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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:

enter image description here

Capacitors in parallel:

enter image description here

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?

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    \$\begingroup\$ 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. \$\endgroup\$
    – Dave Tweed
    Commented Jan 5, 2015 at 13:06
  • \$\begingroup\$ 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. \$\endgroup\$
    – Dave Tweed
    Commented Jan 5, 2015 at 13:14
  • \$\begingroup\$ @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? \$\endgroup\$
    – solid.py
    Commented Jan 5, 2015 at 13:18
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    \$\begingroup\$ 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? \$\endgroup\$
    – tomnexus
    Commented Jan 5, 2015 at 13:45
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    \$\begingroup\$ @Andyaka + Future readers, for the pair of nodes b,c can someone elaborate on the solution? \$\endgroup\$
    – solid.py
    Commented Jan 5, 2015 at 14:39

1 Answer 1

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

schematic

simulate this circuit – Schematic created using CircuitLab

The second simplification

schematic

simulate this circuit

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

schematic

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.

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  • \$\begingroup\$ @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. \$\endgroup\$
    – Warren Hill
    Commented Jan 5, 2015 at 15:23

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