[R-390] 6080 in place of 6802 - RMS ???

[Roger L Ruszkowski]

Barry Ornitz,

I agree with This group has changed. You are right.

Take two 6080's and put them in series at 12 VAC.
They draw 2.5 amps and the heaters dissipate 30 watts.
The effective resistance is 12Vac  /  2.5  amps = 4.8 ohms. OK
The effective power is 30 watts (12 VAC x 5 amps).

Now double the voltage with no diode.
Take two 6080's draw 5 amps and the heaters dissipate 120 watts.
The effective current is 24 vac / 4.8 Ohms = 5 amps
The effective power is 120 watts (24 VAC x 5 amps).

Now add the diode.
What ever happens now happens 1/2 the time.
Barry said so. 1/2 the time it will work. 1/2 the time it will not work.
This is a half wave rectifier. 1/4 the power.
120 watts is back to 30 watts.

The applied voltage is 24 volts  1/2 the time. This averages to 12 volts.
The applied current is 5 amps 1/2 the time. This averages to 2.5 amps.
An average 12 volts time and average 2.5 amps is an average 30 watts.

Before you leap on that RMS average and .707 and PI pile again,
Remember that if you turn that diode around and get a half wave rectifier
of the opposite polarity you do not get more or less power out of the
circuit.
Therefore the final results is either way the diode goes in the circuit the
output must be equal and by logic 1/2 of the input.

Fellows have operated these circuits for years, with or with out math
and understanding. The Provda is and it is us who must come to
understand nature.

Roger KC6TRU