# Equivalent Simplified Capacitor circuit

Schematic created using Multisim

I'm trying to find the equivalent capacitance for the above diagram between nodes b and c but I'm having a lot of trouble understanding which capacitor is in series/parallel to which capacitor.

I know these equations:

Capacitors in series:

Capacitors in parallel:

From looking at the circuit I believe C4 and C3 are in series same as C1 and C2

Would the resulting two equivalent capacitors C4,3 and C1,2 then be parallel with C5 and C6 respectively?

• Equivalent capacitor between which two nodes? Note that the circuit is completely symmetrical -- there are four nodes that form the vertices of a tetrahedron, and there's a capacitor along every edge. But the equivalent capacitance between any pair of nodes depends very much on the actual values of the capacitors. Jan 5, 2015 at 13:06
• For example, if $\frac{C1}{C4} = \frac{C2}{C3}$, then the value of C5 doesn't affect the result at all, since there's never any voltage across it. Jan 5, 2015 at 13:14
• @DaveTweed I never thought that any capacitor would be completely eliminated Thanks. However because of the symmetry of this circuit, I am not sure what pair of capacitors should I examine first? Jan 5, 2015 at 13:18
• This looks like a homework problem designed to break the usual method of series and parallel reduction of a problem. It needs something called nodal analysis, there are four nodes, so four simultaneous equations. Does the problem state between which two nodes you'd like to know the equivalent capacitance? Jan 5, 2015 at 13:45
• @Andyaka + Future readers, for the pair of nodes b,c can someone elaborate on the solution? Jan 5, 2015 at 14:39

Here is some 'simplification of the circuit'
That is the first simplification.

simulate this circuit – Schematic created using CircuitLab

The second simplification

simulate this circuit

The furthest simplification, assuming C1/C4=C2/C3

simulate this circuit

Here are the results C1+C4=C3+C2=C/2

nodes BC= C6+2*(C/2)=C+C=2C

I think I have drawn the circuit correctly. Read this for more information on circuits like these.

• @Wing you don't need all caps to be the same value only $\dfrac{C1}{C4} = \dfrac{C2}{C3}$ so yes you can eliminate C5 as it has the same voltage both sides so will have zero current through it. Jan 5, 2015 at 15:23