# Capacitance: Equivalent C of a Network

#### calvert11

1. The problem statement, all variables and given/known data

A capacitor network is shown below.
http://img99.imageshack.us/img99/9967/79727282.gif [Broken]

What is the equivalent capacitance between
points y and z of the entire capacitor network?

2. Relevant equations

In parallel: C = c1 + c2 + ...
In series: 1/C = 1/c1 + 1/c2 + ...

3. The attempt at a solution

I'm not sure exactly which capacitors are in series and which are parallel.

But I was thinking that the capacitors in series are the ones that form an isolated system. This seems to be all the capacitors with 15 mF, which is C2 = 15/6. The remaining capacitor C1 = 10 mF seems to be parallel with C2 so C = C2 + C1 = 15/6 + 10 = 12.5 mF. But this is the wrong answer.

I'm a bit stumped. I'd appreciate any help.

Last edited by a moderator:

#### mplayer

Start collapsing the circuit from the far right, back towards the nodes x and y. Try re-drawing the circuit after each combination for clarity. If an element shares the same current with another element, it is in series. If an element has the same voltage across it as another element, it is in parallel.

Remember that a series combination for a capacitor is: 1/Ceq = 1/C1 + 1/C2 + 1/C3 + ...
and NOT Ceq = 1/C1 + 1/C2 + 1/C3 + ...

I see that mistake often on here.

#### berkeman

Mentor
Actually, it's a trick question (or a poorly asked question). With the ideal battery or power supply connected in parallel, the circuit is a short circuit for AC between y and z, so the equivilant capacitance is infinite.

The problem probably means to ask what the equivalent capacitance is for the set of capacitors to the right of the battery, not including it. In that case, mplayer has given you the hints.

Last edited:

#### calvert11

Collapsing the circuit step by step worked. Thanks for the help.

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