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Capactor circuit

  1. Dec 4, 2008 #1
    Capacitor circuit

    As shown in the figure, capacitor1 (C1=8.0μF), capacitor2 (C2=6.0μF), and capacito3 (C3=8.0μF) connected to 12.0 V battery. When switch S is closed so as to connect uncharged capacitor4 (C4=6.0μF), a) how much charge passes through point P from the battery and b) how much charge shows up on capacitor4?

    q = CΔV


    I would like to know the concept of this problem. Would you please suggest me the direction to calculate this problem?

    The answer from textbook: a) 7.2μC and b) 18.0μC. I'm trying to find out the solution.

    Thank you.
    Last edited: Dec 4, 2008
  2. jcsd
  3. Dec 4, 2008 #2


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    Homework Helper

    First find the initial equivalent capacitance of C1, C2, C3. Then from this you can find the charge on the equivalent capacitor. Note that the charge on the equivalent capacitor is the same for all capacitors in series since caps in series have the same charge.

    Then after the switch is connected, find the new equivalent capacitance. See what to do from here?
  4. Dec 7, 2008 #3
    Re: Capacitor circuit

    I am really sick of this problem.

    For a), I tried by summing of C1+C2+C3, I get 2.4. Then, why I can't get the right answer from 2.4*12=28.8? (The correct answer is 7.2μC)

    For b) I tried:

    C2+C4=12μF; then the circuit will be series, which q will be equal in each capacitor.
    q = C1V1 = C2V2 = C3V3, but C3V3 = C1V1

    8V1=12V2 ==>>> 2V1 = 3V2

    And for V1+V2+V3 = 12, then 2V1 + 2V2 = 12
    After that, I will get V2 = 3V and I use q = CΔV equation. Then, q = 6*3 = 18μC

    Please help me for a) and is my solution in b) correct?
    Thanks again.
  5. Dec 7, 2008 #4


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    Homework Helper

    Re: Capacitor circuit

    What is the formula for equivalent capacitor for caps in series? Secondly your approach here shouldn't be to simply find the amount of charge on the equivalent capacitor at first. It is the change in the charge on the capacitor that the question is asking for.

    Yes correct.
    Why should C2 have the same charge as C1 and C3 after the switch is closed? Unless you mean to say "Now I denote the equivalent capacitance for C2 and C4 by simply C2", then you're right.

    Why is this true?
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