Re: [EE]: considering charging capacitor losses. When a charged cap charges a discharged cap.

At 12:18 AM 2026-04-02, you wrote:
>When I was at TAFE (not sure what the equiv but its below Uni), a lecturer
>made an off hand comment about when charging one capacitor from another
>that the math does not work out and may have continued to say only half the
>energy remains.  My son is now studying simiar subjects which we have
>discussed and has renewed my curiosity, so I have finally after 30+ years
>decided to get to the bottom of this.
>A simple google search yields a formula where the energy is shared between
>the two which conflicts with there being energy loss.
>Using more specific searches results in half the energy is lost when
>directly charging and can be reduced using inductors and diodes.
>Asking why half results in some answers restating that half is always lost.
>Not helpful.
>Others state heat, but does not explain why half.
>
>I think (and curious what others have to say) the statement should be "When
>direct charging a capacitor a *minimum* of half the energy is lost and can
>be explained considering *Maximum Power Transfer Thereom* "
>
>And I am guessing the issues my lecturer  (and now including myself) was
>having with the math was when considering an ideal capacitor.
>
>So what happens here? As the capacitors tends to ideal, the loss of energy
>remains fixed at 1/2 *but* then jumps to no loss if we consider a
>theoretical ideal capacitor.
>
>That does not seem logical.
>
>Maybe the question lies in the same realm as "does current flow thru a
>capacitor?"
>
>If I was near a pc I would try to ltspice it.

Hi, Justin:-

I am near a PC so I did a quick LTspice simulation.

https://i.sstatic.net/2AUlAmM6.png

This is a 10 second simulation with 2 1F capacitors, one charged to 1V
and one discharged. There is a 1 ohm resistor and a 1H inductor to
represent parasitic inductance and resistance in the connection. At
t = 10ns the switch changes from 100M ohm to 1 micro ohm.

You can see that the energy sloshes back and forth a bit because of
the inductance, but eventually settles down to 0.5V across each
capacitor.

So the original energy was CV^2/2 = 0.5J and ends up with 0.125J
in each capacitor. As the _integral_ of the energy dissipated by the
resistor in the simulation shows, the energy lost in the resistor
makes up all the remainder of 0.25J.

Note that the current in the inductor is initially essentially zero
so there  is no energy stored in the inductor at t=0. If I didn't
put the switch in there there would be an initial current of 1A
and the results would look different due to the additional 0.5J
of energy (but the end result after a long time would be exactly the
same with 0.5V across each capacitor- not that I would do that).

Also if you do a simulation with just an inductor and no series
resistor, LTspice defaults to an internal resistance of 0.001 ohm
so it will eventually settle  down just as with the simulation
I did. If you pick a more realistic inductance of a few
microhenries the results will look qualitatively similar, but on
a much shorter time scale.

An interesting corollary* to this is the efficiency of 50% is
worst-case. If the capacitor voltages are closer to each other
when the switch closes then the efficiency can be much better.
For example, if C1 is at 1V and C2 is at 0.9V then initial
energy is 0.905J total (0.5J + 0.405J) and after the switch
closes and things settle the total energy in the caps is
0.9025J (0.95V across each cap). So 2.5mJ was lost in the
resistor but 46.25mJ was transferred, almost 95% efficiency.


Best regards,
Spehro Pefhany

*dedicated Toyota storage?

>Thoughts?
>
>Justin
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