measuring ac wattage from ampere

[John Fields]

On Sun, 24 Oct 2010 18:24:20 +0200, "F. Bertolazzi"
<TOGLIeset@MAIUSCOLEtdd.it> wrote:

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.

---
I hadn't noticed it.
---

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.

---
Hardly, Since what Mr. Allison was doing was outlining the sequence of
events to get from volt-amperes in to watts out (or by reading it
backwards to get from watts out to VA in) without going through a
detailed analysis, which would have involved making actual
measurements or using real data, which wasn't available.
---

Showing off?

How would you call giving detalied and precise answer to a question that
has not been posed?

---
A question was posed, and what I did was to apprise whit3rd of his
errors regarding the use of an ammeter and the inability of being able
to get adequate voltage data from the mains using a voltmeter, and
then provide him with a clearly worked out, mathematically precise
explanation of how the differences between real and imaginary power
are established.


Here's the exchange, from an earlier post:

whit3rd:
"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)."

Me:
"No, you can't.

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

whit3rd:
"Huh? You can only measure Volt-Amperes that way (if you have
the Volt value from the utility standards)."

I then continued with the explanation.

BTW, notice the "Huh?" two sentences up?

In American English, this: "?" (without the quotes) indicates a
question.

As for the showing off, let's say I was. So what?
---

Without knowing the required precision the answer could have been even
"It's a laptop poer supply? From 50 to 100W".

---
Ah, but notice that in my example I didn't use a laptop power supply,
I used a resistor and a capacitor, which was perfectly adequate for
the task at hand.
---

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?

---
What caused me to call you an asshole.
---

In any case, I don't have the disease

Think what you like. ;-)

---
I sure will, with or without your permission, but tell me, why did you
snip: "but I might be a carrier by association"?
---

Sorry for this exchange. I just meant to let you know what's the impression
a newcomer has of this NG.

---
_One_ newcomer.
---

And was not aimed at you. Peace.

---
OK.

---
JF

 

[John Fields]

On Sat, 23 Oct 2010 14:38:59 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Sat, 23 Oct 2010 12:07:58 -0700 (PDT), whit3rd <whit3rd@gmail.com
wrote:

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.

---
After reading this thread through a couple of times, (after a little
prodding from the new Italian influence) it appears that I misread
what you were saying, and that you were right and I was wrong with
reference to the upper limit of the power measurement.

Of course, with a power factor of 1, EI will yield watts.

However, if the power factor is less than 1, EI will yield
volt-amperes, and the real power will always be less than EI.


---
JF

 

[F. Bertolazzi]

John Fields:

Honoured. Truly.

 

On Sat, 23 Oct 2010 21:12:42 -0700 (PDT), suraj joneja
<suraj.joneja@gmail.com> 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.

So get a 240 volt version:
http://www.maplin.co.uk/Module.aspx?moduleno=38343

Thanks.

 

[suraj joneja]

Thanks!

 

[Jasen Betts]

On 2010-10-23, suraj joneja <suraj.joneja@gmail.com> wrote:

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.

Turn of every other appliance and observe the watt-hour meter on your supply.
it have be written on it how many pulses of the LED (or turns of the rotor) is
one watt hour, so count them for a timed duration and do some arithmetic.

This method is somewhat inconvenient if frequent measurements are needed
but for a one-off on a tight budget it seems ideal.

--
ɹǝpun uʍop ɯoɹɟ sƃuıʇǝǝɹ⅁

 

[Jasen Betts]

On 2010-10-23, Sjouke Burry <burrynulnulfour@ppllaanneett.nnll> wrote:

suraj joneja wrote:
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? 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?

Many thanks.
Suraj.
By measuring volage,current and cosine.

That will get you the wrong answer most of the time, especially when
electronics are involved.

http://en.wikipedia.org/wiki/Power_factor#Non-linear_loads

--
ɹǝpun uʍop ɯoɹɟ sƃuıʇǝǝɹ⅁

 

[suraj joneja]

Thanks,

I tried the power company meter method, I have rather primitive meter
which has a rotating disc. It had a resolution of 100 spins per kwatt
hour. So I connected a 100 w bulb and It spun once in 6 minutes.
Surely the maths adds up. 10 spins for 100w hr. ..so it should spin
once in six mins for a 100 watt load.

This way I could profile a lot of things.

Thanks to all.

Suraj
Www.indiacall.tk