Basic AC wattage question: am I doing my math right?

[The Phantom]

On Thu, 3 Jan 2008 17:01:46 -0800 (PST), HC <hboothe@gte.net> wrote:

Thank you all for your replies. There have been a variety of useful
responses that have helped me. I found the link to Wikipedia about
Power Factor (http://en.wikipedia.org/wiki/Power_factor) provided by
The Phantom to be quite helpful. I am going to check my equipment to
see what it states/claims it can do for AC readings of both voltage
and current (see if it does RMS) and go from there. Thank you,
Phantom, that helped a bunch.

It might be well worth your while to spend $20 and get a Kill-a-watt. It
can measure true watts, volt-amperes, power factor, amps, volts--the whole
menu! The only problem it will have is that it won't have much resolution
at very low power levels (below 1 or 2 watts), but at least it can give you
an estimate even there. And, if it says 1 or 2 watts, then you can figure
you're not paying much per month for that load.

Again, thanks everybody.

Happy New Year.

--HC

 

[Eeyore]

The Phantom wrote:

Eeyore wrote:
The Phantom wrote:
Eeyore wrote:
HC wrote:

Now, I take the amps times voltage to get watts

Incorrect. Amps times Volts gives VA (volt-amperes) which is 'apparent
power'.

As measured with true RMS meters, right?

True RMS meters are only a requirement if the waveform is non-sinusoidal.

As is the waveform associated with essentially all modern power supplies,
including the OP's television.

Very much so. But even then the VA =/= the watts.

Even power factor corrected supplies don't
have a perfectly sinusoidal current draw. One might find a .95 power
factor acceptable in a particular circumstance, and the use of true RMS
meters will be necessary to verify that. Using true RMS meters will always
give the correct result

NOT for watts since the instantaneous power is the PRODUCT of the instantaneous
volts and amps and NO meter measuring only one at a time can do that.

Graham

 

[HC]

On Jan 3, 10:49 pm, The Phantom <phan...@aol.com> wrote:

On Thu, 3 Jan 2008 17:01:46 -0800 (PST), HC <hboo...@gte.net> wrote:

Thank you all for your replies. There have been a variety of useful
responses that have helped me. I found the link to Wikipedia about
Power Factor (http://en.wikipedia.org/wiki/Power_factor) provided by
The Phantom to be quite helpful. I am going to check my equipment to
see what it states/claims it can do for AC readings of both voltage
and current (see if it does RMS) and go from there. Thank you,
Phantom, that helped a bunch.

It might be well worth your while to spend $20 and get a Kill-a-watt. It
can measure true watts, volt-amperes, power factor, amps, volts--the whole
menu! The only problem it will have is that it won't have much resolution
at very low power levels (below 1 or 2 watts), but at least it can give you
an estimate even there. And, if it says 1 or 2 watts, then you can figure
you're not paying much per month for that load.



Again, thanks everybody.

Happy New Year.

--HC

I probably will, but I am still going to try to learn more about this
so I can do it manually if for no other reason than that I would like
to know how to do it, how it works, and why. Ya know? I want the
answer, but I hate to give up on the knowledge of how it works.

Thank you.

--HC

 

[The Phantom]

On Fri, 04 Jan 2008 05:38:49 +0000, Eeyore
<rabbitsfriendsandrelations@hotmail.com> wrote:

The Phantom wrote:

Eeyore wrote:
The Phantom wrote:
Eeyore wrote:
HC wrote:

Now, I take the amps times voltage to get watts

Incorrect. Amps times Volts gives VA (volt-amperes) which is 'apparent
power'.

As measured with true RMS meters, right?

True RMS meters are only a requirement if the waveform is non-sinusoidal.

As is the waveform associated with essentially all modern power supplies,
including the OP's television.

Very much so. But even then the VA =/= the watts.


