measuring ac wattage from ampere

[Rich Grise]

On Sat, 23 Oct 2010 14:38:59 -0500, John Fields wrote:

On Sat, 23 Oct 2010 12:07:58 -0700 (PDT), whit3rd <whit3rd@gmail.com
On Oct 23, 10:50 am, suraj joneja <suraj.jon...@gmail.com> wrote:
On Oct 23, 9:58 pm, "F. Bertolazzi" <TOGLIe...@MAIUSCOLEtdd.it> wrote:

Now I believe one can calculate the wattage VA* PF.

I checked on youtube and saW "kill a watt". Just the thing I need but
its expensive and in the USA.

What if I get an in range ammeter?

As you know by now, the ammeter measures one of three items, A.
Your AC mains might be constant enough to just plug in the V number,
for a rough calculation. The 'power factor', though, is only known
to be 'the cosine of...' and can be any number in the range of
+1, -1.

A negative cosine?

Tricky!

With ammeter only, you can establish an upper limit
of how much power is drawn (the upper limit is achieved if the
power factor is +1).

No, you can't.

With only an ammeter and voltmeter you never know whether you're
measuring Volt-Amperes or watts.

Oh, you're definitely measuring Volt-Amperes. How close that is to

actual watts is when you need the PF correction, which so far in this
case is unknown.

Doesn't the "Kill-a-Watt" have a provision for that? Do they have the
"Kill-a-Watt" or equivalent in India?

Cheers!
Rich

 

[Rich Grise]

On Sat, 23 Oct 2010 15:53:57 -0400, Jamie wrote:

Two Atoms walking down the street talking;

A1. I think I lost an Electron?

A2. Are you sure ?

A1. Yes, I am positive..




Yes, I know, it's a bad one!

The kiddies should love it! ;-)

Cheers!
Rich

 

[John Fields]

On Sat, 23 Oct 2010 13:03:56 -0700 (PDT), whit3rd <whit3rd@gmail.com>
wrote:

On Oct 23, 12:38 pm, John Fields <jfie...@austininstruments.com
wrote:
On Sat, 23 Oct 2010 12:07:58 -0700 (PDT), whit3rd <whit...@gmail.com
wrote:

With ammeter only, you can establish an upper limit
of how much power is drawn (the upper limit is achieved if the
power factor is +1).

No, you can't.

With only an ammeter and voltmeter you never know whether you're
measuring Volt-Amperes or watts.

Huh? You can only measure Volt-Amperes that way (if you have
the Volt value from the utility standards).

---
I don't know what that means.

If you measure RMS volts and RMS amperes and multiply them, the
product will be watts if current and voltage are in phase, or
volt-amperes if they're not.

For example, let's say you measured the voltage and current going into
a black box and found them to be 120V and 8.5A.

The product would yield 1020VA or 1020 watts, maybe, but not knowing
what was in the box you couldn't be sure.

So you look in the box and you find that there's a 10 ohm resistor in
there in series with a 265ľF capacitor.

First, let's find the reactance of the cap:


1 1
Xc = --------- = ----------------------- = 10 ohms
2pi f C 6.28 * 60Hz *2.65e-2F


Next, the impedance of the circuit:


Z = sqrt (R˛ + Xc˛) = sqrt (10˛ + 10˛) = 14.14 ohms


The current through the resistor and capacitor:


E 120V
I = --- = ------- = 8.51 amperes
Z 14.1R


The all-important phase angle:


Xc -10R
tan phi = ---- = ------ = -1
R 10R


phi = arctan -1 = -45°


And, finally, the power factor:


PF = cos phi = 0.707


Now we can calculate the real power dissipated by the components in
the box:


P = EI cos phi = 120V * 8.5A * 0.707 = 721 watts


So, had there been only a 14.1 ohm resistor in the box, we would have
measured 120V across it and 8.5A through it and it would have been
dissipating 1020 watts, while with the RC in the box we still measured
120V across the series combination and 8.5A through it, but it was
only dissipating 721 watts.

So, you see, if you measure a voltage across something and a current
through it, you can't be sure whether you're measuring VA or watts
unless you know something about the phase relationship between the
current and the voltage.
---

The Watts value cannot
exceed the VA value, can it?

---
It cannot.

