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
--

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
--

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

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.

No, I'm sorry, that isn't true. The average power isn't the product of
the product of the RMS voltage times the RMS current, except in the
single circumstance of their being in phase.

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

I don't have this book, but I know that in the past you've quoted from
books without having fully understood the context of the quote. I'm sure
that's the case here. Any electrician or technician should know that for
sinusoidal waveforms, Pavg = Vrms * Irms * cos(theta) where theta is the
phase angle between V and I. And hopefully you can see with a few
moments and a calculator that if theta = 90 degrees, Pavg = zero
regardless of V and I.

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

Perhaps your "usual circumstances" are that the load is purely
resistive. But that's not "usual circumstances" for a host of applications.

Roy Lewallen, W7EL

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

How about P = Square(V) / R watts

or P = Square(I) * R watts

for no phase angles.

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

Reg Edwards wrote:
How about P = Square(V) / R watts

or P = Square(I) * R watts

for no phase angles.

Those are fine by me.

Just for the record, though:

V is the voltage across R, I is the current through R. This is important
when there are other components, hence possibly other values of V and I,
in the circuit. P can be average and V and I RMS; or P, V, and I can all
be functions of time -- the formulas are ok in either case. And finally,
just for the fuss-budgets, we're assuming R isn't varying with time.

Picky as this might seem, it's awfully important to make perfectly clear
just what a formula applies to and what it doesn't. Otherwise it's sure
to be misapplied in situations where it isn't valid. (Clarification
reduces the chance of misapplication from certain to only highly likely.)

Roy Lewallen, W7EL

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

Roy Lewallen, W7EL wrote:
"V is the voltage across R, I is the current through R."

Yes. The voltage drop across a resistor is always in-phase with the
current through the resistor as there is no energy storage in a
resistor.

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 in rec.radio.amateur.antenna,
(Richard Harrison) wrote,
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.

That is true for the Special Case where your load is a pure
resistance. In general, no.

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

Reg Edwards wrote:
How about P = Square(V) / R watts

or P = Square(I) * R watts

for no phase angles.

Actually, the "R" implies an impedance of R+j0,
i.e. a phase angle of zero.
--
73, Cecil http://www.qsl.net/w5dxp

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

Owen Duffy wrote:
"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."

Exactly, except the term "effective value is as old as a-c power
calculations. The value of the a-c volt was chosen to produce the same
effect, lamps as bright, heat as warm, as d-c does.

My electronic dictionary says:
"rms amplitude - Root-mean-square amplitude, also called effective
amplitude. The value assigned to an alternating current or voltage that
results in the same power dissipation in a given resistance as dc
current or voltage of the same numerical value.'

Best regards, Richard Harrison, KB5WZI

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

On Thu, 16 Feb 2006 09:20:34 -0600, (Richard
Harrison) wrote:

Owen Duffy wrote:
"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."

Exactly, except the term "effective value is as old as a-c power
calculations. The value of the a-c volt was chosen to produce the same
effect, lamps as bright, heat as warm, as d-c does.

My electronic dictionary says:
"rms amplitude - Root-mean-square amplitude, also called effective
amplitude. The value assigned to an alternating current or voltage that
results in the same power dissipation in a given resistance as dc
current or voltage of the same numerical value.'

Richard, your discussion here is limited to DC circuits.

Your proposition in another post that Vrms * Irms gives the Real Power
(ie indicates net energy flow over time) does apply to DC circuits,
but it does not apply generally.

You have cited a text book to support your position, however it is
likely that you have misinterpreted the text book.

Work this example through with your textbook:

We have 120Vrms AC 60Hz (sinusoidal) impressed across a load of 85
ohms of resistance and an ideal inductor of 85 ohms reactance in
series. The load impedance is 85+j85. The Circuit current is
120/(|85+j85|) or 1Arms.

The power dissipated in the resistance is 85W, and since it is the
only resistance dissipating power, the Real Power for the entire
circuit is 85W. The circuit Apparent Power is 120 * 1 or 120VA, the
Reactive Power is 85VAR.

The circuit Vrms * Irms does not give the Real Power for this circuit.

Owen
--

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

Owen Duffy wrote:
"Work this example through with your textbook."

Don`t need the textbook. I`ve been working these for over 60 years.

Pythagoras gave us the solution in ancient times when electricity was
produced by rubbing an amber rod with an animal pelt. The impedance is
close enough to 120 ohms.

I=E/Z
E= 1 ampere

Power=Isquared x R
Power = 85 watts

Best regards, Richard Harrison, KB5WZI

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

On Thu, 16 Feb 2006 17:03:29 -0600, (Richard
Harrison) wrote:

Owen Duffy wrote:
"Work this example through with your textbook."

Don`t need the textbook. I`ve been working these for over 60 years.

Pythagoras gave us the solution in ancient times when electricity was
produced by rubbing an amber rod with an animal pelt. The impedance is
close enough to 120 ohms.

I=E/Z
E= 1 ampere

Power=Isquared x R
Power = 85 watts

Nothing like Vrms * Irms though, is it?


Best regards, Richard Harrison, KB5WZI
--

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

Owen Duffy wrote:
"Nothing like Vrms times Irms through it is it?"

Yes it is. Sorry I fat fingered (E=1ampere). It is I=1 ampere.

1 ampere times the 85 volts dropped across an 85-ohm resistor by 1
ampere = 85 watts, same as 1x1x85=85, or 85x85/85=85. All methods of
calculation must give the same answer, and they do if you don`t make a
mistake.

Best regards, Richard Harrison, KB5WZI