Capacitors in a circuit / Calculate total capacitance

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SUMMARY

The total capacitance between terminals A and B for the given circuit configuration is calculated to be 1.625 μF. This is derived from the series and parallel combinations of capacitors C1 (2 μF) and C2 (1 μF). The calculation involves first determining the total capacitance of three C1 capacitors in parallel, yielding 5/3 μF, and then combining this with C2 in series and further calculations leading to the final result. The discrepancy with the book's answer of 1.58 μF indicates an error in the reference material.

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Homework Statement



Capacitors are connected as shown in this diagram:

http://o3xn.files.wordpress.com/2012/04/c4.png

C1 = 2 μF, C2 = 1 μF. What is the total capacitance of the system in between the terminals A and B?

Homework Equations



For capacitors C1 and C2 connected in series, \dfrac{1}{C_\Sigma}=\dfrac{1}{C_1}+\dfrac{1}{C_2}

For capacitors C1 and C2 connected in parallel, C_\Sigma=C_1+C_2

The Attempt at a Solution



The problem here is that my answer doesn't match that in the book. So I just wanted to check if you guys get the same answer as me.

The three rightmost capacitors C1 have a total capacitance of 2/3 μF.
They are connected in parallel to a C2-capacitor, their overall capacitance is 1+2/3=5/3 (μF).
Then we have a C1 in series with what we calculated above in series with yet another C1, which yields 1/(1+3/5)=5/8 (μF).
Last, we have a C2 in parallel with what we calculated above, which yields 1+5/8=13/8=1.625 (μF).

The book gives 1.58 μF.
 
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Looks like your answer is correct and the book's is not.
 

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