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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Gar,
Correction.
The Source "A" feeds two coils (1) and (2).
Observations presented from a "West" perspective,
looking straight through the meter.
 
100814-20387 EST

glene77is:

For a discussion of D'Arsonval see
http://en.wikipedia.org/wiki/Galvanometer a very good reference
http://www.engineersedge.com/instrumentation/electrical_meters_measurement/darsonval_movement.htm

When you replace the permanent magnet part of the D'Arsonval meter with an electromagnet it is no more a D'Arsonval movement.

Electrodynamometer
http://en.wikipedia.org/wiki/Wattmeter this is not a very good reference.

A good discussion on meters as of 1950 is given in "Basic Electrical Measurements", by Melville B. Stout, Prentice-Hall, 1950. See Chapter 17.

I have two patents that I can say resulted directly from the training I received from Stout on bridge circuits.

The Triplett meter you are describing appears to be of the electrodynamometer type.

.
 
gar said:
100814-20387 EST

Electrodynamometer
http://en.wikipedia.org/wiki/Wattmeter
The Triplett meter you are describing appears to be of the electrodynamometer type.

.
Click to expand...

Gar,
Thank you.
The design appears to match
that of the Wiki Wattmeter ElectroDynamoMeter.

I can identify the components and functions.
The doubly-fed Current coils appear to allow for AC-DC operation.
The scaling resistors are present, and switched.
The additional circuitry (DIY) will take some more thinking.
The equations I can follow. This is comfortable. :)

You have been around,
more than any coil I've seen.
 
glene77is said:
Mivey,

Pavg <= 1kW ???

Is there such a thing as Power Average ?
Click to expand...
Yes. For what we are discussing, the rate of energy conversion.
glene77is said:
Should that be something like KWh related to five hours ?
Click to expand...
Yes. Same way a digital revenue meter works (but a much smaller time than 5 hours). Energy converted divided by a time interval. As the time interval approaches zero, we approach the value for instantaneous power.
glene77is said:
This is my comment for you :
If the bulbs filament has resistance ranging over a factor or 10,
and this ramps up in, say 1/20 second,
and we were given that the Frequency of the supplied 'AC' voltage were .1 Hz,
then might we have a power factor of .2 ?

Just ball-parking figures for a hypothetical question
just to see if it flies anywhere at all.
If it does, then I might go further in specifying real figures. :)
Click to expand...
We might have anything but, considering where this thread has gone so far, there is not enough information to answer your question. :)
glene77is said:
The point is that
(1) the frequency of the applied voltage is important,
(2) given that there is a dynamic resistance involved with the filament.

Comments: :)
Click to expand...
Of course the frequency is important. So are a myriad of other things. But where to we draw the line? If the question does not state that anything is possible, are we to assume that? Not normally. We would normally assume normal conditions. If anything is possible, then the question will never be answered because we could have any kind of goofy scenario thrown in like parallel universes, multiple dimensions, miracles, and who knows what else (not that I'm necessarily against that :D).
 
Enough Already!

Enough Already!

mivey said:
Yes. For what we are discussing, the rate of energy conversion. Yes. Same way a digital revenue meter works (but a much smaller time than 5 hours). Energy converted divided by a time interval. As the time interval approaches zero, we approach the value for instantaneous power. We might have anything but, considering where this thread has gone so far, there is not enough information to answer your question. :)
Of course the frequency is important. So are a myriad of other things. But where to we draw the line? If the question does not state that anything is possible, are we to assume that? Not normally. We would normally assume normal conditions. If anything is possible, then the question will never be answered because we could have any kind of goofy scenario thrown in like parallel universes, multiple dimensions, miracles, and who knows what else (not that I'm necessarily against that :D).
Click to expand...

The OP stated 200Vrms and 5Arms. Therefore,

Pavg = 200Vrms x 5Arms = 1kw

This is the average of p(t) over time.