---
JF

 

[John Fields]

On Sat, 23 Oct 2010 15:16:57 -0700, Rich Grise <richgrise@example.net>
wrote:

On Sat, 23 Oct 2010 14:38:59 -0500, John Fields wrote:
On Sat, 23 Oct 2010 12:07:58 -0700 (PDT), whit3rd <whit3rd@gmail.com
On Oct 23, 10:50 am, suraj joneja <suraj.jon...@gmail.com> wrote:
On Oct 23, 9:58 pm, "F. Bertolazzi" <TOGLIe...@MAIUSCOLEtdd.it> wrote:

Now I believe one can calculate the wattage VA* PF.

I checked on youtube and saW "kill a watt". Just the thing I need but
its expensive and in the USA.

What if I get an in range ammeter?

As you know by now, the ammeter measures one of three items, A.
Your AC mains might be constant enough to just plug in the V number,
for a rough calculation. The 'power factor', though, is only known
to be 'the cosine of...' and can be any number in the range of
+1, -1.

A negative cosine?

Tricky!

With ammeter only, you can establish an upper limit
of how much power is drawn (the upper limit is achieved if the
power factor is +1).

No, you can't.

With only an ammeter and voltmeter you never know whether you're
measuring Volt-Amperes or watts.

Oh, you're definitely measuring Volt-Amperes. How close that is to
actual watts is when you need the PF correction, which so far in this
case is unknown.

---
And, since it's unknown, the PF would be one if the load were
resistive, which it could easily be if it were an incandescent lamp
load, a heater...

The point is that by merely measuring voltage and current there's
nothing "definite" about the measurement in terms of how much power
the load's using.


---
---
JF

 

[amdx]

"suraj joneja" <suraj.joneja@gmail.com> wrote in message
news:4dedb226-fbdc-4cfc-8f8f-be44548362eb@x7g2000prj.googlegroups.com...
On Oct 23, 9:58 pm, "F. Bertolazzi" <TOGLIe...@MAIUSCOLEtdd.it> wrote:

suraj joneja:

Now I believe one can calculate the wattage VA* PF. So I believe it
will come out to be 220w . Ok. But I take the point that that is the
peak onsumption. Ok. Now to add more to the confusion is what is
written on the power supply that the input voltage range is 110-250v !

And there is also written that it draws 1A? I don't think so.

Ok let me tell you that my real objective is to measure the real watts
being drawn by the device (and many other devices) . A wattmeter will
be best?

Sure. But it will cost you a bit.

I don't have an AC ammeter but it will be easier for me to
buy one than a wattmeter.

And cheaper.

Can you please look at this product listing:
http://cgi.ebay.in/Digital-Clamp-Meter-1000Amps-AC-Large-LCD-Model-DT...

Probably your laptop consumes around 0,1A, so there's no way you can use
that.

One more ques, with this clampmeter can I measure the current without
breaking the circuit?

Yes, bur it measures currents 10.000 times bigger than the one you are
interested in.

Thanks. Please don't doubt! I checked again! Its the power supply for
a large screen toshiba laptop. But I guess that the one amp rating
must be for 110v?

I checked on youtube and saW "kill a watt". Just the thing I need but
its expensive and in the USA.

Hi Suraj,
I know there is a currency difference, but the "kill a watt" always seemed
to be very in-expensive,
I suspect the cost of shipping is probably more then the cost of the unit.
What did you see for a price of
the "kill a watt"
MikeK

 

[Rich Grise]

On Sat, 23 Oct 2010 18:32:02 -0500, amdx wrote:

"suraj joneja" <suraj.joneja@gmail.com> wrote in message
On Oct 23, 9:58 pm, "F. Bertolazzi" <TOGLIe...@MAIUSCOLEtdd.it> wrote:
suraj joneja:

Now I believe one can calculate the wattage VA* PF. So I believe it
will come out to be 220w . Ok. But I take the point that that is the
peak onsumption. Ok. Now to add more to the confusion is what is
written on the power supply that the input voltage range is 110-250v !

And there is also written that it draws 1A? I don't think so.

Ok let me tell you that my real objective is to measure the real watts
being drawn by the device (and many other devices) . A wattmeter will
be best?

Sure. But it will cost you a bit.

I don't have an AC ammeter but it will be easier for me to
buy one than a wattmeter.

And cheaper.

Can you please look at this product listing:
http://cgi.ebay.in/Digital-Clamp-Meter-1000Amps-AC-Large-LCD-Model-DT...

