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Infinite Ladderof Capacitors

  1. Mar 11, 2010 #1
    Arrangment as is follows a's are there as place holders to keep the diagram intact while posting, all capacitors have = capacitance C

    aaaaaa=aaaaaaa=aaaaa= .........

    The question just asks for the equvilant capacitance.

    The professor hinted at the solution by writing 1/ceq = 1/c + 1/ceq where ceq is the equivelent capacitance and the answeris given as (c/2)((5)^(1/2)-1)
    Last edited: Mar 11, 2010
  2. jcsd
  3. Mar 11, 2010 #2
    Suppose you remove the two leftmost capacitors from the diagram.
    What you have left is the same infinite network of capacitors, wich should
    have the same capacitance.
  4. Mar 11, 2010 #3
    ok I understand that, so that would leave you with 1/ceq=(2/c + 1/ceq)? because the 2 "removed" capacitors are in series
    Last edited: Mar 11, 2010
  5. Mar 11, 2010 #4
    They are not in series. draw a diagram with just the first 2 capacitors and the rest of the network as Ceq
  6. Mar 11, 2010 #5
    so 1/ceq=1/c + 1/ceq is what i got, which is insolvable, circuts are definatly not my strong point i guess
    aaaaaaaa=aaaaa= ceq
  7. Mar 11, 2010 #6
    The capacity of the entire circuit Ceq is equal to the capacity of the circuit you've drawn
    here, wich is one capacitor of value Ceq in parallel with a capacitor of value C, and the
    equivalent capacitor to that in series with another capacitor with capacitance C
  8. Mar 11, 2010 #7
    Just do like willem2 said, and remember: capacitors in series combine like resistors in parallel, and vice versa. I tried out the problem myself, and the answer you have quoted is correct: i.e., Ceq = (c/2)(5^0.5 - 1).
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