How do I calculate the equivalence capacitance of this circuit?

[Bishop556]

Homework Statement

http://loncapa2.fsu.edu/enc/66/45d3639022485d2ba995cf84c579b584bffe5ace9851f3da3fbac5f5a1c7d2ee9dfed8743cf75459ef336ad7a1bbe176f7a0cf1a35a519ce0cef3c31febf9858404cac5a2cd3ace3851dbef61ee03a25f7a5c41a87b32e0f.gif

Calculate the equivalence capacitance of the circuit shown in the diagram above; where C1 = 8.65 μF, C2 = 4.00 μF, C3 = 8.55 μF, C4 = 3.20 μF.

Homework Equations

Capacitance in a series: 1/Ctot = Ʃ(1/C)

Capacitance in parallel: Ctot = ƩC

The Attempt at a Solution

Well, I attempted multiple ways to order each capacitor into a way to solve a formula. In my sixth attempt I presumed that C1 and C2 are in a series and are parallel to C3. I also believe C4 is parallel to C3, so here is my equation:

C = (1/C1 + 1/C2) + C3 + C4

Another idea that came into ferment was the concept of thinking that C2 and C3 are in series and are parallel to C1. C4 would be parallel to C2 and C3 so:

C = C1 + (1/C2 + 1/C3) + C4

I could go on, but since my professor never went over this I am thoroughly confused.

 

[ehild]

Show the circuit diagram, please.

ehild

 

[Bishop556]

If you can't see it on the thread, then here is the link:

http://loncapa2.fsu.edu/enc/66/45d3639022485d2ba995cf84c579b584bffe5ace9851f3da3fbac5f5a1c7d2ee9dfed8743cf75459ef336ad7a1bbe176f7a0cf1a35a519ce0cef3c31febf9858404cac5a2cd3ace3851dbef61ee03a25f7a5c41a87b32e0f.gif

 

[ehild]

I can not open the link. It is not safe.

Anyway, the equation you wrote "C = (1/C1 + 1/C2) + C3 + C4" is wrong.
The resultant of the series capacitors is 1/(1/C1+1/C2).

ehild

 

[CWatters]

I can't see image by either method.

Suggest you upload as an attachment?