Probably your laptop consumes around 0,1A, so there's no way you can use
that.

One more ques, with this clampmeter can I measure the current without
breaking the circuit?

Yes, bur it measures currents 10.000 times bigger than the one you are
interested in.

Thanks. Please don't doubt! I checked again! Its the power supply for
a large screen toshiba laptop. But I guess that the one amp rating
must be for 110v?

I checked on youtube and saW "kill a watt". Just the thing I need but
its expensive and in the USA.

Hi Suraj,
I know there is a currency difference, but the "kill a watt" always seemed
to be very in-expensive,
I suspect the cost of shipping is probably more then the cost of the unit.
What did you see for a price of
the "kill a watt"

I wonder what UPS or FedEx or USPS or DHL or whatever would get to ship
a 5 ounce product to India?

There's a kill-a-watt on the internet for like $16.95; if that's a month's
wage, then clearly it's not feasible. But I don't know the exchange rate
or what pay is like over there or anything.

Cheers!
Rich

 

[Phil Allison]

"John Fields"

No, the one ampere is undoubtedly what the output of the supply is
capable of delivering continuously,

** Nonsense.

The 1 amp rating applies to the AC input current.

It will be the maximum, rms value too - which occurs under full load with
the lowest AC supply voltage of 100 volts.

So the max VA rating is 100VA.

The PF is likely to be about 0.6.

The efficiency is likely to be about 75% at full load.

Makes max the DC output circa 45 watts.


..... Phil

 

[Phantom]

On Sat, 23 Oct 2010 10:50:57 -0700 (PDT), suraj joneja <suraj.joneja@gmail.com>
wrote:

I checked on youtube and saW "kill a watt". Just the thing I need but
its expensive and in the USA.

What if I get an in range ammeter?

Thanks.

The Kill-a-Watt is what you need.

It is not only a wattmeter.

It can measure current, voltage, watts, volt-amperes, PF and line frequency, and
can acccumulate watt-hours.

And, it's probably less expensive than all but the very cheapest DVMs.

 

[Phantom]

On Sun, 24 Oct 2010 12:16:30 +1100, "Phil Allison" <phil_a@tpg.com.au> wrote:

"John Fields"


No, the one ampere is undoubtedly what the output of the supply is
capable of delivering continuously,

** Nonsense.

The 1 amp rating applies to the AC input current.

It will be the maximum, rms value too - which occurs under full load with
the lowest AC supply voltage of 100 volts.

So the max VA rating is 100VA.

The PF is likely to be about 0.6.

The efficiency is likely to be about 75% at full load.

Makes max the DC output circa 45 watts.


.... Phil

Actual measurements made with a Kill-a-Watt on a Compaq wall wart.

Input 100-240VAC, max current 1.7A, 65 watts

With a resistor load:

Input voltage 120.0VAC
Input current .92A
Input watts 62
Input PF .58

Output volts 18.1
Output amps 2.94
Efficiency 85.8%

 

[suraj joneja]

This site states that the PF of modern computers with smps is close to
unity: http://www.generatorguide.net/watt-acpower.html

Now I checked the output of the toshiba power supply in question and
it says the output is 3.4 amps at 19 v DC. That makes it almost 65
watts. Now if assume the efficiency to be 75% then the pf comes out to
be almost .87.

Now one more thing since I know the peak output wattage is 65
therefore I can measure the input wattage by using the ballpark
efficiency? I can do the same for my other power supplies also..this
solves my problem for power supplies which have a mentioned dc
amperage and voltage.DC is just VI right?

Also the kill a watt does only 125 v max so it won't work here.

Thanks.

 

[Phil Allison]

"suraj joneja"

This site states that the PF of modern computers with smps is close to
unity: http://www.generatorguide.net/watt-acpower.html

** Can you read at all ???

Now I checked the output of the toshiba power supply in question and
it says the output is 3.4 amps at 19 v DC. That makes it almost 65
watts. Now if assume the efficiency to be 75% then the pf comes out to
be almost .87.

** Giant HUH ????

Wot absolute drivel.

Now one more thing since I know the peak output wattage is 65
therefore I can measure the input wattage by using the ballpark
efficiency? I can do the same for my other power supplies also..this
solves my problem for power supplies which have a mentioned dc
amperage and voltage.

