Capacitor Puzzle Redeux

My first blog post discussed the following interview question I received many years ago.

You are handed a 1 F capacitor charged to 10 V and two boxes containing uncharged capacitors: one box contains an uncharged 1 F capacitor and the other contains 1 million 1 µF capacitors. You can connect the capacitors from one box, one at a time, across the charged capacitor. Your job is to determine which box will allow you to discharge the voltage on the charged capacitor the most?

I received a comment recently from a reader who had looked at the problem and brought up an interesting aspect of the problem that I had not mentioned – energy does not appear to be conserved. My solution used charge conservation, so I skirted the energy conservation issue. Here is how the question came to me.

Meant to mention that these two problems together beg the question of how much energy is stored in the million charged 1uFs and, added to 6.7668 joules remaining on the 1F, how does the total compare to the original 50 joules. This is a tougher problem because each small cap is charged to a different – though calculable – value. I haven’t tackled that question yet; maybe you have.

Many problem solutions depend on some property remaining true throughout some transformation. A property that remain true throughout a transformation is called an invariant. There were clearly two candidate invariants in this problem: energy conservation and charge conservation. It turns out that the bookkeeping for the charge conservation is easier because this problem presents no opportunity to lose charge. Unfortunately, the bookkeeping for energy conservation is more difficult because energy can be lost in the interconnection between the capacitors. I will illustrate how using energy conservation for this problem presents an issue – how do I determine the losses that occur in the interconnection?

Let's begin my computing the energy on the 1 F capacitor before and after the million 1 microfarad capacitors have been attached. Since I am computationally lazy, I will write a Mathcad program to compute the result.

Figure 1: Energy on the 1 F capacitor before and after one million microfarad capacitors were attached.

Figure 1: Energy on the 1 F capacitor before and after one million microfarad capacitors were attached.

From this calculation we can see that the 1F capacitor loses about 43.2 Joules of energy. We now need to compute how much energy did the million one microfarad capacitors pick up, which is done below.
Figure 2: Energy on the One Million Microfarad Capacitors.

Figure 2: Energy on the One Million Microfarad Capacitors.

The million microfarad capacitors only ended up with 21.6 J of energy. Therefore, we are missing 42.3 J - 21.6 J = 21.7 J of energy.

Energy must be conserved, so where did the energy go? It was lost during the charge transfer through a combination of heat and electromagnetic radiation. This is why I used charge conservation for my invariant. I did not have a way to determine the energy loss that occurs during the connection.

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4 Responses to Capacitor Puzzle Redeux

  1. john says:

    This is a very different problem if ESR/ESL is assumed as zero (ideal cap, superconducting leads), than actually working with real life 1F and 1uF caps with realistic ESR/ESL values.

    The most notable variable, is the power lost in both the charged and uncharged caps, as (I^2)*ESR. For two 1F caps is the integral of the current modulated by ESL, and later ringing as an LC tank circuit till the charge equalizes, damped by (I^2)*ESR*2. This could make the real world solution significantly more difficult to solve, than ESR=ESL=0.

    First consider the case where ESR is large, like with Panasonic EEC-S5R5H105 1F super caps that are spec'd as less than 30 ohms at 1K hz, and a 5.5V charge limit. Assuming ESL = 0, then initial current would be slightly more than 5.5V/(2*30ohms) = .092A. Initial heating losses are slightly less than (0.092^2)*2*30 = 0.51 watts. The RC time constant for this circuit is slower than the 60s of t=RC because the charge/discharge voltage is moving as they equalize, instead of fixed. ESL can be assumed as zero, simply because the time period is so long, and the actual inductance is fairly small. It will take more than several minutes for the caps to equalize, and they will get moderately warm in the process.

    Then consider the case where ESR is small, a few milli-ohms at most, like a black box composite device with 100, Panasonic ECOS1JA103EA 10000uf electrolytic caps inside, wired in parallel, with a 64V charge limit. ESR for each cap is about 30 mOhms, so the effective ESR for the composite capacitor device is 30/100=0.30mOhms. Assuming they are wired as a 10x10 array, with a two layer heavy copper PCB, plus and neg planes, so interconnect resistance will be fairly low. So will the additional inductance, except for the black box "leads" connected to be planes. If charged to 63 volts, the initial current will approach 63/.0003 = 210,000 amps assuming the effective ESL for the device was pretty low. The heating losses at that instant would be (210000^2)*0.0003*2 = 2.65*(10^7) watts. Because the RC time constant is very short (slight over) 1*0.0003=300usec, and the dv/dt is very fast, even a small inductance will have a very noticeable effect, and this LC tank circuit will ring for a while. As a 10x10 array the box is going to be about 14 inches on a side. and about 2 inches high, and probably something like 0 gauge leads a few inches long. Connecting the two, will be very exciting 🙂

  2. john says:

    Hmm ... I forgot about the 10v part of the problem.

    The super cap (Panasonic EEC-S5R5H105) solution would have to be a 2x2 array composite device as well ... wired two in parallel, as pairs in series, to get the voltage up to 10v. Equiv ESR would still be 30 ohms worst case. Initial current would be slightly more than 10V/(2*30ohms) = 0.17A. Initial heating losses would then be (0.17^2)*30*2 = 1.7W.

    At 10V the electrolytic (Panasonic ECOS1JA103EA) solution would have an initial current of 10/.0003 = 33333A. Initial heating losses would then be (33333^2)*0.0003*2 = 667KW.

    RC time constant remains the same, so equalization time remains unchanged, for both.

  3. john says:

    The point is that the charge transfer resistive heating losses are ((V/ESR)^2)*2*ESR which isn't constant, or even linear - both because of the (I^R) function, and because of the ESL current wave form effects for small ESR.

  4. john says:

    dang ... because of the (I^2)*R power losses (be nice to have editing for replies.

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