Even power factor corrected supplies don't
have a perfectly sinusoidal current draw. One might find a .95 power
factor acceptable in a particular circumstance, and the use of true RMS
meters will be necessary to verify that. "Using true RMS meters will always
give the correct result

NOT for watts since the instantaneous power is the PRODUCT of the instantaneous
volts and amps and NO meter measuring only one at a time can do that.

This paragraph is about measuring VA. You said:

"Amps times Volts gives VA (volt-amperes) which is 'apparent
power'."

And I said:

"As measured with true RMS meters, right?"

believing we were discussing measuring VA, which is a part of the power
factor formula.

Then when I said:

"Using true RMS meters will always give the correct result"

I was referring to the measurement of VA since that was what the discussion
had been about.

I never suggested that measuring voltage or amperage one at a time could
give the instantaneous product of volts and amps.

Graham

 

[The Phantom]

On Thu, 3 Jan 2008 21:32:11 -0800 (PST), HC <hboothe@gte.net> wrote:

On Jan 3, 10:49 pm, The Phantom <phan...@aol.com> wrote:
On Thu, 3 Jan 2008 17:01:46 -0800 (PST), HC <hboo...@gte.net> wrote:

Thank you all for your replies. There have been a variety of useful
responses that have helped me. I found the link to Wikipedia about
Power Factor (http://en.wikipedia.org/wiki/Power_factor) provided by
The Phantom to be quite helpful. I am going to check my equipment to
see what it states/claims it can do for AC readings of both voltage
and current (see if it does RMS) and go from there. Thank you,
Phantom, that helped a bunch.

It might be well worth your while to spend $20 and get a Kill-a-watt. It
can measure true watts, volt-amperes, power factor, amps, volts--the whole
menu! The only problem it will have is that it won't have much resolution
at very low power levels (below 1 or 2 watts), but at least it can give you
an estimate even there. And, if it says 1 or 2 watts, then you can figure
you're not paying much per month for that load.



Again, thanks everybody.

Happy New Year.

--HC

I probably will, but I am still going to try to learn more about this
so I can do it manually if for no other reason than that I would like
to know how to do it, how it works, and why. Ya know? I want the
answer, but I hate to give up on the knowledge of how it works.

An admirable attitude, and continued study will get you there. Good luck.

Thank you.

--HC

 

[Eeyore]

The Phantom wrote:

Eeyore wrote:
The Phantom wrote:
Eeyore wrote:
The Phantom wrote:
Eeyore wrote:
HC wrote:

Now, I take the amps times voltage to get watts

Incorrect. Amps times Volts gives VA (volt-amperes) which is 'apparent
power'.

As measured with true RMS meters, right?

True RMS meters are only a requirement if the waveform is non-sinusoidal.

As is the waveform associated with essentially all modern power supplies,
including the OP's television.

Very much so. But even then the VA =/= the watts.


Even power factor corrected supplies don't
have a perfectly sinusoidal current draw. One might find a .95 power
factor acceptable in a particular circumstance, and the use of true RMS
meters will be necessary to verify that. "Using true RMS meters will always
give the correct result

NOT for watts since the instantaneous power is the PRODUCT of the instantaneous
volts and amps and NO meter measuring only one at a time can do that.

This paragraph is about measuring VA. You said:

"Amps times Volts gives VA (volt-amperes) which is 'apparent
power'."

And I said:

"As measured with true RMS meters, right?"

Not required if the waveform is sinusoidal.

believing we were discussing measuring VA, which is a part of the power
factor formula.

Which formula ?

Then when I said:

"Using true RMS meters will always give the correct result"

I was referring to the measurement of VA since that was what the discussion
had been about.

I never suggested that measuring voltage or amperage one at a time could
give the instantaneous product of volts and amps.

When the load current waveform is non-sinusodal, even using a true RMS meter will
not help you get a true wattage figure. Hence whether you use a true RMS meter or
not is irrelevant since with sinusoidal waveforms and a phase shift (a la classic
power factor) the averaging meter will give equally accurate results.

