# Homework Help: Series or parallel wired capacitors for max power storage

1. May 30, 2013

### Birdnals

1. The problem statement, all variables and given/known data
6. Based on what you can tell so far, if you have an N number of capacitors (say at a nuclear fission site) and you need massive power storage, you would place the N number of capacitors in
a. Series
b. Parallel
c. Combination of series and parallel
d. Both choices a and b would work.

2. Relevant equations

U=(cV^2)/2, C eq for parallel = Sum of Cs, C eq for series = (C1*C2)/C1+C2, Q=CV(battery), ???

3. The attempt at a solution

Common sense originally pushed me towards parallel since the equivalent C for them would obviously be higher. But then I researched it and came across an electronic website that said they were the same and that the voltage would differ. Their equation didn't make a lot of sense to me but the underlying theme that energy is the same no matter how you arrange the capacitors makes sense.

I used (the same) arbitrary capacitor ratings to find an equivalent C and then plugged that into the U=(cV^2)/2 formula with an arbitrary voltage and U was higher for the parallel wired caps but then I went back to the aforementioned website that tried to explain that voltage would differ, not total storage so this did nothing to resolve my confusion.

Let me also ask this. If both circuits were connected to the same power supply, would they both store equal amounts of power with one just running at a different voltage than the other one?

The link for the referenced site is here. Any help would be greatly appreciated, I can't seem to wrap my mind around this. Thanks!

Last edited: May 30, 2013
2. May 30, 2013

### haruspex

Consider a different equation, U = QV/2. This tells you that you can store more energy by adding more charge. To determine how much energy a given capacitor can store you have to consider what it is that limits the charge. Is it the source voltage? Is it the breakdown voltage of the capacitor? Something else?
You've got contradictory results so far because the equations you've used have made different implicit assumptions about the limiting factor.

3. May 31, 2013

### Birdnals

Thanks for the thought provoking response. This seems to be the spot I am getting stuck at. Before I posted here, I played around with that formula and its derivatives but its the concept, as you discussed, that is giving me trouble. To me, it seems like the limiting factor would be the voltage. But I don't understand how the voltage would be different if using the same source voltage. What I also looked at was the formula Q=CΔV for a parallel circuit and. But again, without knowing what the limiting factor is, I'm stuck.

I'm starting to think that the charges from all the circuit with parallel capacitors will add up to the charge of the circuit with series capacitors meaning Q would be the same for both circuits and therefor the energy stored would be the same but I can't quite find the missing links to prove it. Am I at least headed in the right direction?

4. May 31, 2013

### Birdnals

Alright, just so we're clear, and so I can answer this question correctly on my take home test, the correct answer would be that it doesn't matter how the capacitors are configured, correct?

Thanks again for your help. You shed new light on the subject so I now have a much better understanding of how these things work!

5. May 31, 2013

### haruspex

Sorry, ignore hat last post. it was nonsense, and I've now deleted it.
More fruitfully, let me expand a little on my original post...
When you charge up a series stack of capacitors with charge Q, induced charges appear through the stack. If each has capacitance C then each gets a voltage Q/C and the stack has voltage NQ/C. The same charge on N in parallel gets spread across them, so you need a charge of NQ to get the same charge per capacitor.
So it comes down to these rules:
- if you are voltage limited (by your source voltage) go for the parallel set-up.
- if you are charge limited, go for the series set-up.
- in general, there being limits on both available charge and available voltage, some other arrangement may give you the best compromise
- if you're not limited on either, it doesn't matter what arrangement you use.

6. Jun 1, 2013

### Birdnals

OK, thank you for clarifying. I now understand completely what you're saying. My only question now is what would cause us to be charge limited?

For my test I answered D (either/or) and it was wrong. However, based on your explanation, it seems as though the question did not provide adequate information to fully answer the question, ie are we limited by voltage or charge (or neither). Would this seem like a viable angle in which to ask my professor about this question?

7. Jun 1, 2013

### rude man

To answer this question intelligently we need to know what preceded question number 6, also what was included under "based on what you can tell so far".

Last edited: Jun 1, 2013
8. Jun 1, 2013

### haruspex

Yes. You could also mention another possibility I alluded to in my first post, that we're limited by the breakdown voltage, Vbd of the individual capacitors. In this case each individual capacitor is limited to storing CVbd2/2 units of energy, and the ensemble to NCVbd2/2 units, regardless of arrangement.
E.g. for the series case, if we want to think of it in terms of the equivalent single capacitor, the breakdown voltage for the stack is NVbd and the capacitance of the stack is C/N, giving stored energy (C/N)(NVbd)2/2 = NCVbd2/2.