Capacitors, charge and energy: a question

[Kalee20]

In order to calculate the loss you would need to know the resistance of the link placed between the capacitors

No you don't.

You could consider 'perfect' capacitors, 1µF, one charged to 10V and one discharged. Energy = ½CV² = 50µJ, all in one capacitor. Then you connect them via a 1MΩ resistor.

After about 2.5 seconds, things will have nearly settled down - each capacitor will be at 5V (one a few mV above, one a few mV below, but that's trivial). Each capacitor will have ½CV² = 12.5mJ in it - that's 25µJ total - to within a few nJ owing to the millivolts of error to satisfy the pedantic.

Half the energy has been lost.

Now use instead of 1MΩ, a 1kΩ resistor to parallel them. The same happens, but in about 25msec this time. The end result is the same even though the resistance is 1000 times less.

Hopefully this will convince you that the energy lost is independent of resistance.

You can do other things like add some inductance in series, add reaslistic values of ESR, and it won't alter things. (Adding ESR will mean some of the energy is lost in the capacitors themselves; adding inductance could mean you get damped oscillations instead of exponential decay, but the quiescent result is the same).

You can reduce resistance to ohms, to milliohms, to microhms - the energy loss is the same. The limiting case is if you make it zero, you are in the irresistable-force-meets-immovable-object situation - you close your knife switch, you get infinite current for zero time - and the switch contacts weld together with precisely 25µJ of energy.