 + C2 . . . [/tex]
    V is same across parallel

    3. The attempt at a solution

    Let 1 and 2 be in Series. Then Capacitance can be added as reciprocals
    [tex] \frac{1}{Ceq} = \frac{1}{18} + \frac{1}{18}[/tex]
    [tex]Ceq = 9μF [/tex] for 1 and 2

    Q1=Q2...because they are in series, Q is same. Q1 = Q2 = 37.4 μC

    [tex]Q = CV.[/tex] So,
    [tex]37.4 = 9*V[/tex]
    [tex]V = 4.155 V[/tex]

    I figured that because Voltage is the same in parallel, then this voltage is the same across the board for 3 and 4. So I found C for 3 and 4

    3 and 4 are in series with each other, so
    [tex] \frac{1}{Ct} = \frac{1}{18} +\frac{1}{28.2} [/tex]
    [tex]Ct = 10.987...[/tex]
    Then [tex]Q = CtV[/tex]
    [tex]= 10.987*4.155 = 45.65 μC[/tex]

    Q3 = Q4 because Q's are equal in series.

    Final Answer: Q1=Q2 = 37.4 and Q3 = Q4 = 45.65 all in microColoumbs (Wrong)

    Although I'm fairly certain Q1 does equal Q2, but since I have to enter all of the answers at once, if one answer is wrong then the rest is wrong.

    Where did I go wrong? Please help, I did my best but I'm pretty shaky on this stuff. Also this is my first time posting so sorry if I did anything weird with the formatting.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 24, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    welcome to pf!

    hi liliacam! welcome to pf! :smile:
    no, they're not in series, there's a junction between them!

    also, what happens to the electrons that leave the top plate of Q4 ?

    they have to spread over the left plates of both Q1 and Q3

    so how could they all be on Q3 ? :wink:
     
  4. Feb 24, 2013 #3
    Hi! Thanks for responding so quickly

    So I tried it again...I guess the diagram confused me. I redrew it here:

    https://dl.dropbox.com/u/47465778/MP24Redrawn.PNG [Broken]

    With 3 still in parallel and 1, 2, 4 in series with each other. My current teacher still hasn't taught us how to redraw capacitors but I hope this is on the right track at least. So the electrons from 4 spread over 1 and 2, then 3?

    If this drawing is correct, then:

    Q1 = Q2 = Q4 = 37.4μC
    Q3 = ?

    [tex] \frac{1}{Ceq} = \frac{1}{C1} + \frac{1}{C2} + \frac{1}{C4} [/tex]
    [tex] \frac{1}{Ceq} = \frac{1}{18} + \frac{1}{18} + \frac{1}{28.2} [/tex]
    [tex] Ceq for 1,2,4= 6.822 [/tex]

    New Capacitance:
    Ceq total = Ceq for 1,2,4 + C3
    [tex] Ceq total = 6.822 + 18 = 24.822 μF[/tex]

    New Voltage:
    [tex] Q = CV [/tex]
    [tex] Q1 = C1V1 [/tex]
    [tex] V1 = 2.077 V [/tex]
    If 1, 2, 4 is in parallel with 3, then V is the same across

    [tex] V3 = 2.077 V[/tex]

    [tex] Q total = Ctotal*Vtotal [/tex]
    [tex] Qt = 24.822*2.077 [/tex]
    [tex] Qt = 51.575 μC [/tex]

    [tex] Qt = Q3 + Q(1,2,4) [/tex]
    [tex] 51.574μC = Q3 + 37.4 [/tex]
    [tex] Q3 = 14.175 μC [/tex]


    Final Answers: Q1 = Q2 = Q4 = 37.4, Q3 = 14.75 all μC


    Is this correct? I haven't had an official lecture on this yet, so it's just hard for me to visualize what exactly is going on.
     
    Last edited by a moderator: May 6, 2017
  5. Feb 27, 2013 #4
    Nevermind, figured it out.
     
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