Equivalent capacitance; complex circuit

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SUMMARY

The discussion focuses on calculating the equivalent capacitance of a complex circuit involving multiple capacitors. The participant utilized the formulas for capacitors in series and parallel, specifically C(series) = (C1^-1 + C2^-1 +...)^-1 and C(||) = C1 + C2 + .... They proposed a systematic approach to simplify the circuit by grouping capacitors into combinations, ultimately leading to the solution. The suggested groupings were CA = C3 + C4, CB = C5 + CA + C2, CC = C6 + C1, and CD = CC + CB, which effectively simplified the calculation process.

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[SOLVED] Equivalent capacitance; complex circuit

Homework Statement



Hello, I was assigned a problem where we are supposed to find the equivalent capacitance of a circuit, and then use that to answer some other questions. However, I'm not sure how to handle this circuit.

http://img206.imageshack.us/img206/3756/circuitph8.th.gif


Homework Equations



We're using the equations where C(series) = (C1^-1 + C2^-1 +...)^-1 and C(||) = C1 + C2 + ...


The Attempt at a Solution



My immediate thought was to find the equivalent capacitance of C2, C3 and C4 using C234 = C2 + (C3^-1 + C4^-1)^-1. But, after that I'm not quite sure what to do, since the capacitors are now connected in a pattern where C1 and C234 connect together on a path that connects between C5 and C6. I didn't post the numbers for each capacitor as I figure I don't need to burden you with number crunching, but any ideas on how to simplify this circuit would be immensely helpful. Thanks.
 
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My suggestion is to group as follows:

1. CA = C3+C4
2. CB = C5 + (CA) + C2
3. CC = C6+C1
4. CD = CC+CB

With the + symbol indicating the equivalent capacitance which is not necessarily an addition.
 
Thank you very much, that did the trick.
 

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