Equivalent Capacitance of a Circuit: Solving a Difficult Capacitance Problem

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Homework Statement


Finding the equivalent capacitance of the supplied circuit.


Homework Equations


For parallel capacitors, C EQ.=C1+C2+C3+..
For serial capacitors, 1/C EQ.=1/C1+1/C2+...


The Attempt at a Solution



The circuit simplification is difficult!
 
Attachments
  • Handwritten circuit schematic of five capacitors C1 to C5 between terminals marked V, with the label Find Ceqv. beside it.
    gview.png
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Here, Look at this diagram I drew for you.

It is actually the same circuit.

What can you say about the various potential differences?

Also, take note that the C=Q/V. Over here, C is specified, V and Q are unspecified. How can you make this property useful?
 
Attachments
  • Circuit schematic of five capacitors between two horizontal wires labelled V at left, with nodes a, b, c, d, e, f, capacitors C1 to C5, a middle capacitor C3 joining b and e, and potential differences Vab, Vbc, Vde and Vef marked by arrows.
    Ceff.png
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Thanks! Just tell me one thing, can C3 be ignored because it carries no charge?
 
That would be the case if all the capacitors except C3 have an equal capacitance. The potential difference between b and e in the diagram will be 0 and so no current will flow through C3.

The circuit will then be simpler.

And yes, C3 is ignored provided whatever is above is true.
 
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