My statements about voltage and current referred to ideal circuits,
which are immensely useful in gaining understanding of fundamental
circuit operation. One can argue for a long time about how perfect a
real circuit can be made, and I won't get into that. I hope I made my
point without the need to.

Roy Lewallen, W7EL

Bill Turner wrote:

On Tue, 19 Oct 2004 06:49:01 -0700, Tim Wescott
wrote:


An electron beam in a vacuum can flow current without any voltage drop,
as can a superconductor.


__________________________________________________ _______

I can't argue the superconductor, but in a vacuum tube there must be a
source of voltage to cause the current to flow. Even if you are talking
about the "contact potential" caused by a heated cathode, there is still
voltage present, isn't there? And of course, the anode voltage causes
current to flow there too.

--
Bill W6WRT

On Tue, 19 Oct 2004 12:20:29 -0700, Roy Lewallen
wrote:

I really appreciate the compliment, and will do my best to try and
deserve it.

Please look very carefully at the diagram at the URL you've posted, and
notice that it's a voltage waveform (see the labeling of the vertical
axis). Then read the text very carefully. Neither the diagram nor the
text contradict what I've said. If you think it does, post the reason
why, and I'll try to clear it up.

Okay, here's the bit that you seem to take exception to (it's spread
over both pages):

"We can define the real power in an AC circuit as the equivalent DC
power that would produce the same amout of heating in a resistive load
as the applied AC waveform. [In a purely resistive load] we can use
the root mean square (RMS) values (Vrms and Irms) to find this
equivalent or RMS power."

Then the equation "P = Vrms x Irms"

Where "P" here explicitly refers to RMS power.

This is something you have stated clearly that you disagree with.
--

"What is now proved was once only imagin'd." - William Blake, 1793.

Paul, I apologize. My browser showed only the first of the two pages,
and I didn't realize that the second was there. While looking for the
quotation I found that the second page was simply scrolled off screen.

Yes, there is one thing (on that second page) I do disagree with the
author on, that the equivalent power, the product of Vrms and Irms, is
"RMS" power. I did a brief web search to find out who the author was so
I could contact him about that, and discovered that it's Joe Carr.
Unfortunately, he died a short time ago.

Maybe my suggestion about looking in non-mathematical texts for an
explanation wasn't such a good idea. It appears that some of the authors
of those texts don't fully understand the math either. I'll have to say
that you certainly have provided some evidence as to how widespread the
misconception is. Next time I'm downtown at Powell's Technical
Bookstore, I'll leaf through a few volumes oriented toward technicians
and see just how bad it is. All I have on my bookshelf in the way of
basic circuit analysis texts is two (Pearson and Maler, and Van
Valkenburg) which are intended for beginning engineering students, and
they of course both have it right. A popular elementary physics text
which I have, Weidner & Sells, _Elementary Classical Physics_, Vol. 2,
succinctly summarizes (p. 913): "Thus, the average rate at which thermal
energy is dissipated in the resistor is the product of the rms voltage
across it and the rms current through it." This follows immediately
below an equation showing the calculation of pav from the classical
definition of average which I posted some time ago.

In response to the question about Vrms, Irms, and equivalent power, I
said that when Vrms = 100 volts and Irms = 1 amp, Pavg = 100 watts and
Prms is about 122.5 watts. I haven't had any previous occasion to
calculate RMS power, so I might have made a mistake. According to my
calculation, for sinusoidal voltage and current, the RMS power equals
the average power (which is Vrms X Irms when the load is resistive)
times the square root of 1.5. Surely some of the readers of this group
can handle the calculus involved in the calculation -- it's at the level
taught to freshman engineering students, and now often taught in high
school. The calculation isn't hard, but it's a little tedious, so
there's ample opportunity to make a mistake. I'd very much appreciate if
one of you would take a few minutes and double-check my calculation. Or
check it with Mathcad or a similar program. I'll be glad to get you
started if you'll email me. And I'll be glad to post a correction if I
did make a mistake.

Roy Lewallen, W7EL

Paul Burridge wrote:

On Tue, 19 Oct 2004 12:20:29 -0700, Roy Lewallen
wrote:


I really appreciate the compliment, and will do my best to try and
deserve it.

Please look very carefully at the diagram at the URL you've posted, and
notice that it's a voltage waveform (see the labeling of the vertical
axis). Then read the text very carefully. Neither the diagram nor the
text contradict what I've said. If you think it does, post the reason
why, and I'll try to clear it up.


Okay, here's the bit that you seem to take exception to (it's spread
over both pages):

"We can define the real power in an AC circuit as the equivalent DC
power that would produce the same amout of heating in a resistive load
as the applied AC waveform. [In a purely resistive load] we can use
the root mean square (RMS) values (Vrms and Irms) to find this
equivalent or RMS power."

Then the equation "P = Vrms x Irms"

Where "P" here explicitly refers to RMS power.

This is something you have stated clearly that you disagree with.

There is a nice simple explanation in the ARRL handbook that I posted
earlier in the thread on "power output formula". Guess nobody read it.