Graham

 

[The Phantom]

On Fri, 04 Jan 2008 06:38:55 +0000, Eeyore
<rabbitsfriendsandrelations@hotmail.com> wrote:

The Phantom wrote:

Eeyore wrote:
The Phantom wrote:
Eeyore wrote:
The Phantom wrote:
Eeyore wrote:
HC wrote:

Now, I take the amps times voltage to get watts

Incorrect. Amps times Volts gives VA (volt-amperes) which is 'apparent
power'.

As measured with true RMS meters, right?

True RMS meters are only a requirement if the waveform is non-sinusoidal.

As is the waveform associated with essentially all modern power supplies,
including the OP's television.

Very much so. But even then the VA =/= the watts.


Even power factor corrected supplies don't
have a perfectly sinusoidal current draw. One might find a .95 power
factor acceptable in a particular circumstance, and the use of true RMS
meters will be necessary to verify that. "Using true RMS meters will always
give the correct result

NOT for watts since the instantaneous power is the PRODUCT of the instantaneous
volts and amps and NO meter measuring only one at a time can do that.

This paragraph is about measuring VA. You said:

"Amps times Volts gives VA (volt-amperes) which is 'apparent
power'."

And I said:

"As measured with true RMS meters, right?"

Not required if the waveform is sinusoidal.

You already said that the first time (and I've never denied it). I replied
that since many loads draw non-sinusoidal current, by using RMS meters
correct results for the VA part of the power factor formula will be
obtained no matter whether the waveforms are sinusoidal or not.

believing we were discussing measuring VA, which is a part of the power
factor formula.

Which formula ?

Power factor = true power/(RMS volts * RMS current)
which is the same as PF = watts/VA. And if VA is measured with true RMS
meters, then this expression is correct no matter what the waveforms. If
average responding meters are used, the correct VA is only given if the
waveforms are all sinusoidal. So why not just use RMS meters and be done
with it? Especially since nowadays so many pre-existing power supplies are
uncorrected switchers, although regulatory agencies are now requiring PF
correction.

Then when I said:

"Using true RMS meters will always give the correct result"

I was referring to the measurement of VA since that was what the discussion
had been about.

I never suggested that measuring voltage or amperage one at a time could
give the instantaneous product of volts and amps.

When the load current waveform is non-sinusodal, even using a true RMS meter will
not help you get a true wattage figure.

As I have said several times now, the use of RMS meters is not intended to
get a true wattage figure. It is intended to get a true VA figure. The VA
number is needed to get the power factor.

As I said in another post, I used a wattmeter to get the true power. That,
together with the VA figure derived from measurements with true RMS meters,
allowed me to compute the power factor for the non-sinusoidal current drawn
by my TV set.

Hence whether you use a true RMS meter or
not is irrelevant since with sinusoidal waveforms and a phase shift (a la classic
power factor) the averaging meter will give equally accurate results.

I've never said that average responding meters won't give accurate results,
IF the waveforms are sinusoidal. But nowadays a substantial fraction of
equipment loads are non-sinusoidal. Someday, when the regulations have
been in effect for years, most switchers will be PF corrected, but that's
not the case now. And accurate measurement of power factor requires the
use of RMS responding meters to get an accurate value for VA.

Graham

 

[redbelly]

On Jan 3, 11:49 pm, The Phantom <phan...@aol.com> wrote:

On Thu, 3 Jan 2008 17:01:46 -0800 (PST), HC <hboo...@gte.net> wrote:

Thank you all for your replies. There have been a variety of useful
responses that have helped me. I found the link to Wikipedia about
Power Factor (http://en.wikipedia.org/wiki/Power_factor) provided by
The Phantom to be quite helpful. I am going to check my equipment to
see what it states/claims it can do for AC readings of both voltage
and current (see if it does RMS) and go from there. Thank you,
Phantom, that helped a bunch.

