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Finding the charge for Capacitor

  1. Apr 13, 2008 #1

    tony873004

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    1. The problem statement, all variables and given/known data
    Find the charge on C4. (That's the 4.0 uF in the middle). C1=15 uF, C2=10uF, C3=15uF, C4=4uF, battery = 10 v

    [​IMG]

    2. Relevant equations
    Q=CV


    3. The attempt at a solution
    The presence of C1 (the one next to the battery) is confusing me. It's in series with C2 and C3. It's also in series with C4, but C4 is parallel to C2,3. I can't find an example on how to do this type of problem.

    The answer is 24mC, but I don't know how to get it.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 13, 2008 #2

    Nabeshin

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    Deconstruct then reconstruct. Combine all resistors in series and parallel. Start with the inner-most part of the circuit (what looks to be the most complicated) and just start combining capacitors.
     
  4. Apr 13, 2008 #3

    tony873004

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    I tried that, but with no luck

    The 10 and 15 are in series, so they make a combined 6uF capacitor
    This new 6 and the 4 are in parallel, so they make a 10
    The new 10 and the bottom 15 are in series, so they make a 6.
    So the whole circuit simplifies to a single 6uF capacitor.

    Q=CV = 6*10 = 60

    The 15 on the bottom and the combined 10 are in series, so the charge must be the same. Is this 60? Or does it get divided?

    Reconstructing it, the 4 and the combined 15 on the top are in parallel. So assuming that the total is 60, how do I compute the charge on each?
     
  5. Apr 13, 2008 #4

    Nabeshin

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    When computing charge, there are two important things to remember: Q is the same for capacitors in series and V is the same for capacitors in parallel. That should be enough ^^
     
  6. Apr 14, 2008 #5

    tony873004

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    I still don't get it. I don't see any opportunity to have a voltage other than 10. When I do it, I don't get 24.
     
  7. Apr 14, 2008 #6

    tony873004

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    Here's my effort showing where I get stuck

    [​IMG]
     
  8. Apr 14, 2008 #7
    If you replace two series capacitors by a single one, and then calculate the charge on the replacement capacitor, both of the original capacitors will have the same charge as the replacement capacitor. (because the same current went through them at all times)
     
  9. Apr 14, 2008 #8

    tony873004

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    In my above diagram, I made a new capacator, c5, from the two series capacators. Assuming that the voltage is still 10, its charge is CV = 6*10 = 60. So c3 and c4 are both 60 also. But I need the charge on c4, so I'm not sure where to go from here.
     
  10. Apr 14, 2008 #9
    you should replace c1, c2 and c4 with a capacitor with the equivalent capacitance, and THAT capacitor is in series with c3, and will have the same charge.
     
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