Reducing Equivalent Resistors and Capacitors

[fatcat39]

Homework Statement

Reduce the circuit as much as possible. All resistors have R and all capacitors have same C.

Homework Equations

The Attempt at a Solution

http://photos-a.ak.facebook.com/photos-ak-sctm/v183/24/75/1238100168/n1238100168_30112052_7338.jpg



And this is my reduced one (I only reduced two capacitors in series and two parallel resistors)

http://photos-b.ak.facebook.com/photos-ak-sctm/v183/24/75/1238100168/n1238100168_30112053_7721.jpg

The reduced capacitors (the ones in the right corner of that little square) =

1/C_eq = 1/C + 1/C = 2/C
C_eq = .5C

1/R_eq = 1/R + 1/R = 2/R
R_eq = .5R


(Ignore the random squiggly at the bottom)

Anyway - that's as far as I got. I'm not sure if the two capacitors in the center are considered parallel? And if so, is that other capacitor on the far left considered in parallel with them too? (even though there are resistors between them?)

 

[quantumdude]

Can you attach the images to your post? I really don't feel like joining facebook just to look at your problem.

 

[fatcat39]

oh, sorry - they were supposed to show up!

 

[Mindscrape]

Yeah, the two caps in the middle are in parallel, but the other one you are asking about would not be. Think of parallel as whether or not the wires are connected at both ends. Think of series as whether or not two elements share a common connection. (A fancy name for connection is a node.)

 

[fatcat39]

Is this as far as it can be reduced?

http://photos-c.ak.facebook.com/photos-ak-sf2p/v164/24/75/1238100168/n1238100168_30112286_5295.jpg

Hmm...is this it? It seems like something simpler could be achieved somehow...*sigh*

 

[BOBAH]

Hello,
Can anyone give me a hint how to calculate the impedance of the following circle ?
Thank you