It might be well worth your while to spend $20 and get a Kill-a-watt. It
can measure true watts, volt-amperes, power factor, amps, volts--the whole
menu! The only problem it will have is that it won't have much resolution
at very low power levels (below 1 or 2 watts), but at least it can give you
an estimate even there.

You can beat the 1W resolution by doing a kWh measurement. At 2W,
you'd need to wait about a day with the Kill A Watt to get to 0.05
kWh, from which you'd calculate the power to +/-20% accuracy
(resolution of the Kill A Watt is 0.01 kWh).

Of course, the OP must be willing to go one day without turning on the
TV.

Mark

And, if it says 1 or 2 watts, then you can figure
you're not paying much per month for that load.



Again, thanks everybody.

Happy New Year.

--HC

 

[John Larkin]

On Tue, 01 Jan 2008 21:04:34 -0800, The Phantom <phantom@aol.com>
wrote:

On Tue, 01 Jan 2008 18:18:03 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 01 Jan 2008 17:57:41 -0800, The Phantom <phantom@aol.com
wrote:

On Tue, 1 Jan 2008 18:25:32 -0500, "Paul E. Schoen" <pstech@smart.net
wrote:


"The Phantom" <phantom@aol.com> wrote in message
news:d1dkn3h5s5giv0n10iq85b6h46fb2rafjo@4ax.com...
On Tue, 01 Jan 2008 11:14:49 +0000, Eeyore
rabbitsfriendsandrelations@hotmail.com> wrote:



HC wrote:

Now, I take the amps times voltage to get watts

Incorrect. Amps times Volts gives VA (volt-amperes) which is 'apparent
power'.

As measured with true RMS meters, right?


Finding true watts (real power) for a complex load is well ..
complicated.
There may be phase angle issues between the volts and the amps (which
make
the watts lower than the VA) and non-sinusoidal waveforms (which make the
watts HIGHER than the VA).

If watts are higher than VA, the power factor is greater than 1, since
power
factor is defined as watts/VA.

According to http://en.wikipedia.org/wiki/Power_factor, power factor is
never
greater than 1.

Can you give a specific example of non-sinusoidal waveforms where the
power
factor would be greater than 1?

If you measure the voltage and current with a true-RMS (AC+DC) meter, I
don't think there is any condition where true power is greater than
apparent power. But if you use other types of meters, particularly with
capacitive coupling which removes the DC component, there might be such
effect.

If you use other types of meters, then you haven't measured apparent power,
have you? You've measured something else.


Whenever there is asymmetrical distortion on a sine wave, phase angle
cannot be measured accurately, or in other words, it becomes undefined.

This is what John Larkin was getting at when he said, '"E I cos theta" is
sort of meaningless for radical waveforms.'

"Power factor" is a definition, and it's somewhat ambiguous. It could
mean...

(True power) / ((RMS volts) * (RMS amps))

or

Cos(E:I phase angle) where just the fundamental frequency components
are considered

or

could be undefined for non-sinusoidal waveforms, as some older
textbooks suggest.

That may have been appropriate for the older times when those textbooks
were current, but the present needs a definition for non-sinusoidal
waveforms.


There's not a lot of point arguing over definitions.

It's important that engineers everywhere mean the same thing when they use
certain defined phrases and words such as ampere, volt and power factor,
for obvious reasons. That makes it worthwhile to discuss definitions.


The first definition is probably the most sensible in most cases
nowadays.

It is the one found in the rules promulgated by regulatory agencies, and
defined by professional societies such as the IEEE. It's not just
sensible, it's mandatory to use in certain scenarios.




