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

For fun, I ran a series of SPICE simulations on what you discussed. It
seems logical to me - the circuit changes if the series resistance is zero
(ie. the resistor is deleted). Deleting a lossy component from an otherwise
ideal circuit tends to produce that sort of change in behavior. However,
everything has resistance and non-resistive materials/connections only
exist as mathematical conveniences/approximations. Therefore, you can
safely ignore the special case were resistance is zero and losses are zero.

Energy in a capacitor is E=1/2*C*V^2

Discharging a 1F capacitor charged to 1V through a 1 ohm resistor into
another 1F capacitor produces a final voltage of 0.5V. The initial system
contained 0.5J of energy, and 0.25J is lost to the resistor. If there were
no resistor (which is nonsense!) then the losses would be 0J and the final
voltage would be 0.707V. We know that regardless of the resistor's value
(so long as it is positive and nonzero) that the system will always reach
the same final voltage eventually. Therefore the losses are fixed at 1/2 of
the available energy.

An analogy might be to ask "what happens when you take the charge on a
parallel plate capacitor and rather than it being evenly distributed, you
move all of the charge to the left half of the plates leaving no charge on
the right half of the plates and then let it come to equilibrium." The
answer to what it does "just depends" on how accurately you decide to model
it (real material properties vs imaginary "ideal" material properties will
produce totally different results). The case with ideal
components/materials is completely made up - it should be unsurprising that
it does not seem to match reality very closely.

In summary: mathematical approximations are just approximations - ignore
them for cases where they produce nonsensical results.

-Jason White