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Infinite ladder's equivalent capcitance

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How do I find the equivalent capcitance of this infinite ladder?

Attempt:

Since $||- ||$ 1-2 system in series is repeating infinitely, I replaced it with $C_{eq}$

Then I got $||-||$ 1-2 system in parallel with $C_{eq}$

$\implies C_{eq} = \dfrac 23 + C{eq} \implies 0 = \dfrac 23$ which is obviously wrong.

Please let me know the flaw in my argument and please provide a hint on how to solve this question.

asked May 22 in Physics Problems by Reststack (404 points)
You have the right idea about what to do. But you need to look more carefully at what is in parallel and what is in series. The $2\mu F$ is in series with $C_{eq}$, and the $1\mu F$ is in parallel with that combination.
Thanks @sammygerbil. Silly mistake on my part.

2 Answers

1 vote
 
Best answer

You have got the right idea.

The capacitance across AB is the same as when the terminals are placed at A'B' just to the right of the $2\mu F$ capacitor and the repeating unit to the left of A'B' is removed (see diagram). Then you have a $1\mu F$ capacitor in parallel with $2\mu F$ and $C_{eq}$ capacitors in series. The equivalent capacitance across AB is $C_{eq}$.

The quadratic equation you should get (in units of $\mu F$) is $$C_{eq}=1+\frac{2C_{eq}}{2+C_{eq}}$$

See similar question resistance independent of number of cells?

answered May 23 by sammy gerbil (26,096 points)
selected May 28 by Reststack
0 votes

If $C$ is eqivalent then, $$\dfrac{2\cdot (C+1)}{2+(C+1)}=C$$

answered May 23 by n3 (318 points)
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