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Equivalent Capacitance (Not in series or parallel)

  1. Jul 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the Equivalent Capacitance:
    Untitled.jpg

    2. Relevant equations

    c=Q/V
    Conservation of energy

    3. The attempt at a solution

    I made Q1 be the capacitor of 1uf, the top 2uf be the capacitor of Q2, bottom capacitor of 2uf be Q'2, capacitor of 4uf be Q4, capacitor of 5uf be Q5.

    I then form the equation -q2/2-q5/5+q1/1=0, -q2/2-q4/4+q'2/2+q1=0, q5/5-q4/4+q2/q=0, -q2/2+q5+q4=0, -q1/1-q5/5-q2/=0
    set them equal and try to solve for a Q but I get Q'=Q'
     
  2. jcsd
  3. Jul 7, 2009 #2

    LowlyPion

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    Homework Helper

    Welcome to PF.

    Since the serial capacitances in the parallel legs are in the same ratio, won't the voltage across the central 5μf capacitor just be 0?

    Shouldn't that simplify your problem?
     
  4. Jul 7, 2009 #3
    Yes, the AC voltage across the center capacitor should be zero volts AC. The DC voltage is...whatever.
     
    Last edited: Jul 7, 2009
  5. Jul 7, 2009 #4
    So that would get you (1/2 + 1/4)^-1 + (1/1+1/2) ^-1 which would get you 2 yes?
     
  6. Jul 7, 2009 #5
    That's a fact, Jack.
     
  7. Jul 7, 2009 #6
    Thanks for the welcome, this forum seems like a good tutor... so going from the 2uf capacitor to the other 2uf capacitor you can cancel out the middle capacitor but what about going from the 1uf cap to the 4uf? Thats not in ratio... I guess a little more explanation on that would be great. Thanks for all the help.
     
  8. Jul 8, 2009 #7

    LowlyPion

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    Simply take the 5μf out of the circuit and figure the C_eq of each leg for the serial capacitors and then add the two legs like parallel capacitors.
     
  9. Jul 8, 2009 #8

    berkeman

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    Staff: Mentor

    Are you two in the same class or something? The values are different, but...

    https://www.physicsforums.com/showthread.php?t=323869

    .
     
  10. Jul 8, 2009 #9
  11. Jul 8, 2009 #10
    Is there any other solution to this problem? I think my professor said there should be 5 unknowns and 5 equations...
     
  12. Jul 8, 2009 #11
    You will need 5 equations if you want to solve for 5 unknowns, but over here you just want the overall capacitance right? You won't need 5 equations for that. Do you understand LowlyPion's solution?

    "Simply take the 5μf out of the circuit and figure the C_eq of each leg for the serial capacitors and then add the two legs like parallel capacitors. "
     
  13. Jul 8, 2009 #12

    berkeman

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    I don't think LP's solution works here. There is not enough symmetry in the values to take out the 5uF cap....
     
  14. Jul 8, 2009 #13
    There isn't...? I thought having the same potential at the two points (assuming 5uF cap wasn't there to start with would justify it...>.<
     
  15. Jul 8, 2009 #14

    berkeman

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    Oh wait, I think I see the symmetry that LP may have been using. I'm a little slow here at the moment.

    Quiz Question -- what symmetry in that circuit would result in you having a zero net AC voltage across the 5uF cap, thus justifying removing it for the overall capacitance calculation?
     
  16. Jul 8, 2009 #15

    berkeman

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    Staff: Mentor

    BTW, in my defense for my slowness, the other thread that looks so much like this one (see my post #8 above) does not have the symmetry that this thread's circuit does. So there!
     
  17. Jul 8, 2009 #16
    There most certainly is.

    In this case, the fact that the ratios of the capacitors on the same parallel branches are equivalent (2/4 = 1/2) means that you can prove that the voltage across the central capacitor must be zero (you can see this by calculating the voltage difference between the two midpoints of the parallel branches two different ways: by taking the path on the left loop and then on the right...the fact that the ratios of the capacitances are equal is precisely what you need for it to come out that the difference in voltage between those points is exactly zero). If there is no voltage across it then it's not really doing anything to help increase the overall capacitance of the circuit, so you might as well just get rid of it since it's not changing the capacitance anyway (it won't charge if there's no voltage across it!)

    However, if you are a masochist (or just didn't see this little trick), you can always break out Kirchoff's Law. This way will get you 5 equations with 5 unknowns. All you have to do is pick five different closed loops around your circuit, sum the voltage drops along each and then set each sum to zero (as Kirchhoff says it should be). Now you will have 5 equations for 5 unknown charges. Now you can break out your algebra to find the charges. In this case, you'll end up with no charge on the middle capacitor. This is grounds for tossing it out (if it's not charged at all, it's not add or subtracting from the total capacitance) and now you can just compute with the regular parallel/serial formulae...fun stuff!
     
  18. Jul 8, 2009 #17
    When I look at your last two equations, I don't see how they form a closed loop? Unless you hook this piece up to a voltage, you can only get three distinct loops from the portion that you have drawn (unless you start tracing back over segments, which I feel is not allowed, though I've never really thought about that).

    Another thing to do might be to use your first 3 equations, and then supplement it with 2 more coming from conservation of charge:

    q2 + q5 = q4 and
    q1 = q5 + q'2

    That will bring you up to a grand total of 5 equations and 5 unknowns (plus these last 2 take on really simple forms).
     
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