But it is still useful if one adopts the modern view and decomposes the
power factor into a "displacement power factor" and a "distortion power
factor".

http://en.wikipedia.org/wiki/Power_factor

http://en.wikipedia.org/wiki/Distortion_power_factor

There was an extensive thread on this topic about 3 years ago:

http://groups.google.com/group/sci.electronics.basics/tree/browse_frm/thread/6382c203016278ec/71654e97add1ae5c?hl=en&rnum=21&_done=%2Fgroup%2Fsci.electronics.basics%2Fbrowse_frm%2Fthread%2F6382c203016278ec%2F68914b13f8fa99d2%3Fhl%3Den%26lnk%3Dst%26q%3D%26#doc_f16b87eafa50bca4

Most phase angle meters use zero crossings to determine the measurement
points, and thus a waveform with zero crossings not exactly spaced at 180
degrees will read differently when the polarity is reversed. Also when
there is crossover distortion, such that there is a flat spot at the zero
crossing, a phase angle meter cannot determine the point accurately.

Power can be measured accurately by using an analog multiplier circuit,
such as the AD534. I used this IC to design a DC wattmeter, which can also
be used on AC and distorted waveforms. The main problem was isolation from
the shunt that was used to measure current. A Hall effect sensor could have
been used, but an isolation amplifier did the trick.

You can also measure power using A/D techniques, especially if you use
simultaneous sampling converters. You must perform instantaneous products
of current and voltage, and then take an average of these sums for true
power.

Measuring power in three phase systems, especially with unbalanced and
non-linear loads, is even more complex. It is not so difficult in
star-connected systems where you can measure each leg and the voltages to
neutral, but a greater challenge in delta systems.

Again, some texts simply consider power factor to be undefined in
unbalanced polyphase systems.

The practicing engineers of that day may have found that a satisfactory
state of affairs. Nowadays, there are working definitions in place to deal
with unbalanced polyphase systems, and I doubt we will ever go back to
saying it's undefined.


John

I sold a few thousand multichannel energy loggers that utilities and
researchers used for end-use load studies. One thing we instrumented
was a chain of fast-food restaurants. Their deep-fat friers are
resistive heaters with zero-crossing burst-mode triac temperature
controllers. When the triac is on, PF is obviously 1. When it's off,
there's no load so pf is undefined. So, what is net power factor? A
number of "experts" could never decide on a definition.

John

 

[The Phantom]

On Fri, 04 Jan 2008 07:08:54 -0800, John Larkin

I sold a few thousand multichannel energy loggers that utilities and
researchers used for end-use load studies. One thing we instrumented
was a chain of fast-food restaurants. Their deep-fat friers are
resistive heaters with zero-crossing burst-mode triac temperature
controllers. When the triac is on, PF is obviously 1. When it's off,
there's no load so pf is undefined. So, what is net power factor? A
number of "experts" could never decide on a definition.

John

Remember this thread from 2004? You were a participant and you mentioned
that same load:
http://groups.google.com/group/sci.electronics.basics/tree/browse_frm/thread/6382c203016278ec/71654e97add1ae5c?hl=en&rnum=21&_done=%2Fgroup%2Fsci.electronics.basics%2Fbrowse_frm%2Fthread%2F6382c203016278ec%2F68914b13f8fa99d2%3Fhl%3Den%26lnk%3Dst%26q%3D%26#doc_f16b87eafa50bca4

I suppose one could ask, "What is the net power factor if I turn on an
incandescent light bulb for 12 hours of the day, and leave it off for the
other 12 hours?"

Some other interesting questions are "How many angels CAN dance on the head
of a pin?" and "Just exactly what is the value of zero divided by zero?"

How about this for a method of MEASURING the net PF in any situation. A
home (or business) already has an accumulating power meter, AKA a
kilowatthour meter. Add another meter, an accumulating volt-ampere meter.
Reset both meters to zero and then let them do their thing for a period of
time for which it is wished to know the net PF. At the end of the time,
take the ratio kilowatthours/kilovoltamperehours. That is the net PF for
that interval of time. The meters won't be befuddled by the time when the
power is off.

