What is the equivalent capacitance for this circuit?

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KeithLucas
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Homework Statement



I'm trying to find the equivalent capacitance for a circuit, but I keep getting stuck.

Homework Equations


C(parallel)=C1+C2+...Cn
C(series)={(1/C1)+(1/C2)+(1/C3)+...(1/Cn)}^-1


The Attempt at a Solution


See the first penciled attachment.
 
Attachments
  • Handwritten capacitor network diagram with values 3 uF, 6 uF, 4 uF, 2 uF, 8 uF, 12 uF and a 72 V source, with stepwise reduction sketches labelled series and parallel.
    CCF11142012_0000.jpg
    11.2 KB · Views: 513
  • Hand-drawn capacitor network diagram with a 72 V battery and capacitors labeled 3 μF, 6 μF, 4 μF, 8 μF, 12 μF and 6 μF; one 12 μF capacitor is circled and annotated 'series'.
    CCF11142012_0002.jpg
    8.6 KB · Views: 499
KeithLucas said:

Homework Statement



I'm trying to find the equivalent capacitance for a circuit, but I keep getting stuck.

Homework Equations


C(parallel)=C1+C2+...Cn
C(series)={(1/C1)+(1/C2)+(1/C3)+...(1/Cn)}^-1


The Attempt at a Solution


See the first penciled attachment.

I would start by solving everything that is in series with no junctions in between. That should clean it up a bit. If you still have trouble, post your drawing again after doing that.
 
superdave said:
I would start by solving everything that is in series with no junctions in between. That should clean it up a bit. If you still have trouble, post your drawing again after doing that.

Which ones are in series with no junctions in between? I've posted the original circuit in one attachment, but another attachment has my attempt where I thought I solved everything in series.
 
KeithLucas said:

Homework Statement



I'm trying to find the equivalent capacitance for a circuit, but I keep getting stuck.

Homework Equations


C(parallel)=C1+C2+...Cn
C(series)={(1/C1)+(1/C2)+(1/C3)+...(1/Cn)}^-1

The Attempt at a Solution


See the first penciled attachment.

your penciled attempt work out to be 3μF, which is same as mine and i think it is correct.
 
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