In a steady state analysis we must assume that impedances are constant so the thermal time constant of the filament is irrelevant. Nuf sed!
 
100815-2035 EST

rattus:

Steady state at 0.1 Hz and you do have a time varying impedance within a time period for averaging that might be used for a 0.1 Hz sine wave.

I stand by my answer of not enough information.

.
 
100815-2048 EST

mivey:

The nature of the question and the answer limitations imposed by only three choices and the meaning of those specific choices takes me to not enough information.

Had there been no preassigned choice of answers then I might have answered 1000 W and maybe added some assumptions. Had the load in the question been a 10,000 W Ohmite fixed resistor, then my likely answer would be 1000 W with no qualifications. Largely because I would not expect the question to be some sort of trick question.

.
 
And, I stand by the argument that we must assume a steady state problem where impedances are constant. The defintion of RMS assumes a constant load resistance does it not?
 
rattus said:
And, I stand by the argument that we must assume a steady state problem where impedances are constant.

The defintion of RMS assumes a constant load resistance does it not?
Click to expand...

Rattus,

You may assume steady state, if you wish.
I don't think that is what the OP proposed, not exactly.

We both know that
RMS is the Root of the Mean of the Squares of the samples taken.

Those samples could be varying values,
representing non-constant load impedances in a real world.

IMO, the OP has presented a logical fallacy of the "complex question",
and is forcing a 'non-sequitar' on any respondent.

I am squirming to get out from between the horns of his disambiguation.
Therefore, I request more information from the OP.

I am aware that the OP's presentation may be complete enough,
assuming steady state, and normal electrician conditions.

But since I am not being paid to get the right answer in 5 seconds,
I like thinking about it longer, and require more information.

:)
 
No Glenn,

RMS means the square ROOT of the MEAN values SQUARED. Says nothing about sampling. The computation of the RMS value of a sinusoid assumes a continous function and also assumes a constant resistance.

The equivalent DC value must drive a constant resistive load in order to make any sense.
 
100815-2242 EST

rattus:

For a mathematical definition of RMS see http://en.wikipedia.org/wiki/Root_mean_square .

The product of an RMS voltage and RMS current to a load does not necessarily provide a correct calculation of the power to the load. If the load is a perfect theoretical ideal invariant resistor, and both RMS values are averaged over the same time period, then Vrms * Irms = Power dissipated in the load resistor.

Note: in non-electrical fields RMS is also used. For example surface finish is usually measured in RMS these days.

Also note what the original poster answered for his question.

.
 
gar said:
100815-2242 EST

rattus:

For a mathematical definition of RMS see http://en.wikipedia.org/wiki/Root_mean_square .

The product of an RMS voltage and RMS current to a load does not necessarily provide a correct calculation of the power to the load. If the load is a perfect theoretical ideal invariant resistor, and both RMS values are averaged over the same time period, then Vrms * Irms = Power dissipated in the load resistor.

Note: in non-electrical fields RMS is also used. For example surface finish is usually measured in RMS these days.

Also note what the original poster answered for his question.

.
Click to expand...

gar, since the computation of RMS values assumes constant loads, RMS values are inapplicable to a varying load.

Now, if we say that the power is VrmsxIrmsxPF, then we have insufficient info.
 
rattus said:
Says nothing about sampling.
... continous function and also assumes a constant resistance.
Click to expand...

Rattus and GAR,

I think we may be on the other sides of the same fence.

Guys,
Following up with Gar's link/URL to Wiki,
since it is usually very clear to read,

quote

In mathematics, the root mean square (abbreviated RMS or rms),
also known as the quadratic mean,
is a statistical measure of the magnitude of a varying quantity.
It is especially useful when variates are positive and negative,
e.g., sinusoids.
RMS is used in various fields, including electrical engineering;
one of the more prominent uses of RMS is in the field of signal amplifiers.

