Equivalent Capacitance of capacitors

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SUMMARY

The equivalent capacitance of the circuit is determined to be 9.26 µF. The discussion clarifies that the 12.0 µF capacitor and the 8.35 µF capacitor are in series, contrary to the initial assumption that they were in parallel. To find the unknown capacitance, C, the user must first combine the series capacitance of the 12.0 µF and 8.35 µF capacitors, and then incorporate the unknown capacitor into the overall calculation. This step-by-step approach is essential for accurately solving for the equivalent capacitance.

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Ballin27
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Homework Statement



The equivalent capacitance of the capacitors shown in the figure is 9.26 uF

http://session.masteringphysics.com/problemAsset/1122585/1/Walker.21.58.jpg

Homework Equations



Cseries = 1/Cs = 1/c1 + 1/c2 + ...
Cparallel = Cp = C1 + C2 + ...


The Attempt at a Solution



Ceq = 9.26uF = Cseries + Cparallel

I am confused as to what I should do with the 12.0uF capacitor and the unknown capacitor. I assume that the 7.22, 4.25, and 8.35 are all in parallel?

Any and all help would be greatly appreciated, thanks in advance.
 
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Ballin27 said:
I am confused as to what I should do with the 12.0uF capacitor and the unknown capacitor. I assume that the 7.22, 4.25, and 8.35 are all in parallel?

No, those capacitors are not in parallel; for capacitors to be in parallel both of their terminal leads need to be directly connected.

You will have to work through the circuit in a series of steps, combining serial and parallel capacitances as they arise. To start off, combine the serial pair 12.0 and 8.35. You'll have to carry the unknown capacitance, C, as a variable as the simplification process incorporates it. Eventually you'll have to solve for it.
 

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