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Easy Capacitor problem

  1. May 19, 2010 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://img37.imageshack.us/img37/2208/43252578.jpg [Broken]

    Ok so
    i did this questions and was satisfied with my answer, but some guy rekons i did it wrong so i just want to double check

    First i found the total capacitance, so i got 1/12 + 1/12 = 1/6 = 6 (for the series in the paralell)
    then added that series to the paralell capacitor, 6+12 = 18
    then added that in series to the final capacitor 1/18+1/12 = blah = 7.2
    so the total capacitance is 7.2, to find the charge i times it by the voltage
    7.2 x 12 = 86.4
    then with this i found the voltage through the paralel circuit, 86.4/18 = 4.8
    then the same with the lone capacitor 86.4 / 12 = 7.2
    7.2 + 4.8 = 12
    so voltage in = voltage out... is that a correct rule ? thats what he said i did wrong
    that voltage in doesnt = voltage out
    but with that
    i halfed 4.8 because they're paralell , so 2.4 on the top part and 2.4 on the bottom part of the paralell circuit
    and the lone capacitor has a V = 7.2

    does that all seem right? or is voltage in not supposed to equal voltage out?
    the guy got like 6 6 3 3 for his voltages which is 18V total
    i dono it doesnt seem right to me at all but im probably wrong
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 19, 2010 #2


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    I am not sure what you mean by "voltage in = voltage out". Nevertheless, you did this right. The 7.2 μF equivalent capacitance will have 4.8 V across it and the lone 12 μF capacitor will have 7.2 V across it. It is correct that the sum of potential differences should add up to the overall difference of 12 V.
    This is where you went wrong. Given that the equivalent, three-capacitor, combination has 4.8 V across it, you can't just say that half the voltage is across each branch. "Parallel" means that each branch has the same voltage across it. In other words, the 12 μF capacitor in the left branch has the same voltage across as the two 12 μF series combination. So how much voltage is across each capacitor in the combination?
    Last edited by a moderator: May 4, 2017
  4. May 19, 2010 #3
    2.4? cause if they have 2.4 it all adds up to 12
    i keep getting confused
    and would the two capcitors in series (of the three combination) have the same potential across them?
  5. May 19, 2010 #4


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    Look at your drawing. Add point "C" between the lone capacitor and the combination. This will help you become unconfused. You have already established that VBC = 7.2 V and that VCA = 4.8 V. What is the potential difference across the one capacitor (left branch) in the combination?
  6. May 19, 2010 #5
    7.2 + 4.8 = 12
    is that what you mean?
    so doesnt the capacitor in the left branch of the combination have to be <4.8 because if it was bigger then the total potential will be >12
    i thought when a voltage goes into a paralel circuit its halved, or 1/3 depending on how many parallells there are
    like equally split across the paralells?
  7. May 19, 2010 #6


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    You are confused. The parallel combination has 4.8 V across it. "Parallel" means that the each branch in the combination has the same voltage across it. So the branch with the one capacitor (left) has 4.8 V across it and the branch with the two capacitors (right) also has 4.8 V across it. So

    The voltage across the single capacitor on the left is (you guessed it) 4.8 V.
    The voltage across the two capacitors combined is also 4.8 V, but the voltage across each of the two is not. Since they are in series they share the same charge. Now Q = CV which means that you can write Q = C1V1 = C2V2. In other words, C1V1 = C2V2. If it so happens that C1 = C2, then you can say that the 4.8 V is equally split between the two because that implies that C1V1 = C2V2[/SUB1 = V2. That's the case here, but in general it is not.

    The voltage is not "equally split between the parallels." If you have to memorize a rule, "the voltage is split equally among the series capacitors but only if they all have the same capacitance."
  8. May 19, 2010 #7
    oh mad and when you do it
    with the math it works out aswell
    q = 4.8 x 6
    28.8/12 = 2.4
    so you wouldnt add it up like 2.4 + 2.4 + 4.8 + 7.2
    itd just be 7.2 + 4.8 because the 4.8 includes everything in that circuit?
    sorry i must of listened to somone say the wrong rule and believed it was true since i heard it
    is current split equally through paralells or same thing?
    thanks alot
  9. May 19, 2010 #8


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    Yes. Or you could go from a to b adding voltages across individual capacitors following one of the following paths.

    Left path 4.8 + 7.2 = 12
    Right path 2.4 + 2.4 + 7.2 = 12

    You see how it works now?
    There is no current here because you have fully charged capacitors, but yes, if you have resistors instead of capacitors, the current is split between two resistors in parallel according to I1R1 = I2R2. It will be equally split only when R1 = R2.
  10. May 19, 2010 #9
    ohh i understand
    so thats what i got it confused with
    thanks alot for all your help
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