Hi
EDIT : Giskard beat me to a ( succinct answer ) but ;
Capacitance :
(A) Does the amplifier have a transformer isolated input with no DC blocking capacitor ?
- If yes, then placing a simple inline .01uf cap will give an @ Fc point of 159 hz . I do say approximate because the 100K is likely not deadon the mark. Anyways, this point Fc point is now @ 2 octaves away from your internal passive point. I'd keep it in that general area. That way the phase-shifts are predictable.
(B)If your amplifier has another type of input topology that includes a DC blocking cap - then that cap ( now electrically end to end with your new cap ) has an effective value that totals the sum of the reciprocals .
An example; Say the existing DC blocking cap is 5 uf . That means adding the recirpocals of 5uf & .01uf. The math is .2 + 100 = 100.2 . Flip that reciprocal total ( 100.2 ) back to a real number and you have a new effective capacitance of .00998004 uf . So, effectively no net change from scenario "A". Certainly not worth spending more money on a bigger capacitor.
Re; attaching an 8 ohm load to the LF side of the crossover .
No resistor;
There is still a highly loaded, series inductor , a parallel capaitor and paralleled zobel (RC ) across the outputs of your amp. That equals a funky series LCR (resonance filter of sorts) with the shunt capacitor "bypassing" the R in the Zobel . This should give an asymetrical notch to the existing crossover slope - somewhere within in the stopband. Whether or not the impedance of this notch ever approaches a low enough Z point to upset the stability of your tube amp is anyones guess.
To calculate the effect of this LCR filter, one would need to know the motional impedance (Z) range down in the stop-band, below the Fc and the HP portion of the crossover. This is very likely a fruitless and fatiguing excercise in math. An advanced Passive Crossover modelling progam ( "Spice ) would be really useful here.
With the 8 ohm load LF ;
- With that 8 ohm load there, the amp will be "encouraged" to deliver a frequency range of (@ 200 hz to 800 hz ) into the load resistor for heat dissipation. That utilizes ( uses up ) some capacity ( current draw ) within the powersupply of the tube amp that otherwise could be reserved for higher frequencies.
iWhat would I do ?
I'd unsolder one end of the series inductor in LF area of the crossover . This makes a clean and utter break in that one area of the crossover. No LCR filter reflected back to the amp - therefore no "Low Z notching" effects.
iWhat should you do ?
Consider the above solution , if you're not comfy with that then add the resistors; And, for the sake of creating a heat-related safety/margin, make it (2) 16ohm resistors paralleled together to derive your 8 ohm load or better yet (4) 32 ohm resistors paralleled .
As I now see in the reply preview, "Like He Said" ( Giskard & Alex )
<> Earl K