What is the equivalent capacitance of the capacitors shown in the figure?

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cjames9001
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Homework Statement


The equivalent capacitance of the capacitors shown in the figure is 9.23 MicroF



Homework Equations



[URL]http://upload.wikimedia.org/math/2/f/5/2f5b978fd081ea63e204909addf3d028.png[/URL]
[URL]http://upload.wikimedia.org/math/f/a/3/fa34ce461e6988fd973081401cc9790c.png[/URL]

The Attempt at a Solution


[URL]http://session.masteringphysics.com/problemAsset/1122585/1/Walker.21.58.jpg[/URL]
So to find C I Attempted to add the Series ones across the top and the left most capacitor with the following calculation:

[1/(1/c+1/12+1/8.35)]=[1/({100.2+20.35c}/100.2c)]=(100.2c)/(100.2+20.35c)

then I took (100.2c)/(100.2+20.35c) and added it to 7.22 microF +4.25 microF =9.23 microF
Then I got :
(100.2c)/(100.2+20.35c)=-2.24

100.2c=-2.24*(100.2+20.35c)

100.2c= -224.448 - 45.584c

145.784c = -224.448

c=-1.540
Which is not correct, and I'm also pretty sure that capacitors can't be negative right?
 
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Components are not in series if there are other components attached to the nodes that join them. In the figure, only the 12 and 8.35 microfarad capacitors are in series.