For Roy Lewallen et al: Re Older Post On My db Question
 

Roy, W7EL wrote:
"The average power is therefore relatively small, much smaller than the
ptoduct of RMS volts times RMS amps."

I have not read the thread, but I recall from some old memory store that
rms volts times rms amps is one of the definitions of "average power".

Best regards, Richard Harrison, KB5WZI

For Roy Lewallen et al: Re Older Post On My db Question
 

On Tue, 14 Feb 2006 15:03:05 -0600, (Richard
Harrison) wrote:


I have not read the thread, but I recall from some old memory store that
rms volts times rms amps is one of the definitions of "average power".

Only in a DC circuit, or a purely resistive load in an AC circuit.

Owen

Best regards, Richard Harrison, KB5WZI
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For Roy Lewallen et al: Re Older Post On My db Question
 

On Tue, 14 Feb 2006 21:25:32 GMT, Owen Duffy wrote:

On Tue, 14 Feb 2006 15:03:05 -0600, (Richard
Harrison) wrote:


I have not read the thread, but I recall from some old memory store that
rms volts times rms amps is one of the definitions of "average power".

Only in a DC circuit, or a purely resistive load in an AC circuit.

I shouldn't use that work ONLY!!!

Only in a DC circuit, or a in an AC circuit (loop) where the current
and voltage measured are in phase. In an AC circuit where the voltage
and current are not in phase you must multiply the product of the RMS
voltage and RMS current by the cosine of the phase difference to get
real power (which is what I think you mean by "average power").

Owen
--

For Roy Lewallen et al: Re Older Post On My db Question
 

Richard Harrison wrote:
Roy, W7EL wrote:
"The average power is therefore relatively small, much smaller than the
ptoduct of RMS volts times RMS amps."

I have not read the thread, but I recall from some old memory store that
rms volts times rms amps is one of the definitions of "average power".

Time to dust off your old circuit analysis text, then. Pay special
attention to the discussion of "imaginary power" or "vars".

Roy Lewallen, W7EL

For Roy Lewallen et al: Re Older Post On My db Question
 

Owen Duffy wrote:

I shouldn't use that work ONLY!!!

Only in a DC circuit, or a in an AC circuit (loop) where the current
and voltage measured are in phase. In an AC circuit where the voltage
and current are not in phase you must multiply the product of the RMS
voltage and RMS current by the cosine of the phase difference to get
real power (which is what I think you mean by "average power").


Of course, that only works when the voltage and current are sinusoidal
and of the same frequency. More generally, the average power is 1/T
times the integral over T of v(t) * i(t) dt, where T is the interval
over which it's being averaged. If the waveforms are periodic, an
interval of one cycle can be used for T.

Roy Lewallen, W7EL

For Roy Lewallen et al: Re Older Post On My db Question
 

On Tue, 14 Feb 2006 14:15:20 -0800, Roy Lewallen
wrote:

Owen Duffy wrote:

I shouldn't use that word ONLY!!!

Only in a DC circuit, or a in an AC circuit (loop) where the current
and voltage measured are in phase. In an AC circuit where the voltage
and current are not in phase you must multiply the product of the RMS
voltage and RMS current by the cosine of the phase difference to get
real power (which is what I think you mean by "average power").


Of course, that only works when the voltage and current are sinusoidal
and of the same frequency.

Yes, implied by the "in phase" condition.

Thinking that through further brings a third case to the "ONLY"
conditions, and that is if the circuit is entirely resistive (eg real
power is the product of Vrms and Irms if the waveform is square and
the circuit contains only resistances).

More generally, the average power is 1/T
times the integral over T of v(t) * i(t) dt, where T is the interval
over which it's being averaged. If the waveforms are periodic, an
interval of one cycle can be used for T.

Roy Lewallen, W7EL
--

For Roy Lewallen et al: Re Older Post On My db Question
 

Owen Duffy wrote:
. . .
Thinking that through further brings a third case to the "ONLY"
conditions, and that is if the circuit is entirely resistive (eg real
power is the product of Vrms and Irms if the waveform is square and
the circuit contains only resistances).

If you look at the definition of average (as in my previous posting),
you'll see that when the load is purely resistive, average power = 1/T *
the integral over T of v^2(t) / R dt or 1/T * the integral over T of
i^2(t) * R dt, for any waveform. And using the definition of RMS(*), you
can see that this is exactly Vrms^2 / R or Irms^2 * R respectively,
again for any waveform. So Pavg = Vrms * Irms for any waveform, as long
as (and only as long as) the load is purely resistive. Again, the
average and RMS values can be calculated for any interval (as long as
they're the same), but a single cycle is adequate to determine the
long-term average and RMS values of periodic waveforms.

(*) frms = Sqrt(avg(f^2(t))) = Sqrt(1/T * integral over T of f^2(t) dt)

Roy Lewallen, W7EL

For Roy Lewallen et al: Re Older Post On My db Question
 

This is probably a good place to mention that people interested in the
relationship between RMS voltage and current and average power (and the
uselessness of the RMS value of power) can find an explanation at
http://eznec.com/Amateur/RMS_Power.pdf. It doesn't use any mathematics
more advanced than a square and square root, so any amateur should be
able to understand it.

Some time ago I was surprised to find this to be one of the most
frequently visited pages at my web site, apparently due to a link in the
Wikipedia entry for RMS.

Roy Lewallen, W7EL

For Roy Lewallen et al: Re Older Post On My db Question
 

Roy Lewallen, W7EL wrote:
"The average power is therefore relatively small, much smaller than the
product of RMS volts times RMS amps."

RMS is short for root-mean-square. RMS is synonymous with the "effective
value" of a sinusoidal waveform.

Therefore, the average power for the time period of one complete cycle
or any number of complete cycles is the product of the effective volts
times the effective amperes.

See page 19 of "Alternating Current Fundamentals" for derivations of the
proof.

Average power is exactly the product of rms volts times rms amps in
usual circumstances.

Best regards, Richard Harrison, KB5WZI

For Roy Lewallen et al: Re Older Post On My db Question
 

On Wed, 15 Feb 2006 17:13:45 -0600, (Richard
Harrison) wrote:

Roy Lewallen, W7EL wrote:
"The average power is therefore relatively small, much smaller than the
product of RMS volts times RMS amps."

RMS is short for root-mean-square. RMS is synonymous with the "effective
value" of a sinusoidal waveform.

Therefore, the average power for the time period of one complete cycle
or any number of complete cycles is the product of the effective volts
times the effective amperes.

Leaving aside your new confusing term "effective value", if you
multiply Vrms by Irms in an AC circuit you get Apparent Power (units
are Volt Amps or VA).

Apparent Power is the vector sum of two quadrature components Real
Power (Watts) and Reactive Power (VAR).

Real Power is the thing you describe when you talk about average
power. It is Real Power that is the rate of flow of energy averaged
over a complete AC cycle.


See page 19 of "Alternating Current Fundamentals" for derivations of the
proof.

Average power is exactly the product of rms volts times rms amps in
usual circumstances.

Depends on what you mean by "usual circumstances". Your rule does not
apply if there is a phase difference between V and I, which is
commonly the case in power distribution, and is commonly the case in
RF where loads circuit impedances may have a reactive component.

To an electrician, (Real) Power = Vrms * Irms * PF where PF (the power
factor) is the cosine of the phase angle between V and I. This isn't
engineering stuff, sparkies know and apply it every day.

Owen
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