The power company would find that they never had to supply your deep-fat
heaters with any VARs, so they would think your net PF was zero. The time
when the PF is undefined is irrelevant and its contribution to the net PF
has measure zero in the sense of Lebesgue integration, so don't even
consider it. That's my "expert" opinion.

 

[Paul E. Schoen]

"redbelly" <redbelly98@yahoo.com> wrote in message
news:aeae9a31-d820-4d8d-a30d-2a4e5f15c554@f3g2000hsg.googlegroups.com...

You can beat the 1W resolution by doing a kWh measurement. At 2W,
you'd need to wait about a day with the Kill A Watt to get to 0.05
kWh, from which you'd calculate the power to +/-20% accuracy
(resolution of the Kill A Watt is 0.01 kWh).

Of course, the OP must be willing to go one day without turning on the
TV.

This may not really help, if the basic resolution and zero offset errors
compromise accuracy. The integral of a large number of bad readings will
just give a bigger bad reading. If the system is analog, multiplier
circuits are often non-linear at low levels and have offsets. A digital
multiplier is limited by the resolution of the A/D components. When you are
operating at two bits, you can have an error of -50% or +33%.

An alternative way to measure the actual power consumption would be to
enclose the TV in a well-sealed styrofoam box, and measure the temperature
rise above ambient to a point where it reaches a steady state. Then repeat
the test with a resistor, and adjust the power until it produces the same
temperature.

This will work with any waveform. But perhaps if there is an RF component
that is radiating energy through the enclosure, it may draw more power than
is measured by heat generation. The heat is simply wasted energy, assuming
you don't want a heater. Along the same line of reasoning, if you use
electric heat, and it is being used, the TV does not add anything to your
utility bill. It may even reduce it if you have the TV under the thermostat
and you don't change the setting.

Paul

 

[John Larkin]

On Fri, 4 Jan 2008 13:33:53 -0500, "Paul E. Schoen" <pstech@smart.net>
wrote:

"redbelly" <redbelly98@yahoo.com> wrote in message
news:aeae9a31-d820-4d8d-a30d-2a4e5f15c554@f3g2000hsg.googlegroups.com...

You can beat the 1W resolution by doing a kWh measurement. At 2W,
you'd need to wait about a day with the Kill A Watt to get to 0.05
kWh, from which you'd calculate the power to +/-20% accuracy
(resolution of the Kill A Watt is 0.01 kWh).

Of course, the OP must be willing to go one day without turning on the
TV.

This may not really help, if the basic resolution and zero offset errors
compromise accuracy. The integral of a large number of bad readings will
just give a bigger bad reading. If the system is analog, multiplier
circuits are often non-linear at low levels and have offsets. A digital
multiplier is limited by the resolution of the A/D components. When you are
operating at two bits, you can have an error of -50% or +33%.

Ac energy measurement is interesting. If you noise-dither the adc and
do the math right, you can get accuracy far, far below 1 adc lsb.

John

 

[Paul E. Schoen]

"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in message
news:k3fvn3p09t6vo52m08anp8jlllf050hdbh@4ax.com...

On Fri, 4 Jan 2008 13:33:53 -0500, "Paul E. Schoen" <pstech@smart.net
wrote:


"redbelly" <redbelly98@yahoo.com> wrote in message
news:aeae9a31-d820-4d8d-a30d-2a4e5f15c554@f3g2000hsg.googlegroups.com...

You can beat the 1W resolution by doing a kWh measurement. At 2W,
you'd need to wait about a day with the Kill A Watt to get to 0.05
kWh, from which you'd calculate the power to +/-20% accuracy
(resolution of the Kill A Watt is 0.01 kWh).

Of course, the OP must be willing to go one day without turning on the
TV.

