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

[rattus]

What would the NEC say the power of a lamp at 200v and 5a is?

Can't answer that one, but I can agree with gar that Vrms*Irms does not necessarily equal Pavg. For example, consider a stiff sinusoidal voltage driving a non-linear load. The effective value of the load current could be computed, but it would be meaningless because we would not have a constant resistance in the formula,

Pavg = (Vrms^2)/R = Irms^2*R

As has been said, the v(t)i(t) product could be averaged to compute Pavg.

 

[Besoeker]

RMS means the square ROOT of the MEAN values SQUARED.

RMS is the square root of the mean of the squares, not the square root of the mean values squared.

The computation of the RMS value of a sinusoid assumes a continous function

Well, v=Vm*sin(wt) would, by definition, be a continuous function. You'd have to set limits to the value t to make it discontinuous.

and also assumes a constant resistance.

You can measure or calculate the RMS current taken by a power factor correction capacitor. I don't think a capacitor can be considered a resistance.

Think about the input current to a lower end variable frequency inverter. It isn't feeding a constant resistance load, it isn't sinusoidal, and it has discontinuities. Yet it is entirely possible to measure the RMS value of that current. Or calculate it from waveform measurements.

 

[rattus]

RMS is the square root of the mean of the squares, not the square root of the mean values squared.

Right! I misspoke.

You can measure or calculate the RMS current taken by a power factor correction capacitor. I don't think a capacitor can be considered a resistance.

For that matter, all sinusoidal currents have the same RMS to peak ratio namely 0.707. But, the derivation of that ratio assumes a resistance. And, if you are computing power, you must assume a constant load resistance.

Think about the input current to a lower end variable frequency inverter. It isn't feeding a constant resistance load, it isn't sinusoidal, and it has discontinuities. Yet it is entirely possible to measure the RMS value of that current. Or calculate it from waveform measurements.

Yes, you can compute RMS values for most anything even "RMS power", but that is meaningless. But why do it unless you wish to compute power. You cannot perform steady state analyses with non sinusoids.

 

[Electric-Light]

Light,
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.

Your problem is that you took assumption too far. I never stated 60Hz, but apparently you assumed 60Hz.

 

[Electric-Light]

And, if you are computing power, you must assume a constant load resistance.

If you're computing power in reality, you do it by not making huge assumption and try to get away with it, like assuming the load is resistance simply because you only have a clamp meter that can't sample voltage and current at the same time and integrate to calculate power.

 

[glene77is]

Glen,
That is, we must limit this discussion to linear circuits.

Rattus,

Granted, for this discussion.
This is an electrical forum.

I am curious how Electric Light is going to dig his way
out of the hole I think he is standing in.

I simply maintain that he is not providing sufficient data
for a proper 'single' solution,
and maybe not an intelligent discussion.

I think he will gradually widen the boundaries of inquiry.

 

[gar]

100816-12503 EST

Any sinusoidal waveform whether it is voltage, current, power, light intensity, displacement of a moving mass, or anything, etc., has the relationship that the RMS value is one-half the sq-root of two times the peak value. This is a mathematical relationship and has nothing to do with the physical entity that is following a sinusoidal variation.

Within certain constraints I can build a sinusoidal current generator and connect it to some nonlinear loads and have the current thru the load sinusoidal.

.

 

[rattus]

If you're computing power in reality, you do it by not making huge assumption and try to get away with it, like assuming the load is resistance simply because you only have a clamp meter that can't sample voltage and current at the same time and integrate to calculate power.

In a power circuit one would expect an incandescent lamp filament to be resistive. if the load is non-linear, you cannot compute the RMS values of currents driven by a stiff source, but this value has been given which implies a linear circuit in which Vrms and Irms are proportional and perhaps separated by a phase angle.

If one does have a non-linear circuit, then indeed integration is required, and we cannot use steady state methods.

Not worth fighting about though, I think we all understand the arithmetic!

 

[rattus]

100816-12503 EST

Any sinusoidal waveform whether it is voltage, current, power, light intensity, displacement of a moving mass, or anything, etc., has the relationship that the RMS value is one-half the sq-root of two times the peak value. This is a mathematical relationship and has nothing to do with the physical entity that is following a sinusoidal variation.

Within certain constraints I can build a sinusoidal current generator and connect it to some nonlinear loads and have the current thru the load sinusoidal.

.

Sure gar, but the load voltage will not be sinusoidal.

 

[Besoeker]

For that matter, all sinusoidal currents have the same RMS to peak ratio namely 0.707. But, the derivation of that ratio assumes a resistance.

It requires no assumptions about the load at all. If you have a sinusoidal voltage the ratio of RMS to peak will always be 0.707 regardless of load. The following shows the derivation for a Vp=1.

Edit:
It's not as neatly formatted as I would wish - I'm still finding my way round the new version of Equation Editor.
And the third from last line should be Vrms^2 = 1/2 It just didn't show as entered...........

 

[rattus]

It requires no assumptions about the load at all. If you have a sinusoidal voltage the ratio of RMS to peak will always be 0.707 regardless of load. The following shows the derivation for a Vp=1.

Edit:
It's not as neatly formatted as I would wish - I'm still finding my way round the new version of Equation Editor.
And the third from last line should be Vrms^2 = 1/2 It just didn't show as entered...........

Sure Bes, but RMS is applicable because the RMS voltages and currents produce the same heating power as a DC voltage or current across or through a resistor.