It can be calculated for a series of discrete values
or for a continuously varying function.
The name comes from the fact that it is the square root of the mean of the squares of the values.
It is a special case of the generalized mean with the exponent p = 2.
end

end quote (Bolds are mine)

IMO, this thread is concerned with some 'normal' electrical information,
which is assumed to be static, non-varying.
Which, I think, Rattus and others have pointed out.
Good point, from the "Electricians" viewpoint.

My point,
since I don't think the OP provided sufficient information ,
is that there is a larger world of math, which the OP ignores.

IMO, this thread may be over-analyzing a simple electrical question.
On the other hand, I wonder why the OP provided so little info,
UNLESS he was 'teasing' the group of experts that frequent this forum.

IMO, this is an interesting (if stretched out) discussion.
To me, it seems that talented people like to "stretch" a discussion.
 
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rattus said:
gar, since the computation of RMS values assumes constant loads, RMS values are inapplicable to a varying load.

Now, if we say that the power is VrmsxIrmsxPF, then we have insufficient info.
Click to expand...

But, I almost forgot. Incandescent lamps exhibit a PF of one.
 
rattus said:
gar, since the computation of RMS values assumes constant loads, RMS values are inapplicable to a varying load.

Now, if we say that the power is VrmsxIrmsxPF, then we have insufficient info.
Click to expand...

Then you're claiming RMS values are inapplicable to anything but a passive device powered on clean DC with no ripple, but there would be no point in using RMS, either by discrete value or continuous integration.

Power factor is not 1.00 if the current and voltage are out of phase or show any sign of non-linearity, such as an incandescent lamp. Just because it doesn't show a significant amount of non-linearity at 60Hz sine wave doesn't mean that its a linear load and anything other than 60Hz sine wave doesn't mean abnormal electrical conditions.
 
glene77is said:
Guys,
Following up with Gar's link/URL to Wiki,
since it is usually very clear to read,

quote

RMS is used in various fields, including electrical engineering;
one of the more prominent uses of RMS is in the field of signal amplifiers.

It can be calculated for a series of discrete values
or for a continuously varying function.
Click to expand...

Glen,
However one computes the RMS values, Pavg = Irms x Vrms. My books say that a graphical approach may be used if the waves cannot be expressed as equations, however the emphasis is on sinusoids, Certainly a wave of any shape may be sampled, but however you do it, Vrms = Irms x R, and R must be constant That is, we must limit this discussion to linear circuits. From Wiki:

"Informally, a linear circuit is one in which the values of the electronic components, the resistance, capacitance, inductance, gain, etc. don't change with the level of voltage or current in the circuit"
 
rattus said:
Glen,
However one computes the RMS values, Pavg = Irms x Vrms. My books say that a graphical approach may be used if the waves cannot be expressed as equations, however the emphasis is on sinusoids, Certainly a wave of any shape may be sampled, but however you do it, Vrms = Irms x R, and R must be constant That is, we must limit this discussion to linear circuits. From Wiki:
Click to expand...
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.
 
Electric-Light said:
Then you're claiming RMS values are inapplicable to anything but a passive device powered on clean DC with no ripple, but there would be no point in using RMS, either by discrete value or continuous integration.

Power factor is not 1.00 if the current and voltage are out of phase or show any sign of non-linearity, such as an incandescent lamp. Just because it doesn't show a significant amount of non-linearity at 60Hz sine wave doesn't mean that its a linear load and anything other than 60Hz sine wave doesn't mean abnormal electrical conditions.
Click to expand...

Light,

Few things are pure, therefore we must assume that our sinusoids are pure, and that the values of the circuit elements are constant. In most cases the error is miniscule.

Furthermore, the variation of filament resistance at 60hz is so small that we can ignore it, and the effects of stray inductance and capacitance on PF of an incandescent lamp can also be ignored.

We must make these assumptions in order to perform a linear circuit analysis. If the assumptions are not valid, we must use other techniques.

Fellow told me once that all engineering calculations had to be as exact as possible. I told him, "No, you have to know the degree of approximation which is acceptable".
 
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