** Only an actual measurement carries any meaning.

God you are a stupid shit.


...... Phil

 

[John Fields]

On Sun, 24 Oct 2010 12:16:30 +1100, "Phil Allison" <phil_a@tpg.com.au>
wrote:

"John Fields"


No, the one ampere is undoubtedly what the output of the supply is
capable of delivering continuously,

** Nonsense.

The 1 amp rating applies to the AC input current.

It will be the maximum, rms value too - which occurs under full load with
the lowest AC supply voltage of 100 volts.

So the max VA rating is 100VA.

The PF is likely to be about 0.6.

The efficiency is likely to be about 75% at full load.

Makes max the DC output circa 45 watts.

---
Makes sense, especially in the light of the OP's latest post re the
power supply input specs.

Thanks,

---
JF

 

[F. Bertolazzi]

suraj joneja:

I am in india where the AC mains supply is 220 v 50hz.

Where that kill-a-watt device is unavailable, unless you order it from the
States, but the courier is very likely to charge you $80 for the delivery.

Now I have a device which has a rating of 1 amp.

That is a SMPS for a laptop. You should have stated that.

How do measure its wattage?

With which precision? I'm amazed that none of the hair-splitters here have
asked you this question.

On the net I read that one has to consider the power factor which is the
cosine of the angle between the current and the voltage. How do I do
that?

If you place a simple AC ammeter in series with your power supply you will
get a reading that, multiplied by 220, will give you the power, probably
underestimated by less than 10%



Which sort of NG is this? One asks for help and all he receives are
lectures he will not understand (it's clear by the way he posed the
question), insults and solutions he cannot afford.

Are you here just to vent your frustrations?

 

[John Fields]

On Sun, 24 Oct 2010 15:21:50 +0200, "F. Bertolazzi"
<TOGLIeset@MAIUSCOLEtdd.it> wrote:

suraj joneja:

I am in india where the AC mains supply is 220 v 50hz.

Where that kill-a-watt device is unavailable, unless you order it from the
States, but the courier is very likely to charge you $80 for the delivery.

Now I have a device which has a rating of 1 amp.

That is a SMPS for a laptop. You should have stated that.

How do measure its wattage?

With which precision? I'm amazed that none of the hair-splitters here have
asked you this question.

On the net I read that one has to consider the power factor which is the
cosine of the angle between the current and the voltage. How do I do
that?

If you place a simple AC ammeter in series with your power supply you will
get a reading that, multiplied by 220, will give you the power, probably
underestimated by less than 10%

---
It's impossible to _underestimate_ that way since if the load is
resistive the product of voltage and current will be watts, while if
the load is complex the product will be volt-amperes and will always
be _higher_ than the true power dissipated by the load.
---

Which sort of NG is this? One asks for help and all he receives are
lectures he will not understand (it's clear by the way he posed the
question), insults and solutions he cannot afford.

---
Actually, with the exception of your "solution", the OP was given
quite a bit of good help.
---

Are you here just to vent your frustrations?

---
Geez, just four days on the grop and you've already turned into an
asshole.

Must be some kind of record...


---
JF

 

[Jamie]

suraj joneja wrote:

This site states that the PF of modern computers with smps is close to
unity: http://www.generatorguide.net/watt-acpower.html

Now I checked the output of the toshiba power supply in question and
it says the output is 3.4 amps at 19 v DC. That makes it almost 65
watts. Now if assume the efficiency to be 75% then the pf comes out to
be almost .87.

Now one more thing since I know the peak output wattage is 65
therefore I can measure the input wattage by using the ballpark
efficiency? I can do the same for my other power supplies also..this
solves my problem for power supplies which have a mentioned dc
amperage and voltage.DC is just VI right?

Also the kill a watt does only 125 v max so it won't work here.

Thanks.
Efficiency and PF is not the same thing in this case..

 

[F. Bertolazzi]

John Fields:

It's impossible to _underestimate_ that way since if the load is
resistive the product of voltage and current will be watts, while if
the load is complex the product will be volt-amperes and will always
be _higher_ than the true power dissipated by the load.

Right. Sorry for the inversion.

Actually, with the exception of your "solution", the OP was given
quite a bit of good help.

Excellent help, but absolutely worthless for him.
It would rather call it a show off.

Geez, just four days on the grop and you've already turned into an
asshole.

