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Circuits and capacitors

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data
    In the arrangement shown in the figure below, a potential difference DeltaV is applied, and C1 is adjusted so that the voltmeter between points b and d reads zero.
    http://capa.physics.mcmaster.ca/figures/sb/Graph26/sb-pic2667.png [Broken]

    This "balance" occurs when C1 = 3.56 µF. If C3 = 9.50 µF and C4 = 12.3 µF, calculate the value of C2

    2. Relevant equations
    CΔV=Q


    3. The attempt at a solution
    I know that capacitor 1-2, 3-4 is connected in series and the charge on capacitors 1-2, 3-4 are the same because the voltmeter reads zero which mean there's no transfer of electrons between the plates so the charge of the capacitors connected in series should be the same. I came up with the following attempt:

    1: C(equivalent 1-2)=(1/C1)+(1/C2)
    2:C(equivalent 3-4)=(1/C3)+(1/C4)
    3:C(total)= C(equivalent 1-2)+C(equivalent 3-4)
    4:Qtotal=Q1+Q2
    5:Qtotal= ΔV(C(equivalent 1-2)+C(equivalent 3-4))
    When i got step 5, i realized everything just cancels out... lol
    so where did I go wrong in my approach?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 31, 2012 #2

    gneill

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    Staff: Mentor

    If the voltmeter reads zero, what does that say about the potential difference between nodes b and d?
     
  4. Jan 31, 2012 #3
    o i see...the voltmeter is zero cause the potential difference is the same in both wires?
     
    Last edited: Jan 31, 2012
  5. Jan 31, 2012 #4

    gneill

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    Staff: Mentor

    Correct. That being so, what can you say about the potentials across capacitors C4 & C1, and capacitors C3 & C2?
     
  6. Jan 31, 2012 #5
    Well the potential are the same but when we say potential we're talking about V and not ΔV right?
     
  7. Jan 31, 2012 #6

    gneill

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    Staff: Mentor

    It is the voltage that exist on the capacitors, measured from terminal to terminal.

    So if the voltages across those capacitor pairs are the same, if you know the voltage on one you automatically know the other.

    Now consider just the two capacitors on the left side of the bridge, C3 and C4. What must be the sum of the voltages of those capacitors?
     
  8. Jan 31, 2012 #7
    is it equal to that of the potential difference applied?
     
  9. Jan 31, 2012 #8

    gneill

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    Clearly, yes, since KVL around the battery-C4-C3-battery loop demands it :smile:

    Since you know the values of C3 and C4 you should therefore be in a position to determine the voltages on both C3 and C4.
     
  10. Jan 31, 2012 #9
    But the only equation with i know is ΔV=Q/C but Q is unknown?
     
  11. Jan 31, 2012 #10

    gneill

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    Staff: Mentor

    The charges on capacitors connected in series, such as C3 and C4, are equal. It's also the same as the charge on the equivalent capacitance of C3 in series with C4. So if you calculate that equivalent capacitance and place the source voltage across that, you'll know the charge that must be on both capacitors. The charge then gives you the individual voltages.

    Another approach, and one I'd recommend learning about, is the voltage divider relationship for capacitors. In the same way that you can determine the voltage that appears across two resistors in series when you known the total voltage across them, you can determine the voltages that appear on individual series-connected capacitors. For two capacitors in series, the voltage divides in inverse ratio to their capacitance values. So for two capacitors, say Ca and Cb, with total voltage V across them, the voltage on Ca is:

    [itex] Va = V \frac{Cb}{Ca + Cb} [/itex]
     
  12. Jan 31, 2012 #11
    My friend broke it don into a simple ratio equation, C2/C3 = C1/C4 and solved for C2 and it works! but i'm not sure why lol
     
  13. Jan 31, 2012 #12

    gneill

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    Staff: Mentor

    It works because of the voltage ratios imposed on the capacitors by nodes b and d being at the same potential (which we discussed above), in combination with the charges on series capacitors being equal (which we discussed above). You would have got there eventually via the voltage divider equation route I was taking above.
     
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