Find Total Capacitance of Cube w/12 4.71pF Capacitors

[eltel2910]

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. How would I even begin to go about finding the total capacitance?

 

[berkeman]

I assume you mean the total capacitance between diagonal corners.

Start with a little easier problem -- a cube of resistors. There may be a shortcut way, but write the node equations for each of the 8 vertexes, and take advantage of the fact that the resistors have equal value. After you see how that works out, do the cube of capacitors.

Problems like this get harder when the components are of different values.

 

[kyle8921]

To find the total capacitance of the cube, we can use the formula for capacitors in parallel, which states that the total capacitance is equal to the sum of the individual capacitances. In this case, we have 12 capacitors in parallel, so the total capacitance can be calculated as:

C_total = C1 + C2 + C3 + ... + C12

Substituting in the value for each capacitor (C=4.71 pF), we get:

C_total = 4.71 pF + 4.71 pF + 4.71 pF + ... + 4.71 pF (12 times)

C_total = 12 x 4.71 pF

C_total = 56.52 pF

Therefore, the total capacitance of the cube with 12 4.71 pF capacitors is 56.52 pF. It is important to note that this calculation assumes that the capacitors are perfectly identical and arranged in a symmetrical manner within the cube. Any variations in the capacitance or arrangement may affect the overall value.