This may not really help, if the basic resolution and zero offset errors
compromise accuracy. The integral of a large number of bad readings will
just give a bigger bad reading. If the system is analog, multiplier
circuits are often non-linear at low levels and have offsets. A digital
multiplier is limited by the resolution of the A/D components. When you
are
operating at two bits, you can have an error of -50% or +33%.

Ac energy measurement is interesting. If you noise-dither the adc and
do the math right, you can get accuracy far, far below 1 adc lsb.

Please elaborate on that. I can understand that a multiplying A/D might be
linear enough at low levels to produce an output that could be integrated
over time to give a better accuracy than 1 LSB, but a system that uses
digital processing for both current and voltage needs more bits to produce
a meaningful output. This is even more so for complex waveforms, or where
the voltage is AC and the current is DC. But the main practical limiting
factors at low levels are nonlinearity, phase shift, distortion, and
offset, especially in the current measuring components.

Paul

 

[John Larkin]

On Sat, 5 Jan 2008 18:45:06 -0500, "Paul E. Schoen" <pstech@smart.net>
wrote:

"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in message
news:k3fvn3p09t6vo52m08anp8jlllf050hdbh@4ax.com...
On Fri, 4 Jan 2008 13:33:53 -0500, "Paul E. Schoen" <pstech@smart.net
wrote:


"redbelly" <redbelly98@yahoo.com> wrote in message
news:aeae9a31-d820-4d8d-a30d-2a4e5f15c554@f3g2000hsg.googlegroups.com...

You can beat the 1W resolution by doing a kWh measurement. At 2W,
you'd need to wait about a day with the Kill A Watt to get to 0.05
kWh, from which you'd calculate the power to +/-20% accuracy
(resolution of the Kill A Watt is 0.01 kWh).

Of course, the OP must be willing to go one day without turning on the
TV.

This may not really help, if the basic resolution and zero offset errors
compromise accuracy. The integral of a large number of bad readings will
just give a bigger bad reading. If the system is analog, multiplier
circuits are often non-linear at low levels and have offsets. A digital
multiplier is limited by the resolution of the A/D components. When you
are
operating at two bits, you can have an error of -50% or +33%.

Ac energy measurement is interesting. If you noise-dither the adc and
do the math right, you can get accuracy far, far below 1 adc lsb.

Please elaborate on that. I can understand that a multiplying A/D might be
linear enough at low levels to produce an output that could be integrated
over time to give a better accuracy than 1 LSB, but a system that uses
digital processing for both current and voltage needs more bits to produce
a meaningful output. This is even more so for complex waveforms, or where
the voltage is AC and the current is DC. But the main practical limiting
factors at low levels are nonlinearity, phase shift, distortion, and
offset, especially in the current measuring components.

Paul

Sample the voltage and current waveforms with, say, an 8 or 10-bit
ADC, preferably simultaneously, but at least not many microseconds
apart. The voltage waveform is usually pretty constant in amplitude in
regular AC systems, so adc resolution is not a problem there. The
current waveform can have a huge dynamic range, 10,000:1 or so for a
metering-quality measurement. So add a few lsb's of noise to the
current signal before digitizing it; that smears out the quantization
errors. Now software autozero the current samples to take out any dc
offsets, to a fraction of an lsb, say, 16 bits or so. Now multiply the
zeroed current sample with the paired voltage sample and average the
product points to get power, integrate to get energy. The statistics
are great if you do the math properly. A few watts resolution out of,
say, 20 kw full-scale is possible.

Actually, DC offset in the current signal washes out when it's
multiplied by the voltage sine wave samples and averaged to make
power, as long as the voltage signals doesn't have offset too.
Software autozering both the voltage and current data allow you to use
unipolar, unsigned adc's and not wory about residual (or huge) dc
offsets in the signal conditioning or the adc itself.

The worst low-power error will be crosstalk between the voltage and
current signals, which can happen magnetically, or in the adc mux, or
any number of other interesting places. Any crosstalk or correlated
noise does produce a power offset, some of which can be software
fudged out.

John