Capacitors problem

[ShawnD1]

I got this question in my book and I have absolutely no clue how to solve it.

Three capacitors of 4, 8, and 16 uF (microfarad), respectively, are connected so that each capacitor forms the side of a triangle. Point A is the connection between the 4 and 8 uF sides; point B between the 8 and 16 uF sides; point C between the 4 and 16 uF sides. Find the capacitance between points A and B, B and C, and C and A.

The answers are:
AB = 11.2 uF
BC = 18.67 uF
CA = 9.33 uF


Does anybody know how to get these answers?
All the books says is this. For series circuits, the capacitance is 1/C = 1/C1 + 1/C2 + 1/C3...... For parallel, capacitance is C = C1 + C2 + C2....

[nochay]

yikes. good thing I don't engineer stuff. Excuse me, I need to pop a few tylenol.

Err, E=MC2?

Sorry dude. Not really sure what the answer would be.



Dane

Unfortunately I don't know much about how capacitance works, but I can tell you that there's no way to reach those answers with those formulas. Not to mention the problem itself doesn't make too much in the way of sense. If they want the capacitance between two points on the triangle, it's simply whatever the capacitor is between those two points.

AB = 8
BC = 16
CA = 4

Gotta be something missing.

[Joost]

Try to look at it this way.

Hold on please......

I made some errors

Joost

[Joost]

Here it is:

Edit: Of course, all numbers need to be multiplied with 10^-6

Joost

[ShawnD1]

Wow joost you're awesome. Are you an engineer or something?

Ahhhhhhhhhhhhhh, now it makes sense.

[originel]

awwww...beat me too it (and i'm an electrical engineering major)

[Joost]

Yes, I am an engineer or at least I have e bachelors degree Applied Physics.

Joost