2*sqrt(2)*32/3.14159265=28.8101221484

Integral from 0 to pi of Sin(theta)/pi = Average voltage of Sin(theta) = 2/pi
therefore 1/sqrt(2) rms - 2/pi
therefore 1 - 2*sqrt(2)/pi
therefore 32 - 2*sqrt(2)*32/pi

John Fields wrote:

"John Fields" wrote in message
...
Does that meter on the left measure rms?
What's the waveform?
If you had a sinewave and two true rms measurements, then obviously the
theory predicts the rectified voltage will be about .707 of the ac less
diode drops...
Maybe the meter on the left measures average voltage and corrects it by a
factor of about 11% to give an artificial rms reading. But that's accurate
only for a pure sine wave. Maybe you hooked up a transformer to line power,
which typically gets flat-topped by all the filter caps in power supplies
for electronic devices.
You left an awful lot out of your post. Want to come clean?

On Fri, 22 Jan 2010 19:29:24 -0500, "Michael Robinson"
wrote:


"John Fields" wrote in message
.. .
Does that meter on the left measure rms?

---
Blow up the picture, get the manufacturer and model number, Google
around for the manual, etc...
---

What's the waveform?
If you had a sinewave and two true rms measurements, then obviously the
theory predicts the rectified voltage will be about .707 of the ac less
diode drops...
Maybe the meter on the left measures average voltage and corrects it by a
factor of about 11% to give an artificial rms reading. But that's accurate
only for a pure sine wave. Maybe you hooked up a transformer to line power,
which typically gets flat-topped by all the filter caps in power supplies
for electronic devices.
You left an awful lot out of your post. Want to come clean?

---
Geez, it seems to me that with what's in the photo you ought to be able
to answer your own questions instead of asking me to revisit an old
thread just to do all your work for you.


JF

On Fri, 22 Jan 2010 19:29:24 -0500, "Michael Robinson"
wrote:


"John Fields" wrote in message
.. .
Does that meter on the left measure rms?
What's the waveform?
If you had a sinewave and two true rms measurements, then obviously the
theory predicts the rectified voltage will be about .707 of the ac less
diode drops...
Maybe the meter on the left measures average voltage and corrects it by a
factor of about 11% to give an artificial rms reading. But that's accurate
only for a pure sine wave. Maybe you hooked up a transformer to line power,
which typically gets flat-topped by all the filter caps in power supplies
for electronic devices.
You left an awful lot out of your post. Want to come clean?


Maybe you should learn how to post a reply to Usenet before re-invading
the forum again.