Jim candela jcandela at prodigy.net
Sun May 7 19:01:42 EDT 2006

```Rick, what kind of ammeter did you use to measure the current? Jim

-----Original Message-----
[mailto:amradio-bounces at mailman.qth.net]On Behalf Of Rick Brashear
Sent: Sunday, May 07, 2006 1:47 PM
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:

>
>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...
>
>
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