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Capacitive Ciruit

  1. Jan 13, 2013 #1
    Reference to question 2 part (ii).

    I found out that:

    Qt =V x Ct
    Qt=50 x 1.67 x 10-6
    Qt=83.5 x 10-6

    I understand that in a series circuit for capacitor Qt=Q1=Q2=Q3

    So if that's the case, looking at the circuit, since C2 and C3 are in series, their charge should be 83.5 x 10-6 since Qt=Q1=Q2=Q3

    Hence, Qc3=VC
    83.5 x 10-6=V x 10 x 10-6
    V=8.35V

    But how come my answer is wrong? Please advise, thank you.
     

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    Last edited: Jan 13, 2013
  2. jcsd
  3. Jan 13, 2013 #2

    mfb

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    Those 3 are not in series, don't forget C4.

    And how do you get that number?
     
  4. Jan 13, 2013 #3
    My working:

    Qt=83.5μC
    C2+C3+C4=10μF

    Qc234=83.5μC
    Q=VC
    83.5μC=V/10μF
    Vab=8.35v

    Using Voltage Divider Rule:
    V3=CtE/C3
    =6μF x 8.35/10μF
    = 5.01V Answer

    I am curious is there another way to solve this question besides the above method?


    Thanks.
     
  5. Jan 13, 2013 #4

    mfb

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    I don't think your Qt-value is correct.
    Apart from that, the solution should be fine.
     
  6. Jan 13, 2013 #5
    Ct=1.67μF
    Qt=Vt x Ct
    =50 x 1.67μF
    =83.5μC

    May I know which part is wrong?


    Thank you.
     
  7. Jan 13, 2013 #6

    mfb

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    I get a different result here.
    If you just present some parts of your calculations, it is hard to find the specific location of the error.
     
  8. Jan 13, 2013 #7
    C2+C3=6uF
    C23+C4=10uF
    C1+C5+C234=5/3 (1.666666666666667)

    Hence Ct=1.67uF
     
  9. Jan 13, 2013 #8

    mfb

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    Oh sorry, was my error.
    Ok, I get the same capacitance now.
     
  10. Jan 13, 2013 #9
    Thank you mfb for the help.
     
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