Thanks Jim...
I am deriving the power rating using a known resistance and the current
drawn by that resistance when placed across the output of the supply.
R = 8.57 ohms
I = 2.9 amps
P = I²R
P = 8.41 x 8.57
P = 72.0737
E(rms) = square root of (P x R)
E(rms) = 24.853 volts
Oddly enough, I measured 24.7 vdc on my Fluke 189 multimeter.
Rick
Jim Candela wrote:
My best answer:
E^2=PR, so E = sq root (PR)
Yes, but how do you measure P?
Another idea:
If you have an RF ammeter, this will measure rms
current regardless of ac waveform or even dc, then
I=E/R so E = IR. Pick a known resistor...