Do you think you're that infectious?

 

[John Fields]

On Sun, 24 Oct 2010 16:56:55 +0200, "F. Bertolazzi"
<TOGLIeset@MAIUSCOLEtdd.it> wrote:

John Fields:

It's impossible to _underestimate_ that way since if the load is
resistive the product of voltage and current will be watts, while if
the load is complex the product will be volt-amperes and will always
be _higher_ than the true power dissipated by the load.

Right. Sorry for the inversion.

Actually, with the exception of your "solution", the OP was given
quite a bit of good help.

Excellent help, but absolutely worthless for him.

---
It doesn't always have to be just for the OP, but what was worthwhile
for the OP was the excellent suggestion to use a kill-a-watt, (whether
he can readily get one or not is a differennt issue) and this, from
from Mr. Allison, is first-rate and right on the money:

"The 1 amp rating applies to the AC input current.

It will be the maximum, rms value too - which occurs under full load
with the lowest AC supply voltage of 100 volts.

So the max VA rating is 100VA.

The PF is likely to be about 0.6.

The efficiency is likely to be about 75% at full load.

Makes max the DC output circa 45 watts."
---

It would rather call it a show off.

---
If you're talking about my article explaining the difference between
real and imaginary power, it wasn't targeted specifically for the OP
but it was on topic and was helpful, I'm sure, for at least _one_ of
the posters.

Showing off?

Think what you like.
---

Geez, just four days on the group and you've already turned into an
asshole.

Do you think you're that infectious?

---
It would have been nice if you hadn't taken my statement out of
context, but oh, well...

In any case, I don't have the disease, but I might be a carrier by
association.

---
JF

 

[F. Bertolazzi]

John Fields:

It doesn't always have to be just for the OP, but what was worthwhile
for the OP was the excellent suggestion to use a kill-a-watt, (whether
he can readily get one or not is a differennt issue)

Excuse my smile.

and this, from
from Mr. Allison, is first-rate and right on the money:

Which money? Given the fact that nobody cared to ask wich precision was
needed, that could have been perfect or dead wrong.

Showing off?

How would you call giving detalied and precise answer to a question that
has not been posed?
Without knowing the required precision the answer could have been even
"It's a laptop poer supply? From 50 to 100W".

Think what you like.

I sure will, with or without your permission.

It would have been nice if you hadn't taken my statement out of
context, but oh, well...

I call it good quoting. Tell me, what was missing of the original context?

In any case, I don't have the disease

Think what you like. ;-)

Sorry for this exchange. I just meant to let you know what's the impression
a newcomer has of this NG. And was not aimed at you. Peace.

 

[F. Bertolazzi]

suraj joneja:

Hi

I am in india where the AC mains supply is 220 v 50hz. Now I have a
device which has a rating of 1 amp. How do measure its wattage?

Use the power meter you already have, courtesy of your electrical company.

Turn off the lights, unplug the refrigerator and all the devices that may
consume power, plug in your laptop, and take two readings, one minute
apart, from the utility meter. Multiply by 60 and you have the consumption.

Beat this money, John.

Unless he still has an electromechanical counter, but, as we know,
availability is not a problem. ;-)

 

[John Fields]

On Sun, 24 Oct 2010 18:33:14 +0200, "F. Bertolazzi"
<TOGLIeset@MAIUSCOLEtdd.it> wrote:

suraj joneja:

Hi

I am in india where the AC mains supply is 220 v 50hz. Now I have a
device which has a rating of 1 amp. How do measure its wattage?

Use the power meter you already have, courtesy of your electrical company.

---
I believe that's already been suggested.
---

Turn off the lights, unplug the refrigerator and all the devices that may
consume power, plug in your laptop, and take two readings, one minute
apart, from the utility meter. Multiply by 60 and you have the consumption.

---
That's difficult to do with any degree of precision since the LSD on
most meters is usually in units of 1kWh.

Some meters do go down to 100Wh however, but it would take an hour
just to traverse one digit if it was a 100 watt (input) supply.

However, all is not lost, as explained here:

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


Look for:

3600*Kh
P = ---------
t

There are also meters which output pulses, but I fear going that way,
if that's what the OP has, would be overwhelming.
---

Beat this money, John.

Unless he still has an electromechanical counter, but, as we know,
availability is not a problem. ;-)

---
JF