This might shed a little light on the discussion.

Consider a diode, forward biased by a 10 mA DC current source connected
directly across it. The characteristics of this particular diode are
such that the voltage is exactly 0.7 volt.

What's the resistance of the diode?

Is the diode's resistance V/I = 0.7/.01 = 70 ohms? Well, yes. The diode
is dissipating power, equal to V*I = 7 mW. If we removed the diode and
replaced it with a 70 ohm resistor, the voltage and current would be
exactly the same as before, and the resistor would dissipate exactly as
much as the diode. If the resistor's resistance is 70 ohms, then the
diode's resistance is surely 70 ohms also.

Is the diode's resistance "non-dissipative"? No, it dissipates exactly
as much as the equivalent resistor.

Now connect another current source across the biased diode, but make it
an AC source and connect it through a capacitor. Make the AC source
current 0.1 mA RMS, which is much less than the bias current of 10 mA.
What AC voltage should we expect? Well, it's not 0.1 mA * 70 ohms,
because 70 ohms is the DC resistance V/I, and the AC current is
encountering a resistance of dV/dI, where dV/dI is the ratio of the
*change* in voltage to the *change* in current. This is the dynamic, or
AC resistance, and it equals the slope of the V/I curve of the diode.
The AC and DC resistance are the same for a simple resistor, but not the
nonlinear diode. The dynamic resistance actually changes for each
instantaneous value of the AC waveform, but as long as the AC current
waveform is much less than the DC bias current, the change won't be
much. It turns out that the AC resistance for a diode is about 0.026/Idc
where Idc is the bias current. So in our case, the AC resistance is
about 0.026/0.01, or about 2.6 ohms. Therefore we'd see an AC voltage
across the diode of 0.1 mA * 2.6 = 0.26 volt.

The AC voltage is in phase with the AC current through the capacitor, so
the AC current source is supplying power to the diode, an amount equal
to Iac * Vac = 0.1 mA * 0.26 V = 0.026 mW. What happens to this power?
The answer is that the diode turns it into heat. Is the AC resistance
"dissipative"? Of course it is, it's dissipating the AC power supplied
by the AC current source. As before, we can replace the resistance, this
time the AC resistance, with a resistor -- we'll just have to connect it
through a capacitor so it doesn't see any of the DC bias. Now the
equivalent diode is a 70 ohm resistor with another 2.6 ohm resistor in
parallel through a capacitor. This will behave just the same as the
diode, with respect to voltage, current, and power dissipation, both AC
and DC, at the specified operating point.

The lengthy discussion about tube plate resistance has muddled the DC
operating point (equivalent to the diode DC bias) and the plate
resistance, which is the plate's AC or dynamic resistance. They're
related just the same as for the diode. And just like the diode, if you
were to send an AC signal to the plate of a biased tube through a
capacitor, you'd find an AC voltage and current which are in phase
resulting in power being delivered to the tube, and an increase in plate
dissipation in the amount of that power.

Roy Lewallen, W7EL