Rick, what kind of ammeter did you use to measure the current? Jim -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Rick Brashear Sent: Sunday, May 07, 2006 1:47 PM To: Discussion of AM Radio Subject: Re: [AMRadio] Ferroresonant transformer revisited
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... > > -- No virus found in this outgoing message. Checked by AVG Free Edition. Version: 7.1.392 / Virus Database: 268.5.3/331 - Release Date: 5/3/2006
