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Homework Help: Infinite capacitors

  1. Jan 24, 2008 #1
    [SOLVED] Infinite capacitors

    1. The problem statement, all variables and given/known data
    Consider the infinite chain of capacitor problem:
    Each of the capacitors (C) below in the infinite series circuit has a capacitance of 6.34 mF. What is the capacitance of a single capacitor that can be connected between points A and B to replace the “chain”? (The picture of the capacitor should be attached to this thread)

    Then the real question:
    suppose that each of the capacitors in the chain has a capacitance of 21-mfarads. What is the equivalent capacitance of the infinite chain?

    2. Relevant equations
    a hint: 5 = [x + {x + (x + . . . )1/2}1/2]1/2
    Solve for x: x = 20

    Q=CV and CT=C1+C2+C3 for parallel chains but capacitors in series
    add as 1/CT=1/C1 + 1/C2 + 1/C3

    3. The attempt at a solution
    For the first part, it would have to be equal to 3 parallel capacitors in series with 4 capacitors, which I assume is where the hint comes from.
    I am assuming that the x=20 means that there are 20 xs in the equation? and that you solve for x, but then why do you never take into consideration the 6.34mF?
    or is the hint for the second part of the problem?

    Attached Files:

  2. jcsd
  3. Feb 11, 2008 #2
    We have the same problem. i think i got it. take two capacitors on end, in parallel with each other, so they add, c + c = 2c [they are all the same c]. now the parallel ones become one capacitor of 2c. now take the next two [opposite each other and in parallel], and do the same thing, you now have another equivalent capacitor value of 2c. now these two are in series, so to get series equivalent, 1/c = you add 1/2c + 1/2c = 2/2c or 1/c, so c = c, which they told us [a hint] in our problem was the answer. in other words the equivalent capacitance of an infinite series of capacitors of 6.34 mF is 6.34 mF.
  4. Feb 12, 2008 #3
    This doesn't make sense at all. There are no 2 capacitors that are parallel in this circuit. Parallel doesn't mean drawn parallel, it means that both sides of each capacitor are connected to each other.

    In order to compute this: suppose that the capacitance of the whole circuit is X. this capacitance doesn't change if you add one more section at the beginning. compute the capacitance of the circuit in wich you replace all but the first section of 3 capacitances with a capacitance X. This capacitance should be equal to X.
    You'll end up with a quadratic equation for X, wich has one positive root. (X depends on c of course)
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