Capacitance of an infinite series

[jumpman11372]

Homework Statement

The capacitance of each capacitor of the infinite series shown in the picture is C = 1$$\mu$$F. Find the total capacitance between points a and b. IMAGE: http://img61.imageshack.us/img61/3674/pic002311.jpg" (continues to infinity)

Homework Equations

In series, (1/Ceq) = (1/C1)+...(1/Cn)
In parallel, Ceq = C1+...Cn

The Attempt at a Solution

Well, since the series extends to infinity, it seems that it is just a repeat of the first branch/loop (A--B, 3 capacitors only), and I thought that the Capacitance between A&B would be infinity. After thinking this over, I doubt this would be the case, seeing as that is TOO easy.

After, I thought that it could be possible for the capacitors in the top and bottom rows to be in parallel with each other, but then I saw there is a break in between by the 3rd capacitor in between.

I'm totally lost on how to handle this :/

 

[Avodyne]

Cute problem.

Let D be the capacitance between points a and b. Think of the whole infinite chain as a single capacitor with capacitance D.

Now consider adding another link of the chain in front of it; that is, connect a and b with a new capacitor with capacitance C, and then create new endpoints a' and b' by adding a capacitor between a and a' and another one between b and b' (each with capacitance C). Now, the whole thing will have a capacitance E between a' and b'; you should be able to calculate E in terms of C and D.

But, since the chain was infinite to begin with, adding another link should not change its total capacitance; that is, we should have E=D.

This is enough info to compute D.

 

[jumpman11372]

Thanks for the quick reply!

Well, I've made this quick picture of what you've typed out:
http://img158.imageshack.us/img158/3674/pic002311.jpg"

So adding this link in this infinite chain will not affect the total capacitance D, where E=D.

Correct me if I am wrong, but the capacitance between a' & b' is the equivalent capacitance of the 3 capacitors, which are in series, so (1/Ceq) = (1/C + 1/C + 1/C), where C = 1 microFarad

So the equivalent capacitance of this is equal to C/3, which, is equal to E, which is also equal to D?

So the capacitance of this infinite series of capacitors (D), is equal to C/3, or 1/3 microFarad?

 

[jumpman11372]

sorry for this...

bump perhaps? :(

 

[willem2]

Thanks for the quick reply!

Well, I've made this quick picture of what you've typed out:
http://img158.imageshack.us/img158/3674/pic002311.jpg"

So adding this link in this infinite chain will not affect the total capacitance D, where E=D.

Correct me if I am wrong, but the capacitance between a' & b' is the equivalent capacitance of the 3 capacitors, which are in series, so (1/Ceq) = (1/C + 1/C + 1/C), where C = 1 microFarad

So the equivalent capacitance of this is equal to C/3, which, is equal to E, which is also equal to D?

So the capacitance of this infinite series of capacitors (D), is equal to C/3, or 1/3 microFarad?

No. D is parallel with one of the C's and the result of that is in series with two more C's
The result of that is equal to D again.

 

[jumpman11372]

Ok, so we have one Capacitor with Ceq = C+D (since one capacitor and D are in parallel)

Then, there are 3 capacitors which are in series with each other. Then here is where confusion steps in...

(1/C) + (1/C) + (1/C+D) = 1/D

or

(1/C) + (1/C) + (1/C+D) = D

I see that this will yield a quadratic equation.

 

[willem2]

(1/C) + (1/C) + (1/C+D) = D

I see that this will yield a quadratic equation.

you mean 1/((1/C) + (1/C) + (1/(C+D))) = D?

You indeed get a quadratic. The final answer for D is quite simple.

 

[jumpman11372]

I got an answer of 1 microFarad, or 1 x 10^-6 Farads.

It seems that this is a quite simple answer, but WHY is this the answer? What leads the capacitance of this whole series to be 1 microF?

 

[jumpman11372]

bump please :]