What is the power of an incandescent lamp at 200v 5A?

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What is the power of an incandescent lamp at 200v 5A?

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rattus said:
Not quite, the area under the v*i curve represents energy. You must divide by time to obtain average, real power.
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You're right. Fine, the area under the curve of |V|*|I| divided by whatever the integration period is in seconds gives Pavg in watts for whatever the duration of integration period used.
 
rattus said:
Mivey, it irritates my tender psyche to hear expressions such as,

"The watts is equal to the volts times the amperes."
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What you call mnemonics I was reading as a typo or the use of the label. The label W would use "carries" while the rate it stands for (watts) would not. "The numerator carries...", "The denominator or product carries", etc.

It did not occur to me that anyone reading this phrase "...W/VA which is power factor" would say "W" stood for work. Why would you be mixing units with abbreviations?
 
Jack, welcome to the forum! :)

jacktar said:
This is basic Ohm's law. E(I)=P
200 (5)=1KW
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That's the obvious answer, and why it is or isn't correct is the whole point of this thread.

You'll find out that questions of mere fact have short threads. The rest, well . . .
 
mivey said:
What you call mnemonics I was reading as a typo or the use of the label. The label W would use "carries" while the rate it stands for (watts) would not. "The numerator carries...", "The denominator or product carries", etc.

It did not occur to me that anyone reading this phrase "...W/VA which is power factor" would say "W" stood for work. Why would you be mixing units with abbreviations?
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Good question Mivey. In this imperfect world, VA is used both as a quantity and as a unit, e.g., "VA rating" to a purist should be "apparent power rating", but let's not argue that pointless point.

My point was though that the average of the vi product in this case is real power into the device, not apparent power. Divide that by real power our to obtain efficiency.
 
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Before anyone skewers me, efficiency is the ratio of power out to power in.
 
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rattus said:
No argument there, but how do you calculate apparent power?
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Apparent power is Vrms * Irms
Instead of multiplying them simultaneously and integrating the product, each RMS values are calculated separately, then multiplied together.

It's literally Vrms * Irms
 
Yes, but.

Yes, but.

Electric-Light said:
Apparent power is Vrms * Irms
Instead of multiplying them simultaneously and integrating the product, the RMS values are calculated separately, then multiplied together.

It's literally Vrms * Irms
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No argument here either, but the average of v(t)*i(t) is average real power. Compute real power out and real power in this way; then take the ratio to obtain efficiency. The efficiency of a fluorescent tube and ballast for example will be something less than 100% as you know.

In one case we have the ratio of real to apparent power.

In the other we have the ratio of real to real power.
 
Electric-Light said:
That's the only way you can get away with calculating the power from only one measurement. If you can't assume a perfectly linear load with no phase shift(an ideal resistor with no reactance or capacitance) , you can't get away from having to simultaneously collect V*I and integrate the product.
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Electric,
Rattus was probably responding to me.

So, Electric,
what is wrong with reading the RMS of both V and I and logging it.
I have designed instruments used for that very purpose.

Collect the data at very small intervals, log the data, calculate the Power at each point, then integrate for a final and singular answer,
if that is the target.

But, . . . This is not what electricians do for a living.

I am still wondering this :
Is your original question intended for blue collar electricians,
who can assume steady state sine waves,
or this a question for engineers who may assume very little ?

When you are cornered, you broaden the playing field, to Engineers.
When you get a broader answer, then you narrow the focus to Electricians.

Anyway, I like the game,
and it is fun to see what the various experts are saying about the topic, from their own perspective.
I like pic-nics. :)
 
100906-2110 EST

glene77is:

When you say measure the RMS value for a very short time it appears to me you are saying measure the instantaneous value at a point in time. Certainly one can do an RMS measurement for a very short time such that its value is equal to the instantaneous value. However, most readers will read RMS value to be a measurement over an integral number of cycles of a periodic waveform or if starting and stopping the the measurement at random times, then over enough cycles that the start and stop times do not contribute substantial error.

For a random signal with no specific period, then the averaging time has to be long enough to result in a stationary process.

.
 
I hate to see this thread die, so I have a little twist...

I pulled a 40 watt 120v incandescent lamp off the shelf today and read through it with an ohmmeter.

how many ohms would you expect to read?
 
100909-2119 EST

realolman:

A ballpark figure. From memory a 100 W bulb filament at about 70 deg F is about 10 ohms. This means no current thru the filament and the bulb is at room temperature. It might be 9 or 11 or thereabouts.

At 120 V and with ambient room temperature about 70 deg F the nominal resistance is about 14,400/100 = 144 ohms.

The hot to cold ratio is about 14 to 1.

For your 40 W bulb I expect the cold resistance to be about (14,400/40)/14 = 360/14 = 26 ohms.

Assumed above is that the filament is tungsten.

Take your bulb out in bright sunlight and measure the resistance. Next obstruct the sun light from the bulb and measure the resistance.

.
 
100909-2342 EST

Just measured a 25 W at 45.2 ohms with a Fluke 27. From my 100 W calculation above I would predict 4 * 10 or 40 ohms.
Also measured a 100 W at 9.7 ohms.

The manufacturing criteria for different wattages may be somewhat different.

.
 
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Gar,
Good point.
I was thinking about an audio waveform, wherein we would measure at a rate greater than twice the max frequency. Thinking of a Complex waveform and collecting data in small segments. There would be two (time) matching streams of data, (1) Voltage and (2) Current.
I am thinking that the RMS calculation would occur after data collection at these many points.
 
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Electric-Light said:
You're right. Fine, the area under the curve of |V|*|I| divided by whatever the integration period is in seconds gives Pavg in watts for whatever the duration of integration period used.
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Electric,

So,
if I collect two data streams (Volts and Amps) of instantanous (peak) values, and then perform the P=V*I calculations on these two data streams, producing a third data list, and then perform the RMS calculation on this list of Power points, what will I end up with.

I see that we have the RMS of the "area under the curves"
from the original two data streams (V and I).

Since the 'curve' was the result of the P=V*I,
then this becomes a statement of Power in RMS terms,
which is a single summation of the 'area under the curves' first collected.

I like this because it looks valid for a complex audio signal,
but . . .
this is not a concept that any electrician uses.
Electricians use steady state sine waves,
and that means that we must broaden the base of inquiry.
 
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