As serious as the mathematics says:
Let’s do Eh (heater voltage) of an amplitude Ep (peak voltage) and frequency f, then ω = 2 * Π * f
Eh = Ep sin(ωt)
is the instant voltage between heater pins, and ground.
Let’s do a Chk = cathode-filament capacity, then a current Ihk between heater and cathode may flow of a value:
Ihk = Chk * dVEh/dt (read as Current = Cap times derivate voltage respect the time), so
Ihk = Chk * d[Ep sin(ωt)]/dt =
= Chk * ω * Ep cos(ωt)
That is, it is advanced 90° respect the voltage, as in any capacitive-resistive AC circuit. As this current flows between cathode and ground, and the grid is at ground potential for this current, though the unbypassed cathode bias resistor will appear a voltage
Vkg = Ihk * Rk = Rk * Chk * ω * Ep cos(ωt).
This voltage will be amplified by the tube, as it is injected in the cathode signal path, and is amplified as the tube is grounded grid, then at the plate a voltage of magnitude:
Vhum = Vkg * (µ+1) = (µ+1)* Rk * Chk * ω * Ep cos(ωt),
will exist, and finally, the voltage at the speaker terminals supposing n = turns ratio of the output transformer,
Vsp = n * Vhum = n *(µ+1) * Rk * Chk * ω * Ep cos(ωt)
Which is the voltage hum at the moving coil because of the capacitive coupling between cathode and heater. And another component in phase can be found of a value:
Vsp´= n *(µ+1) * Rk / (Rhk + Rk) * Ep sin(ωt)
Being Rhk the cathode-filament resistance (internal of the tube, and all external resistances added: socket, wiring, etc).
Can you be more serious than me?
Briliant!
No, I can't. I am too lazy even to add something to your equations.
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