Equivalent Capacitors and their charges:

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SUMMARY

The equivalent capacitance (C-eq) of the given circuit is 9.22 microfarads, as established in the discussion. The correct approach involves recognizing that the 12μF and 8.35μF capacitors are in series, and their equivalent capacitance is in parallel with a 4.5μF capacitor. This combination is then in series with capacitor C, and the entire arrangement is parallel to a 7.22μF capacitor. Properly applying the formulas for series and parallel capacitors will yield the value of C.

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Homework Statement


Hi, I'm trying to solve this problem on MasteringPhysics:

This circuit is given, and it says C-eq for the circuit is 9.22 microfarads:

http://img143.imageshack.us/img143/3475/walker4ech21pr118.jpg

The question is to find C in the figure knowing C-eq is 9.22 microfarads.

Homework Equations


C-eq series = 1/C-eq = 1/C1 + 1/C2 + ...
C-eq parallel = C1 + C2 + C3 ...

The Attempt at a Solution



At first I tried to take the three center capacitors to be parallel (The 7.22, 4.25, and 8.35) and find the Ceq of them and then take the other two to be in a series with this C-eq. But when I do that, I get a negative value for C.

Then I tried to take the C, 12.0 uF, and 8.35 uF to be in a series, combine them and then have that C-eq be parallel to the other two, but that also gives me a negative value.


Thank you.
 
Last edited by a moderator:
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In the circuit diagram no three capacitors are either series or parallel.
Here 12μF and 8.35μF are in series. Equivalent of these capacitors is in parallel with 4.5 μF.
Combination of these capacitors is in series with C. Finally all the above is in parallel with 7.22 μF. The whole combination is equal to 9.22 μF. Write down the correct equations and solve for C.
 

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