 

[Electric-Light]

In a power circuit one would expect an incandescent lamp filament to be resistive. if the load is non-linear, you cannot compute the RMS values of currents driven by a stiff source, but this value has been given which implies a linear circuit in which Vrms and Irms are proportional and perhaps separated by a phase angle.

If one does have a non-linear circuit, then indeed integration is required, and we cannot use steady state methods.

Not worth fighting about though, I think we all understand the arithmetic!

Say what? I'm not sure if I understood you, but you're claiming that ability to measure and calculate I rms is contingent on stiffness of power source?

In attached photo, voltage and current are not phase shifted, yet they both have their own RMS values.
http://ecmweb.com/mag/606ecmINSIDEPQfigB.jpg
The Pavg/(Vrms * Irms) is something in the neighorhood of 0.5 to 0.7 though. Many parameters affect the power factor value, so this is why you can't just measure Vrms * Irms and use a coefficient to calculate Pavg. It must be sampled or integrated using both inputs. (V & I ) This is an extreme yet real example of non-linear load. The resistance is almost infinite at the lower end of wave cycle, but suddenly drops with when a certain voltage value is reached. This is your average CFL and electronic power supplies.

 

[Besoeker]

Sure Bes, but RMS is applicable because the RMS voltages and currents produce the same heating power as a DC voltage or current across or through a resistor.

I was responding to your point in post #143:

For that matter, all sinusoidal currents have the same RMS to peak ratio namely 0.707. But, the derivation of that ratio assumes a resistance.

If you have a sinusoidal quantity (of anything as GAR has noted), the 1/sqrt(2) relationship applies for RMS to peak. If it's surface imperfections for example, resistance isn't a relevant quantity.

 

[rattus]

I was responding to your point in post #143:

If you have a sinusoidal quantity (of anything as GAR has noted), the 1/sqrt(2) relationship applies for RMS to peak. If it's surface imperfections for example, resistance isn't a relevant quantity.

This being a forum for electricians, et al, it is reasonable to assume that we are speaking of effective values of voltage and current rather than surface finish.

 

[gar]

100817-1253 EST

There are many viewers of this forum that have a rather incomplete understanding of electrical instruments, devices, circuits, and the use of equations to solve problems with respect to these areas.

Using terms correctly, having an understanding of the words, and being flexible in understanding different perspectives is important in effective communication.

Having an understanding of the meaning of RMS, how it evolved, and how it can be effectively used is important to obtaining reasonably correct answers to some problems.

I can not find a precise trail on the evolution of the use of RMS measurements in the electrical field.

I believe the trail on the concept of RMS starts in the field of statistics with Francis Galton conceiving the standard deviation function in the 1860s. This certainly predates electrical usage. See section "Statistics, standard deviation, regression and correlation" at http://en.wikipedia.org/wiki/Francis_Galton. There are many other search results that fill in some of the holes.

At some point the name RMS became associated with standard deviation. Then at some other point the correlation of an RMS calculation with the calculation of power in a pure resistance like, Pave = summation from time 1 to time 2 of v^2 / R divided by (time 2 - time1), produced a result of Vrms^2 / R. So the sq-root of (Vrms^2) or just Vrms of any wave shape and a pure resistance could predict the power dissipated in that resistance.

Note: R = Vrms^2 / P and R = P / Irms^2. Combining these produces Vrms^2 / P = P / Irms^2 and rearranging results in Vrms^2 * Irms^2 = P^2 or Vrms * Irms = P. But this equation only applies in certain special cases, albeit, these may be quite common circuits.

The important point of this whole thread is that the user must understand their instruments, equations, and circuits or they may get an incorrect answer.

.

 

[Besoeker]

This being a forum for electricians, et al, it is reasonable to assume that we are speaking of effective values of voltage and current rather than surface finish.

It was to make a point.
But, even in the electrical context, you still don't need resistance for the relationship to work.

 

[LarryFine]

Here's a semi-related question: does a motor "respond" to the RMS value of its supply?

That's sort-of related to my earlier assertion about the moving-coil meter doing so.

 

[rattus]

It was to make a point.
But, even in the electrical context, you still don't need resistance for the relationship to work.

Yes Bes, but the derivation of effective values is based on the fact that,

p = i^2*R = (v^2)/R

where the value of R is constant.

Then,

Pavg = Ieff^2*R =(Veff^2)/R

where the load R is constant.

It turns out that effective and RMS values are equivalent.

My contention is that it is improper to use RMS values to compute power in a non-linear load. You would have to compute Reff!

 

[rattus]

Well yes:

Well yes:

Here's a semi-related question: does a motor "respond" to the RMS value of its supply?

That's sort-of related to my earlier assertion about the moving-coil meter doing so.

You may as well ask if the motor responds to the peak value of the supply.

 

[Electric-Light]

Here's a semi-related question: does a motor "respond" to the RMS value of its supply?

That's sort-of related to my earlier assertion about the moving-coil meter doing so.

No, it does not. The inductance make it behave completely differently.

Vrms is useful in determining equivalent DCV to produce the same heating value in a resistor. If you have a 120.00ohm fixed value ideal resistor, the heating value will always be 120.00W regardless of input type as long as it is 120.00v RMS.

A steady DC 120v voltage source and a 120v @ 60Hz sinusoidal power both have the same heating value on a resistor. The steady state of DC current is solely determined by winding resistance(but the rise time, dI/dt is affected by reactance) while AC power current is affected by impedance (resistance + reactive resistance)

Both DC 120v and AC 120v 60Hz sine wave can have RMS values.