Infinite Ladderof Capacitors

  • Thread starter cgjones
  • Start date
  • #1
3
0
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|aaaaaa|
aaaaaa=aaaaaaa=aaaaa= .........
aaaaaa|aaaaaaa|aaaaaa|
____________________________

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)
 
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Answers and Replies

  • #2
2,073
332
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.
 
  • #3
3
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ok I understand that, so that would leave you with 1/ceq=(2/c + 1/ceq)? because the 2 "removed" capacitors are in series
 
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  • #4
2,073
332
ok I understand that, so that would leave you with 1/ceq=(2/c + 1/ceq)? because the 2 "removed" capacitors are in series

They are not in series. draw a diagram with just the first 2 capacitors and the rest of the network as Ceq
 
  • #5
3
0
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|
aaaaaaaa=aaaaa= ceq
aaaaaaaa|aaaaa|
______________|
 
  • #6
2,073
332
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|
aaaaaaaa=aaaaa= ceq
aaaaaaaa|aaaaa|
______________|

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
 
  • #7
4
0
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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