Equivalent Capacitance in a circuit

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The discussion centers on calculating the equivalent capacitance of four capacitors in a circuit with a 15.0 V battery. The capacitors have values of C1 = 2.32 mF, C2 = 2.25 mF, C3 = 4.85 mF, and C4 = 4.89 mF. Participants clarify that capacitors in parallel add directly, while those in series require the reciprocal formula. The initial calculation provided was incorrect, leading to confusion about the series and parallel arrangements. The correct equivalent capacitance is suggested to be approximately 3.11 mF.
carpelumen
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Homework Statement


In the circuit shown in the figure the voltage of the battery is 15.0 V, and the capacitors have the following capacitances: C1 = 2.32 mF, C2 = 2.25 mF, C3 = 4.85 mF, C4 = 4.89 mF
What is the equivalent capacitance of the four capacitors?

http://tinypic.com/r/24dn5oi/8
Two sets of series capacitors that are parallel to each other

Homework Equations


Ceq=(1/C1+1/C2+1/C3...)^-1 ---- series
Ceq=(C1+C2+C3...) --- parallel

The Attempt at a Solution


Ceq=(C1+C3)(C2+C4)/(C1+C2+C3+C4)
=(7.17)(7.14) / 14.31
= 3.58 mF

According to online homework system, it's wrong. Is my equation incorrect?
 
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I think you've got the series/parallel actions backwards. Do capacitors add directly when in series or when in parallel?
 
Perhaps show us how you got...

Ceq=(C1+C3)(C2+C4)/(C1+C2+C3+C4)
 
I think you've got the series/parallel actions backwards. Do capacitors add directly when in series or when in parallel?

I don't think that's his problem because his relevant equations..

Ceq=(1/C1+1/C2+1/C3...)^-1 ---- series
Ceq=(C1+C2+C3...) --- parallel

... are correct. Capacitors add in parallel.
 
CWatters said:
... are correct. Capacitors add in parallel.

Right, which is what led me to wonder how he got a term (c1+c3)
 
Is the answer near 3.13 mF
 
harley davidson said:
Is the answer near 3.13 mF
Very close to that, yes.
 
Even closer to 3.11 :)
 

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