2000 ARRL handbook 6.6 chapter 6, RMS VOLTAGES AND CURRENTS.

73
Gary K4FMX


On Tue, 19 Oct 2004 16:33:53 -0700, Roy Lewallen
wrote:

Paul, I apologize. My browser showed only the first of the two pages,
and I didn't realize that the second was there. While looking for the
quotation I found that the second page was simply scrolled off screen.

Yes, there is one thing (on that second page) I do disagree with the
author on, that the equivalent power, the product of Vrms and Irms, is
"RMS" power. I did a brief web search to find out who the author was so
I could contact him about that, and discovered that it's Joe Carr.
Unfortunately, he died a short time ago.

Maybe my suggestion about looking in non-mathematical texts for an
explanation wasn't such a good idea. It appears that some of the authors
of those texts don't fully understand the math either. I'll have to say
that you certainly have provided some evidence as to how widespread the
misconception is. Next time I'm downtown at Powell's Technical
Bookstore, I'll leaf through a few volumes oriented toward technicians
and see just how bad it is. All I have on my bookshelf in the way of
basic circuit analysis texts is two (Pearson and Maler, and Van
Valkenburg) which are intended for beginning engineering students, and
they of course both have it right. A popular elementary physics text
which I have, Weidner & Sells, _Elementary Classical Physics_, Vol. 2,
succinctly summarizes (p. 913): "Thus, the average rate at which thermal
energy is dissipated in the resistor is the product of the rms voltage
across it and the rms current through it." This follows immediately
below an equation showing the calculation of pav from the classical
definition of average which I posted some time ago.

In response to the question about Vrms, Irms, and equivalent power, I
said that when Vrms = 100 volts and Irms = 1 amp, Pavg = 100 watts and
Prms is about 122.5 watts. I haven't had any previous occasion to
calculate RMS power, so I might have made a mistake. According to my
calculation, for sinusoidal voltage and current, the RMS power equals
the average power (which is Vrms X Irms when the load is resistive)
times the square root of 1.5. Surely some of the readers of this group
can handle the calculus involved in the calculation -- it's at the level
taught to freshman engineering students, and now often taught in high
school. The calculation isn't hard, but it's a little tedious, so
there's ample opportunity to make a mistake. I'd very much appreciate if
one of you would take a few minutes and double-check my calculation. Or
check it with Mathcad or a similar program. I'll be glad to get you
started if you'll email me. And I'll be glad to post a correction if I
did make a mistake.

Roy Lewallen, W7EL

Paul Burridge wrote:

On Tue, 19 Oct 2004 12:20:29 -0700, Roy Lewallen
wrote:


I really appreciate the compliment, and will do my best to try and
deserve it.

Please look very carefully at the diagram at the URL you've posted, and
notice that it's a voltage waveform (see the labeling of the vertical
axis). Then read the text very carefully. Neither the diagram nor the
text contradict what I've said. If you think it does, post the reason
why, and I'll try to clear it up.


Okay, here's the bit that you seem to take exception to (it's spread
over both pages):

"We can define the real power in an AC circuit as the equivalent DC
power that would produce the same amout of heating in a resistive load
as the applied AC waveform. [In a purely resistive load] we can use
the root mean square (RMS) values (Vrms and Irms) to find this
equivalent or RMS power."

Then the equation "P = Vrms x Irms"

Where "P" here explicitly refers to RMS power.

This is something you have stated clearly that you disagree with.

Paul Burridge wrote:
If this is a misconception it must be an extremely widespread one. I've
found a reference to RMS power and how it's calculated in 'Practical
Radio Frequency Test & Measurement' (Newnes) and have posted the
relevant page he

http://www.burridge8333.fsbusiness.co.uk/cvcvcv.gif

If we are wrong, it appears we're not alone...


Any fool can write a book that includes the words "RMS power". I've done
it myself.

I have known since age 15 how to calculate the average power or an
arbitrary AC waveform: "Take the root-mean-square averages of the
voltage and current first, and then multiply those results together.
Doing it the other way around gives the wrong answer, stupid boy!"

My particular mistake was to try to use "RMS power" as a shorthand
label. I naively assumed that everybody would understand that it was
short for:

"This is power that has been calculated by a *correct* use of RMS
averaging, which we all know means taking the RMS average of the voltage
or current first, and obviously it does not mean taking the RMS average
of the power, because that would *not* be correct, and now that we've
wasted half a paragraph on this point, and completely derailed the
original train of thought, which was mostly about something else, I'd
rather like to get back to our main topic."

Just writing "RMS power" seemed so much more appealing...

BIG mistake! It fell right into the gap between people who don't know
how to calculate power correctly; and people who do know, but were all
too eager to assume they'd found a fault.

I learned from that mistake... but the mistake itself is immortalised
6000 times over, in cold type.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

I've found a reference to RMS power and how it's calculated in 'Practical
Radio Frequency Test & Measurement' (Newnes)

=============================

Yet another Beta-tested bible is discredited! ;o) ;o)

Paul,

I've been following the thread still. Let me go back to the original
question for a moment. So is the 4W maximum power of a CB radio is actually
average power, not RMS, right? If it is modulated at 100% with a sine wave,
what wil the PEP be? Is 16W the correct answer?

