1. Suppose the a 12 V bulb, used to discharge C, is rated 6 W. Then at that
voltage I=0.5A and R=24 ohms. Leigh is correct, R decreases rapidly with
the temperature. I would not be surprized, by measuring V and I, to find
that R is only 2 or 3 ohms at T at which the filament becomes dark. This
is still large in comparison with about 0.1 ohms we attributed to the
capacitor itself.
2. The term "internal R" of a capacitor bothers me. We say that "R of a wire"
is due to random collisions between drifting electrons and atoms. Consider
a vacuum capacitor. Here we have a "displacement current" due to net
changing charges on the opposite plates. I know this current is real; it
creates magnetic field. But the vacuum has no ohmic resistance. Neither
atoms nor electrons are present; nothing flows through the vacuum. There
may a leakage resistance of many gigaohms, for example along the walls of
the vacuum vessel, but this is not what we were referring to as "the
internal R" of 0.1 ohms or so.
3. Now consider a dielectric made of polarized molecules. Here the leakage
may be through the dielectric, but again, the leakage resistance is very
large. What causes our small "equivalent series resistance"? The dominant
effect, when a capacitor is being charged, is the displacement current.
Is it possible that no heat is produced in the dielectric and that only
conducting leads are responsible for our small R? The other extreme is
that the resistance of leads is negligible and that the dissipation of
energy does take place in the dielectric, for example through "internal
friction" between the alligning molecules in the increasing electric field.