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Charge on capacitor

  1. Sep 13, 2011 #1
    1. The problem statement, all variables and given/known data
    Using the figure, calculate the charge on capacitor 4.

    2. Relevant equations

    3. The attempt at a solution
    The C[itex]_{eq}[/itex] for the entire system is 5[itex]\mu[/itex]F
    C[itex]_{14}[/itex] = 10[itex]\mu[/itex]F

    Since C1 and C4 are in a series, wouldn't they receive the same 9 volts? If so, then would it be C[itex]_{14}[/itex] times the voltage (9 V) resulting in a charge of 90 [itex]\mu[/itex]C?

    That seems a little too simple, but I am wanting to be sure. It is not online grading so I am wanting to be completely sure before I turn in the paper.
  2. jcsd
  3. Sep 13, 2011 #2


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    Not correct.
    They are not in series.

    Only if they were in parallel with the battery would they receive the same 9 volts.
  4. Sep 13, 2011 #3
    I understand how I am wrong on the last bits but how am I wrong on the Ceq? I have gone through it multiple times and continue to come up with 5 as the overall capacitance.

    C123=1/(1/30 + 1/15 + 1/30)

    C1234=1/(1/C123 + 1/15)

    I get 5.

    If I am wrong then I do not quite understand how these special setups work, such as the order you have to go in.
  5. Sep 13, 2011 #4


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    This would be appropriate if C1, C3, and C2 were in series, but they are not in series. C4 hanging off of the junction of C1-C3 and C3-C2 makes C1, C3, and C2 not be in series. Remember, 2 things are in series if current flowing between them has no choice. C4 provide "an other choice" at said junctions.

    [edit-1] As far as order goes, it does not matter what you look for first-- series or parallel components. Just make sure they are truly in series or in parallel.

    [edit-2] What do you think about C3 and C4?
    Last edited: Sep 13, 2011
  6. Sep 13, 2011 #5


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    Your Ceq looks okay. EDIT: Actually, it's NOT okay! See following posts.

    If you take C3 and C4 to be in parallel (call the net C34), then C34 will be in series with C1 and C2. Series connected capacitors all receive the same charge, and it will be the same as the charge on the net capacitance (your Ceq). You should be able to determine the voltages that will appear across C1 and C2 from that, and thus the voltage that must be across C34. Can you find the charge on C3 and C4 from that?
    Last edited: Sep 13, 2011
  7. Sep 13, 2011 #6


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    gneill, wouldn't the Ceq be 1 / (1/C1 + 1/(C3 + C4) + 1/C2)?
  8. Sep 13, 2011 #7


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    Yes... my bad. I saw C3 and C4 with the same values of capacitance as being in parallel and committed an unthinking brainfart -- I just wrote down "15/2" for their combined value! Argh. Silly.

    Good catch lewando.

    My previous comments will apply with the correct Ceq used.
  9. Sep 14, 2011 #8
    That us a wierd way of thinking to me because we have yet to learn about optional directions and whatnot. I was not sure how to go about combining c3 and c4. Thanks a lot.

    Now concerning the actual calulation. Would I have to find The voltage that would reach c4 or will the full 9 volts reach it? I believe it is the former which means I will have to wait until I get home where my stuff is.
  10. Sep 14, 2011 #9


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    Do you know the formula that relates capacitance, charge, and voltage? That is, given the charge on a capacitor, can you determine the voltage across it?

    If you have determined the equivalent capacitance of the network as a whole and the charge that this equivalent capacitance will hold, then you can take advantage of the property of series connected capacitors to determine the individual voltages across each of them; Each capacitor in a series connected set always shares the same current, so their charges must be equal. The charge on each of the three capacitances will equal the net charge on the equivalent capacitance.


    In this case you're interested in the charge that will be on C4. Since C4 is in parallel with another capacitor, C3, that charge Q is going to be split between them depending upon their individual capacities. You have a choice as to how you work out what this split will be. You can determine the voltage across the pair and then determine the individual charges, or you can take advantage of the fact that in this case the capacities are equal, so the charges must also be equal (each gets half of Q).

    In general parallel capacitors are not always of equal value, so it's easier to determine the shared voltage across the pair and use the capacitor/charge/voltage relationship to determine the individual charges. In this problem it's easy to determine the voltages across C1 and C2 from the net charge, and thus the voltage that must appear across the parallel pair (the sum of all three must equal the battery voltage).

    Attached Files:

  11. Sep 14, 2011 #10
    So the I found that the voltage on C1 is 3 V and since it is in parallel with C2, then C2 has 3V as well. So that means that 3V remains to be used, which is then split between C3 and C4, having Q4 being 1.5V*15μF which would come out to be 22.5 μC.

    I actually want to understand it fully because I have a feeling this will be on the exam next week.
  12. Sep 14, 2011 #11


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    C1 is not in parallel with C2. The fact that pictorially they lie in the same direction does not mean that they are in parallel -- what counts is their terminal connections. Their terminals are separated by other components and/or connections to other components, so they cannot be in parallel. In order for two components to be in parallel their two terminals must be paired (share the same nodes).

    You want to determine the voltages on C1 and C2 separately, and that is due to the same charge, Q, being "pushed" onto them as they are charged by the battery driving current through the series-connected capacitors.

    An equivalent circuit for the problem would have three capacitors in series, as I showed in the figure in my previous post. Since C3 and C4 are in parallel, their capacities sum, yielding a value of 30μF. So all three of the capacitors of the equivalent circuit are 30μF. Intuitively, then, the battery voltage should split evenly across the three, 1/3 apiece.

    If you start out by determining the net capacitance of the network, you can then determine the net charge that must be on the net capacitance due to the voltage impressed upon it. You've done this. Now, for series connected capacitors, the same charge is pushed onto each of them since any current that flows must be the same for each. That tells you that the charge on each of the capacitors is the same and equal to the net charge that you calculated.

    Suppose that C1 was not the same capacitance as C2 (this could be a variation that shows up on an exam). Then you would want to determine the individual voltages on C1 and C2 due to the charge Qnet that you determined, in order to find out the "remainder" voltage that must be on the C3/C4 pair.
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