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Finding the Charge on a Capacitor in a Circuit

  1. Apr 21, 2013 #1
    1. The problem statement, all variables and given/known data

    nodi10.jpg

    2. Relevant equations

    q=CV
    C=C1+C2...(parallel)
    1/C=1/C1+1/C2...(series)

    3. The attempt at a solution

    I'm not completely understanding these problems on capacitors in a circuit, as the textbook was pretty brief when explaining these particular concepts. I've only tried part a and b, but I've gotten neither right. This was my train of thought:

    C(top branch)=(1/(1x10^-6)+1/(3x10^-6))^-1 = 7.5x10^-7 Farads
    C(bottom branch)=(1/(2x10^-6)+1/(4x10^-6))^-1 = 1.33 x 10^-6 Farads
    C(net)=2.08x10^-6 Farads
    using q=CV, q=2.496x10^-5 Coulombs.

    In the top branch of the circuit, the charges on C1 and C3 are equal. The voltage is divided equally between the two branches, so 6 V goes to the top branch and 6 V to the bottom branch. Using q=CV, q1+q3=C(top)*V(top)=4.5x10^-6 coulombs. Since in a series, the charges are the same, q(top)=q1=q3, so q1=q3=4.5x10^-6 coulombs. However, the answer is 9.0x10^-6 coulombs. Where did I go wrong? Thanks in advance!
     
  2. jcsd
  3. Apr 21, 2013 #2

    gneill

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    Noooo. Both branches have 12V across them, since they are both in parallel with the battery. Parallel branches have the same potential drop.
     
  4. Apr 21, 2013 #3
    Wait, I'm rather confused. If both branches have 12 V potential difference, doesn't that mean that the total voltage is 24 V? Also, how are the capacitors in parallel with the battery? Aren't the branches themselves in series with the battery?
     
  5. Apr 21, 2013 #4

    gneill

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    You have a battery and two parallel branches. C1 and C3 comprise one branch, C2 and C4 comprise the other. The battery fixes the potential across each branch that it parallels.
     
  6. Apr 21, 2013 #5
    I understand that there are two parallel branches, and I understand that the voltage in each branch is equal, but I still don't quite understand why that voltage is 12 and not 6. Is the voltage of the battery simply the voltage through every branch, and if so, why?
     
  7. Apr 21, 2013 #6

    gneill

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    Voltage goes "across", current goes "through". The 12V of the battery is across both branches, so both branches "see" the 12 V of the battery.
     
  8. Apr 21, 2013 #7
    Okay, I think I've been confusing voltage with current, since current does divide when it goes through branches. However, because of this explanation, a previous problem doesn't make sense now:

    2djzcrq.jpg

    For point B, the answer stated that the voltage across C1 and C2 was q/C, or 72x10^-6/12x10^-6, or 6 V, not the 12 V like the battery. Why is this case different than that of the previous problem?
     
  9. Apr 21, 2013 #8

    gneill

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    Because in this case C3 forms a series connection with the net capacitance of C1 || C2. In effect, branch that parallels the battery is comprised of C3 in series with C1 || C2.
     
  10. Apr 21, 2013 #9
    So am I right in thinking that if the capacitors are in series with the battery, then the voltages across the capacitors aren't necessarily the same as across the battery?

    I also tried thinking it out this way (unsuccessfully). Since the charge is the same through C3 and C1+C2, and since C3 has a charge of C3*12, and since C1+C2=C3, then the net charge through C1+C2 is C3*12 also. Then according to that logic, the voltage across C1, for example, is C3*12/C1, or 36 V, which isn't right.
     
  11. Apr 21, 2013 #10

    gneill

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    Correct. The voltages across series capacitors will sum to the voltage across the series string. Or in other words, series capacitors create a form of voltage divider. Call the parallel combination of C1 and C2 Cp. That is, Cp = C1 || C2. Then the potential across Cp plus the potential across C3 must equal the potential of the battery, 12V.
    Yes, your thinking is incorrect because you have assumed that there is 12V across both C3 and Cp. This is not the case, as the 12V will be divided between the two capacitances. How the potential divides depends upon the capacitances, C3 and Cp. But they must sum to 12V.
     
  12. Apr 21, 2013 #11
    Wait, so for number 23, the potential across C3 is also 6 V? With that diagram, how does one even tell how the voltages are broken up. Is the reason why C3 has a potential difference of 6 V and Cp has a potential difference of 6 V because C3=Cp?
     
  13. Apr 21, 2013 #12

    gneill

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    Yup. Equal capacitances with equal charges yield equal potential differences (voltages).

    The capacitive voltage divider is a circuit you should learn to recognize. It comes up fairly often, and it helps if you can write down the voltage split by inspection. There's a simple formula that tells you how the voltage splits, and it's based upon the individual capacitor values.

    You can derive it yourself by considering two capacitors in series with a given total potential across them. You know what the total equivalent capacitance is, so you know the charge q that will be on the equivalent capacitor, and therefore on both of the capacitors (series connection so same charge). knowing the charge and capacitance values, you can find the individual potentials.
     
  14. Apr 21, 2013 #13
    Alrighty, I think I got it. Thanks for all the help and being so patient!
     
  15. Apr 21, 2013 #14

    gneill

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    Glad to be of help!
     
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