If I have a cube made out of capacitors, that's one capacitor for everyside for a total of 12 capacitors. Each capacitor is C=4.71 pF. What is the effective capacitance of the cube?
2006-09-20 13:40:34 · 5 answers · asked by eltel2910 1 in Science & Mathematics ➔ Physics
The electrical connections are coming from opposite ends of the box
2006-09-20 13:59:52 · update #1
opposite corners of the cube
2006-09-20 14:00:19 · update #2
I assume what you are asking is that you have a capacitor along each edge (where 2 of the 6 sides meet) and there are then 3 tied together at each of the 6 vertices. Finally, you want the capacitance measured between two of the opposite vertices.
This problem does not yield to typical series parallel combinations but by using symmetry and some logic it is sovlveable. If you unwrap the cube and draw the circuit you will find 3 caps in parallel followed by 6 in parallel and the another group of 3 in parallel. The problem is that there are multiple nodes at the junctions between parallel sets. As long as all the caps are equal, you can use symetry to show that the potential is the same even though the nodes are not copnnected. Doing this you have three series groups of capacitors at 3C, 6C and 3C. Adding them in parallel, 1/(1/(3C) + 1/(6C) + 1/(3C)) = (6/5) C.
2006-09-20 14:01:42 · answer #1 · answered by Pretzels 5 · 0⤊ 0⤋
Your description doesn't describe how they are connected.
I can tell you that capacitence adds in the opposite manner of resistance. Resistors in series add. Resistors in parrallel use the harmonic mean or 1/(1/r1 + 1/r2)
OH, I think I just clicked to the picture. You mean that each edge of the cube is formed from a capacitor?
Then, I would need to know where the leads coming in and going out of this box are attached. Are they attached at opposite corners of the cube?
2006-09-20 13:49:44 · answer #2 · answered by tbolling2 4 · 0⤊ 0⤋
At two adjacent corners there are 12 caps in paralell and 12 caps in series.
Sounds like a wash to me.
Effective capacitance = 4.71 pF
2006-09-20 14:18:29 · answer #3 · answered by LeAnne 7 · 0⤊ 0⤋
Due to symmetry, we know that several points have same potential, namely the three points neighboring the input (V1); the three neightboring the output (V3); and the six in between (V2).
The problem reduces to a network of three parts in series, and each part consists of a number of parallel capacitors:
(3 C) in series with (6 C) in series with (3 C).
1/C' = 1/(3C) + 1/(6C) + 1/(3C)
... = 5/(6C)
so C' = 6C/5 = 5.63 pF.
2006-09-20 18:20:48 · answer #4 · answered by dutch_prof 4 · 0⤊ 0⤋
Hi. Sorry, you must know the polarity of each.
2006-09-20 13:43:21 · answer #5 · answered by Cirric 7 · 0⤊ 0⤋