Solving Capacitor Circuit: 3uF, 6uF, 2uF, 4uF & 90V

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SUMMARY

The discussion focuses on solving a capacitor circuit involving capacitors of 3uF, 6uF, 2uF, and 4uF in series and parallel configurations, with a total voltage of 90V. The user initially miscalculated the equivalent capacitance for the series capacitors, leading to incorrect voltage calculations. After reevaluating the approach using the formula for capacitors in series (1/C_total = 1/C1 + 1/C2), the user corrected their calculations, determining the equivalent capacitance to be approximately 3.3uF. The discussion emphasizes the importance of unit consistency and proper application of circuit equations.

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brittydagal
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Homework Statement



the question - http://tinypic.com/r/14l37yv/6
its a little hard to see but the top left capacitor is 3 uF, the one to its right *(in series) is 6 uF, the one on the bottom left is 2 uF, the one in series with the 2uF one is 4 uF and on the bottom is 90V.

Homework Equations



V = IR
C = Q/V
Series = same Q
Parallel = Same V

The Attempt at a Solution



part 1 of my work - http://tinypic.com/r/jtpd0p/6
part 2 of my work - http://tinypic.com/r/2lkbdww/6
 
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How did you calculate (a)? It is wrong. I think you forgot to take inverse values somewhere. Working with units would help here.
As additional check, you could calculate the potential difference at the top left capacitor. The sum of that and your result in (b) should add up to 90, but it does not.
 
UGHH i totally see what i did. dangit
i was doing
1/3 + 1/6 = .5
but I am pretty sure its suposed to be
1/3 + 1/6 = 1/x ----> which would make the combination of the top to uF equal 2 instead
 
..so would A) equal 3.3 uF then?
 
And was I doing B on the right track than? just my numbers were messed up?
 
10/3µF, which is about 3.3µF, right.
The concepts for b-d are fine.
 
It worked out perfectly :) thanks
 

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