measuring ac wattage from ampere

The kiddies should love it! ;-)

Cheers! Rich

--- 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 265µF 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 (R² + Xc²) = sqrt (10² + 10²) = 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.

---

--- It cannot. :-)

--- JF

--
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.

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

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

"John Fields"

** 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

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.

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%

This site states that the PF of modern computers with smps is close to unity:

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.

"suraj joneja"

** Can you read at all ??? ** Giant HUH ????

Wot absolute drivel.

** Only an actual measurement carries any meaning.

God you are a stupid shit.

..... Phil

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

Thanks,

suraj joneja:

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.

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

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

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?

--
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.

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

John Fields:

Right. Sorry for the inversion.

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

Do you think you're that infectious?

--- 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."

---

--- 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.

---

--- 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

John Fields:

Excuse my smile.

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

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".

I sure will, with or without your permission.

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

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.

suraj joneja:

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. ;-)

--
I believe that's already been suggested.