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Capacitance of an infinite series

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data

    The capacitance of each capacitor of the infinite series shown in the picture is C = 1[tex]\mu[/tex]F. Find the total capacitance between points a and b. IMAGE: http://img61.imageshack.us/img61/3674/pic002311.jpg" [Broken] (continues to infinity)

    2. Relevant equations

    In series, (1/Ceq) = (1/C1)+....(1/Cn)
    In parallel, Ceq = C1+...Cn

    3. The attempt at a solution

    Well, since the series extends to infinity, it seems that it is just a repeat of the first branch/loop (A--B, 3 capacitors only), and I thought that the Capacitance between A&B would be infinity. After thinking this over, I doubt this would be the case, seeing as that is TOO easy.

    After, I thought that it could be possible for the capacitors in the top and bottom rows to be in parallel with each other, but then I saw there is a break in between by the 3rd capacitor in between.

    I'm totally lost on how to handle this :/
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 13, 2009 #2

    Avodyne

    User Avatar
    Science Advisor

    Cute problem.

    Let D be the capacitance between points a and b. Think of the whole infinite chain as a single capacitor with capacitance D.

    Now consider adding another link of the chain in front of it; that is, connect a and b with a new capacitor with capacitance C, and then create new endpoints a' and b' by adding a capacitor between a and a' and another one between b and b' (each with capacitance C). Now, the whole thing will have a capacitance E between a' and b'; you should be able to calculate E in terms of C and D.

    But, since the chain was infinite to begin with, adding another link should not change its total capacitance; that is, we should have E=D.

    This is enough info to compute D.
     
  4. Oct 14, 2009 #3
    Thanks for the quick reply!

    Well, I've made this quick picture of what you've typed out:
    http://img158.imageshack.us/img158/3674/pic002311.jpg" [Broken]

    So adding this link in this infinite chain will not affect the total capacitance D, where E=D.

    Correct me if I am wrong, but the capacitance between a' & b' is the equivalent capacitance of the 3 capacitors, which are in series, so (1/Ceq) = (1/C + 1/C + 1/C), where C = 1 microFarad

    So the equivalent capacitance of this is equal to C/3, which, is equal to E, which is also equal to D?

    So the capacitance of this infinite series of capacitors (D), is equal to C/3, or 1/3 microFarad?
     
    Last edited by a moderator: May 4, 2017
  5. Oct 17, 2009 #4
    sorry for this...

    bump perhaps? :(
     
  6. Oct 17, 2009 #5
    No. D is parallel with one of the C's and the result of that is in series with two more C's
    The result of that is equal to D again.
     
    Last edited by a moderator: May 4, 2017
  7. Oct 24, 2009 #6
    Ok, so we have one Capacitor with Ceq = C+D (since one capacitor and D are in parallel)

    Then, there are 3 capacitors which are in series with each other. Then here is where confusion steps in...

    (1/C) + (1/C) + (1/C+D) = 1/D

    or

    (1/C) + (1/C) + (1/C+D) = D

    I see that this will yield a quadratic equation.
     
  8. Oct 24, 2009 #7
    you mean 1/((1/C) + (1/C) + (1/(C+D))) = D?

    You indeed get a quadratic. The final answer for D is quite simple.
     
  9. Oct 25, 2009 #8
    I got an answer of 1 microFarad, or 1 x 10^-6 Farads.

    It seems that this is a quite simple answer, but WHY is this the answer? What leads the capacitance of this whole series to be 1 microF?
     
  10. Nov 3, 2009 #9
    bump please :]
     
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