Equivalent capacitance; complex circuit

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The discussion focuses on finding the equivalent capacitance of a complex circuit involving multiple capacitors. The user initially attempts to simplify the circuit by calculating the equivalent capacitance of capacitors C2, C3, and C4, but struggles with the subsequent connections. Suggestions are made to group the capacitors into manageable sections, leading to a clearer path for calculating the overall equivalent capacitance. The proposed grouping effectively simplifies the circuit analysis. Ultimately, the user successfully resolves the problem with the provided guidance.
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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.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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