Infinite ladder's equivalent capcitance

Question

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.

Comments

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.

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Thanks @sammygerbil. Silly mistake on my part.

Answer 1

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?

Answer 2

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