Chris
"Roy Lewallen" wrote in message
...
| Paul, I apologize. My browser showed only the first of the two pages,
| and I didn't realize that the second was there. While looking for the
| quotation I found that the second page was simply scrolled off screen.
|
| Yes, there is one thing (on that second page) I do disagree with the
| author on, that the equivalent power, the product of Vrms and Irms, is
| "RMS" power. I did a brief web search to find out who the author was so
| I could contact him about that, and discovered that it's Joe Carr.
| Unfortunately, he died a short time ago.
|
| Maybe my suggestion about looking in non-mathematical texts for an
| explanation wasn't such a good idea. It appears that some of the authors
| of those texts don't fully understand the math either. I'll have to say
| that you certainly have provided some evidence as to how widespread the
| misconception is. Next time I'm downtown at Powell's Technical
| Bookstore, I'll leaf through a few volumes oriented toward technicians
| and see just how bad it is. All I have on my bookshelf in the way of
| basic circuit analysis texts is two (Pearson and Maler, and Van
| Valkenburg) which are intended for beginning engineering students, and
| they of course both have it right. A popular elementary physics text
| which I have, Weidner & Sells, _Elementary Classical Physics_, Vol. 2,
| succinctly summarizes (p. 913): "Thus, the average rate at which thermal
| energy is dissipated in the resistor is the product of the rms voltage
| across it and the rms current through it." This follows immediately
| below an equation showing the calculation of pav from the classical
| definition of average which I posted some time ago.
|
| In response to the question about Vrms, Irms, and equivalent power, I
| said that when Vrms = 100 volts and Irms = 1 amp, Pavg = 100 watts and
| Prms is about 122.5 watts. I haven't had any previous occasion to
| calculate RMS power, so I might have made a mistake. According to my
| calculation, for sinusoidal voltage and current, the RMS power equals
| the average power (which is Vrms X Irms when the load is resistive)
| times the square root of 1.5. Surely some of the readers of this group
| can handle the calculus involved in the calculation -- it's at the level
| taught to freshman engineering students, and now often taught in high
| school. The calculation isn't hard, but it's a little tedious, so
| there's ample opportunity to make a mistake. I'd very much appreciate if
| one of you would take a few minutes and double-check my calculation. Or
| check it with Mathcad or a similar program. I'll be glad to get you
| started if you'll email me. And I'll be glad to post a correction if I
| did make a mistake.
|
| Roy Lewallen, W7EL
|
| Paul Burridge wrote:
|
| On Tue, 19 Oct 2004 12:20:29 -0700, Roy Lewallen
| wrote:
|
|
| I really appreciate the compliment, and will do my best to try and
| deserve it.
|
| Please look very carefully at the diagram at the URL you've posted, and
| notice that it's a voltage waveform (see the labeling of the vertical
| axis). Then read the text very carefully. Neither the diagram nor the
| text contradict what I've said. If you think it does, post the reason
| why, and I'll try to clear it up.
|
|
| Okay, here's the bit that you seem to take exception to (it's spread
| over both pages):
|
| "We can define the real power in an AC circuit as the equivalent DC
| power that would produce the same amout of heating in a resistive load
| as the applied AC waveform. [In a purely resistive load] we can use
| the root mean square (RMS) values (Vrms and Irms) to find this
| equivalent or RMS power."
|
| Then the equation "P = Vrms x Irms"
|
| Where "P" here explicitly refers to RMS power.
|
| This is something you have stated clearly that you disagree with.

Reg Edwards wrote:

I've found a reference to RMS power and how it's calculated in 'Practical
Radio Frequency Test & Measurement' (Newnes)

=============================

Yet another Beta-tested bible is discredited! ;o) ;o)

I wouldn't trust anything from that publisher.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

On Wed, 20 Oct 2004 12:08:59 GMT, "Chris"
wrote:

Paul,

I've been following the thread still. Let me go back to the original
question for a moment. So is the 4W maximum power of a CB radio is actually
average power, not RMS, right? If it is modulated at 100% with a sine wave,
what wil the PEP be? Is 16W the correct answer?

AIUI the specified 4 Watts is the maximum*average* power allowed (in
the UK, anyway). When you modulate it 100% AM., it's still 4W average
power. If you fully modulate it with FM., it's *still* 4W average
power. But as you've seen here, for every assertion, there's a
contradiction. ;-)
--

"What is now proved was once only imagin'd." - William Blake, 1793.

On Wed, 20 Oct 2004 13:17:28 +0100, "Ian White, G3SEK"
wrote:

Reg Edwards wrote:

I've found a reference to RMS power and how it's calculated in 'Practical
Radio Frequency Test & Measurement' (Newnes)

=============================

Yet another Beta-tested bible is discredited! ;o) ;o)

I wouldn't trust anything from that publisher.

Who was *your* publisher, Ian?
--

"What is now proved was once only imagin'd." - William Blake, 1793.