Why this resistor value?

[Lectroid]

Currently building a one-channel AB763.  I got the Hoffman schematic from this site, drawing credited to  sluckey. Most of the component values on that schematic look correct for the Deluxe Reverb/Pro Reverb ballpark that I'm shooting for.

But one resistor doesn't make sense to me.  One the original Fender AB763 drawing, (Pro Reverb below) the Normal channel's second stage triode, (post tone stack,) shares its cathode resistor and bypass cap with the Vibrato channel's second triode.  On the ProReverb schematic below you can see them joined by the "A" linkage. Both tubes share the 820 ohm resistor and the 25uF cap.

So far so good.

BUT. On the Hoffman 1-channel schematic, that single remaining second-stage triode has a cathode resistor of 1.5K ohms. That value is what I don't get.  I've worked some math using sample schematics with one triode and two triodes but to me it looks like the value should be 410 ohms, not 1.5K (roughly 1640, which is 820 doubled.)

I accept that the 1.5K value is correct but I'd like to know how to look at this via Ohm's law  I hope someone can explain why that two-channel value is (roughly) doubled for one channel instead of halved?

[vampwizzard]

shared cathodes vs unshared. When theres two triode halves used you need halve the resistor to bias the tube to the same level. The pro reverb shares cathodes, the one channel does not.

[tubeswell]

V=I x R
So
1V= 2I x 0.5R
Or
1V= 0.5I x 2R
And so on

[Lectroid]

vampwizzard,

Thanks for laying that out.  I knew my math had to be wrong, just couldn't see how.

tubeswell,
The Ohm's law example was exactly what I needed.  As per usual, I was making it harder than it is.